Aimathic
Login | English | Deutsch

Free math worksheets

Build your own math worksheets from 30,000+ problems for grades 3 to 12, from fractions to AP Calculus. Every problem comes with step-by-step solutions.

Build linear models from descriptions

Click problems to add them to your worksheet.

5546018
A plant is modeled as \(12\,\text{cm}\) tall at the start of an observation period and growing at a constant rate of \(1.5\,\text{cm}\) per week. Let \(h(w)\) be its height, in centimeters, after \(w\) weeks. Write a linear function for the model.

Hints

- In a linear model, the initial value is the output when the input is \(0\). - The constant weekly growth rate is the coefficient of the time variable.

Solution

1. The initial value is \(12\), so the constant term is \(12\). 2. The rate of change is \(1.5\) centimeters per week, so the coefficient of \(w\) is \(1.5\). 3. Therefore, the linear model is \(h(w)=1.5w+12\).

Answer

\(h(w)=1.5w+12\)
5119118
A swimming pool is being prepared for summer. At the start, there are already \(4\,\text{in.}\) of water in the pool. Water is then added at a constant rate. The table shows the water depth at different times: <table> <tr><td>Time \(t\) (h)</td><td>\(0\)</td><td>\(1\)</td><td>\(2\)</td><td>\(3\)</td><td>\(4\)</td><td>\(5\)</td></tr> <tr><td>Water depth \(h\) (in.)</td><td>\(4\)</td><td>\(10\)</td><td>\(16\)</td><td>\(22\)</td><td>\(28\)</td><td>\(34\)</td></tr> </table> a) Describe the graph formed by the data points, with time on the x-axis and water depth on the y-axis. b) How many inches per hour does the water depth increase? c) Write a function for the water depth \(h\) after \(t\) hours. d) Use your function to find the water depth after \(2.5\) hours.

Hints

- Compare the water-depth values from one hour to the next. Is the change constant? - What water depth is already present at \(t = 0\)? - Combine the initial value with the hourly increase to write the function. - Substitute \(2.5\) for \(t\) in your function.

Solution

1. The points \((0, 4)\), \((1, 10)\), \((2, 16)\), \((3, 22)\), \((4, 28)\), and \((5, 34)\) lie on a straight increasing line. 2. The water depth increases by \(10 - 4 = 6\) inches from one hour to the next, and the same increase continues throughout the table. The rate of change is \(6\,\text{in./h}\). 3. The initial value is \(4\) inches, so the function is \(h(t) = 6t + 4\). 4. \(h(2.5) = 6 \cdot 2.5 + 4 = 19\).

Answer

a) The data points lie on a straight increasing line. b) \(6\,\text{in./h}\) c) \(h(t) = 6t + 4\) d) \(19\,\text{in.}\)
5126018
A gym offers two monthly plans. Plan A costs a flat \(\$45.00\). Plan B costs \(\$10.00\) plus \(\$3.50\) per visit. Let \(v\) be the number of visits in a month. Write a cost function for each plan. Then evaluate this claim: “I work out every other day in a \(30\)-day month, so Plan B is definitely less expensive.” Justify your conclusion and state the difference in cost.

Hints

- Identify the fixed part and the per-visit part of each plan. - Convert “every other day in a 30-day month” into a number of visits. - Evaluate both functions at the same visit count before comparing the costs.

Solution

1. Plan A has the constant cost function \(A(v)=45\). Plan B has the linear cost function \(B(v)=10+3.50v\). 2. Working out every other day in a \(30\)-day month gives \(30\div2=15\) visits. 3. Evaluate the functions: \(A(15)=45\) and \(B(15)=10+3.50\cdot15=62.50\). 4. Since \(62.50>45\), Plan B is not less expensive. It costs \(62.50-45=17.50\) dollars more.

Answer

\(A(v)=45\) and \(B(v)=10+3.50v\). For \(15\) visits, Plan A costs \(\$45.00\) and Plan B costs \(\$62.50\), so the claim is false. Plan B costs \(\$17.50\) more.
5128778
A line \(g\) is parallel to the x-axis and passes through \(A(-4, -7)\). 1. Find the equation of line \(g\). 2. For each point, determine whether it lies above, below, or on line \(g\): \(B(-4, -5)\), \(C(10, -7)\), and \(D(0, -10)\). Justify each answer by comparing y-coordinates.

Hints

- What form does the equation of a horizontal line have? - A point's y-coordinate tells you its vertical position. - Compare each y-coordinate with \(-7\).

Solution

1. A line parallel to the x-axis is horizontal, so it has the form \(y = c\). Since it passes through \(A(-4, -7)\), its equation is \(y = -7\). 2. Point \(B\) has y-coordinate \(-5\), and \(-5 > -7\), so \(B\) is above the line. Point \(C\) has y-coordinate \(-7\), so it is on the line. Point \(D\) has y-coordinate \(-10\), and \(-10 < -7\), so \(D\) is below the line.

Answer

1) \(y = -7\) 2) \(B\) is above the line; \(C\) is on the line; \(D\) is below the line.
5130398
In a spring-stretching experiment, \(x\) is the attached mass in grams and \(y\) is the spring's total length in centimeters. <table> <tr><td>Mass \(x\) (\(\text{g}\))</td><td>\(0\)</td><td>\(50\)</td><td>\(100\)</td><td>\(150\)</td><td>\(200\)</td></tr> <tr><td>Length \(y\) (\(\text{cm}\))</td><td>\(12.0\)</td><td>\(14.5\)</td><td>\(17.0\)</td><td>\(19.5\)</td><td>\(22.0\)</td></tr> </table> a) Find the linear function \(f(x) = mx + b\) that exactly models the data. b) Interpret the slope \(m\) and y-intercept \(b\) in the context of the experiment.

Hints

- What does the table show when no mass is attached? - How much does the spring length change over a known change in mass? - Use change in output divided by change in input to find the slope.

Solution

1. When \(x = 0\), the spring length is \(12.0\,\text{cm}\), so \(b = 12\). 2. Using \((0, 12)\) and \((50, 14.5)\), \(m = \frac{14.5 - 12}{50 - 0} = \frac{2.5}{50} = 0.05\). 3. Therefore, \(f(x) = 0.05x + 12\). 4. The y-intercept means the unloaded spring is \(12\,\text{cm}\) long. The slope means the spring length increases by \(0.05\,\text{cm}\) for each additional gram of mass.

Answer

a) \(f(x) = 0.05x + 12\) b) \(b = 12\) is the spring's length with no attached mass, in centimeters. \(m = 0.05\) is the increase in length, in centimeters per gram.
5241978
A candle is \(8\,\text{in.}\) tall when it is lit. It burns at a constant rate, and its height decreases by \(1\,\text{in.}\) each hour. a) Write a linear function for the candle's height \(h\), in inches, after \(t\) hours. b) After how many hours will the candle be half its original height? c) When will the candle be completely burned down?

Hints

- What is the candle's starting height, and how much does it change each hour? - What sign should the rate of change have when a quantity is decreasing? - What height represents a candle that has completely burned down? - For each target height, set the function equal to that height and solve for time.

Solution

1. The initial height is \(8\) inches, and the height decreases by \(1\) inch per hour. Therefore, \(h(t) = -t + 8\). 2. Half the original height is \(4\) inches. Solve \(4 = -t + 8\), which gives \(t = 4\). 3. The candle is completely burned down when \(h(t) = 0\). Solve \(0 = -t + 8\), which gives \(t = 8\).

Answer

a) \(h(t) = -t + 8\) b) \(4\) hours c) \(8\) hours
5241998
When a freezer is turned on, its inside temperature is \(72\,^\circ\text{F}\). The temperature decreases at a constant rate of \(8\,^\circ\text{F}\) per hour. 1) Write a linear function for the temperature \(y\), in degrees Fahrenheit, after \(x\) hours. 2) Find the temperature after \(5.5\) hours. 3) After how many hours will the freezer reach its target temperature of \(0\,^\circ\text{F}\)?

Hints

- Identify the starting temperature and the amount it changes each hour. - Because the temperature is decreasing, what sign should the rate of change have? - To find a time, set the model equal to the target temperature and solve for the time variable.

Solution

1. The initial value is \(72\), and the rate of change is \(-8\). Therefore, \(y = -8x + 72\). 2. Substitute \(x = 5.5\): \(y = -8 \cdot 5.5 + 72 = 28\). The temperature is \(28\,^\circ\text{F}\). 3. Set the model equal to the target temperature: \(0 = -8x + 72\). Solving gives \(x = 9\).

Answer

1) \(y = -8x + 72\) 2) \(28\,^\circ\text{F}\) 3) \(9\) hours
5242018
Paul is saving for a new bike. He already has \(\$45.00\) saved, and he adds \(\$7.50\) each week. 1) Write a linear function for the amount \(y\), in dollars, after \(x\) weeks. 2) Find the points on the graph after \(0\), \(2\), \(4\), \(6\), and \(8\) weeks. 3) How many weeks will it take Paul to have exactly \(\$120.00\)?

Hints

- Which amount is already there at week \(0\), and which amount changes each week? - How can you relate the number of weeks to the amount saved? - What does a specified target balance mean for the equation you need to solve? - Which quantity belongs on the x-axis, and which belongs on the y-axis?

Solution

1. The initial amount is \(45\) and the weekly increase is \(7.50\), so \(y = 7.50x + 45.00\). 2. Substitution gives \((0, 45)\), \((2, 60)\), \((4, 75)\), \((6, 90)\), and \((8, 105)\). 3. Solve \(120.00 = 7.50x + 45.00\). Then \(75.00 = 7.50x\), so \(x = 10\).

Answer

1) \(y = 7.50x + 45.00\) 2) \((0, 45)\), \((2, 60)\), \((4, 75)\), \((6, 90)\), \((8, 105)\) 3) 10 weeks
5321918
The graphs of three linear functions \(f\), \(g\), and \(h\) are shown. a) Write an equation in the form \(y = mx + b\) for each line. b) Which function represents a proportional relationship? Justify your answer using both the graph and the equation. c) Describe how the three lines are positioned relative to one another. Explain how you can tell directly from their equations.
Figure for problem 532191

Hints

- Where does each line cross the y-axis? - Use two grid points on each line to calculate rise over run. - What must be true about the graph and equation of a proportional relationship? - What do parallel nonvertical lines have in common?

Solution

1. For each line, read the y-intercept and use two grid points to find the slope. For \(f\), the y-intercept is \(1\). Using \((0, 1)\) and \((2, 4)\), \(m = \frac{4 - 1}{2 - 0} = \frac{3}{2}\), so \(f(x) = \frac{3}{2}x + 1\). For \(g\), the y-intercept is \(-2\). Using \((0, -2)\) and \((2, 1)\), \(m = \frac{1 - (-2)}{2 - 0} = \frac{3}{2}\), so \(g(x) = \frac{3}{2}x - 2\). For \(h\), the y-intercept is \(0\). Using \((0, 0)\) and \((2, 3)\), \(m = \frac{3 - 0}{2 - 0} = \frac{3}{2}\), so \(h(x) = \frac{3}{2}x\). 2. The function \(h\) is proportional because its graph passes through the origin and its equation has the form \(y = mx\). 3. All three lines are parallel because they have the same slope, \(\frac{3}{2}\), and different y-intercepts.

Answer

a) \(f(x) = \frac{3}{2}x + 1\), \(g(x) = \frac{3}{2}x - 2\), \(h(x) = \frac{3}{2}x\) b) \(h\). Its graph passes through \((0, 0)\), and its equation has no nonzero constant term. c) The three lines are parallel because they all have slope \(\frac{3}{2}\).
5322098
The graph shows two linear functions, \(f\) and \(g\). a) For \(f\), find the y-intercept \(b_1\) and slope \(m_1\). Then write the function equation. b) For \(g\), find the y-intercept \(b_2\) and slope \(m_2\). Then write the function equation.
Figure for problem 532209

Hints

- Where does each line cross the y-axis? - Choose two grid points on each line to form a slope triangle. - Find slope as vertical change divided by horizontal change, including the correct sign. - Substitute the slope and y-intercept into \(y = mx + b\).

Solution

1. The graph of \(f\) crosses the y-axis at \((0, -1)\), so \(b_1 = -1\). Using \((0, -1)\) and \((2, 0)\), the slope is \(m_1 = \frac{0 - (-1)}{2 - 0} = \frac{1}{2}\). Therefore, \(f(x) = \frac{1}{2}x - 1\). 2. The graph of \(g\) crosses the y-axis at \((0, 3)\), so \(b_2 = 3\). Using \((0, 3)\) and \((2, 0)\), the slope is \(m_2 = \frac{0 - 3}{2 - 0} = -\frac{3}{2}\). Therefore, \(g(x) = -\frac{3}{2}x + 3\).

Answer

a) \(b_1 = -1\), \(m_1 = \frac{1}{2}\), and \(f(x) = \frac{1}{2}x - 1\) b) \(b_2 = 3\), \(m_2 = -\frac{3}{2}\), and \(g(x) = -\frac{3}{2}x + 3\)
5322208
At the start of a fire-department training exercise, a water tank contains \(3000\) gallons. Water is pumped out at a constant rate of \(375\) gallons per hour. The graph shows the amount of water remaining as a function of time. a) Use the graph to estimate how many hours pass before \(1125\) gallons remain. b) Write a linear function \(V\), where \(t\) is time in hours and \(V(t)\) is the remaining volume in gallons. State the contextual domain, and use the function to find exactly when the tank is empty.
Figure for problem 532220

Hints

- Use the vertical axis for the amount of water and the horizontal axis for time. - A constant decrease produces a linear function with a negative rate of change. - The contextual domain begins at the start of the exercise and ends when the tank is empty. - The tank is empty when its volume equals \(0\).

Solution

1. Locate \(1125\) gallons on the vertical axis and move horizontally to the graph. The corresponding time is \(5\) hours. 2. The initial value is \(3000\), and the volume decreases by \(375\) gallons each hour. Therefore, \(V(t)=3000-375t\). The model applies from the start until the tank is empty, so its contextual domain is \(0\le t\le8\). The tank is empty when \(V(t)=0\): \(3000-375t=0\), so \(375t=3000\) and \(t=8\).

Answer

a) \(5\) hours b) \(V(t)=3000-375t\), with \(0\le t\le8\); the tank is empty after \(8\) hours.
5324068
The figure shows two linear functions, \(f\) and \(g\). Two points are marked on each line. For each line, find the slope \(m\), the y-intercept \(b\), and the complete equation in the form \(y = mx + b\).
Figure for problem 532406

Hints

- Read the coordinates of the two marked points on each line. - Use change in y divided by change in x to calculate the slope. - Where does each line cross the y-axis? - Substitute the slope and y-intercept into \(y = mx + b\).

Solution

1. For \(f\), the marked points are \((0, -1)\) and \((3, 1)\). The slope is \(m = \frac{1 - (-1)}{3 - 0} = \frac{2}{3}\). Since \((0, -1)\) is on the y-axis, \(b = -1\). Therefore, \(f(x) = \frac{2}{3}x - 1\). 2. For \(g\), the marked points are \((-1, 3)\) and \((1, -1)\). The slope is \(m = \frac{-1 - 3}{1 - (-1)} = -2\). Using \((1, -1)\), \(-1 = -2 \cdot 1 + b\), so \(b = 1\). Therefore, \(g(x) = -2x + 1\).

Answer

\(f\): \(m = \frac{2}{3}\), \(b = -1\), \(f(x) = \frac{2}{3}x - 1\) \(g\): \(m = -2\), \(b = 1\), \(g(x) = -2x + 1\)
5324108
Find the equation of each line \(f\), \(g\), and \(h\) shown in the coordinate plane. Write each equation in the form \(y = mx + b\).
Figure for problem 532410

Hints

- First read the y-intercept of each line. - Choose two grid points on each line to form a slope triangle. - Compute slope as \(\frac{\Delta y}{\Delta x}\). - A line that falls from left to right has a negative slope.

Solution

1. Line \(f\) crosses the y-axis at \((0, 1)\). Using \((0, 1)\) and \((2, 4)\), the slope is \(\frac{4 - 1}{2 - 0} = \frac{3}{2}\). Therefore, \(f(x) = \frac{3}{2}x + 1\). 2. Line \(g\) crosses the y-axis at \((0, -2)\). Using \((0, -2)\) and \((2, -3)\), the slope is \(\frac{-3 - (-2)}{2 - 0} = -\frac{1}{2}\). Therefore, \(g(x) = -\frac{1}{2}x - 2\). 3. Line \(h\) crosses the y-axis at \((0, 3)\) and the x-axis at \((3, 0)\). Its slope is \(\frac{0 - 3}{3 - 0} = -1\). Therefore, \(h(x) = -x + 3\).

Answer

\(f(x) = \frac{3}{2}x + 1\), \(g(x) = -\frac{1}{2}x - 2\), and \(h(x) = -x + 3\).
5332018
The line \(g\) shown in the coordinate plane passes through points \(P_1\) and \(P_2\). a) Use the coordinates of \(P_1\) and \(P_2\) to calculate the slope \(m\). b) Calculate the y-intercept \(b\). c) Write the complete function equation in the form \(g(x) = mx + b\).
Figure for problem 533201

Hints

- How can two points be used to calculate a slope? - Once you know the slope, substitute one point into \(y = mx + b\). - Pay attention to the scale when reading the point coordinates. - Check that your equation gives the correct y-value for both points.

Solution

1. The points are \(P_1(1, 5)\) and \(P_2(4, -1)\). 2. The slope is \(m = \frac{-1 - 5}{4 - 1} = -2\). 3. Substitute \(P_1(1, 5)\) into \(y = -2x + b\): \(5 = -2 \cdot 1 + b\), so \(b = 7\). 4. Therefore, \(g(x) = -2x + 7\).

Answer

a) \(m = -2\) b) \(b = 7\) c) \(g(x) = -2x + 7\)
5332038
The coordinate plane shows four lines, \(g_1\) through \(g_4\). Find the equation of each line. What do you notice when you compare the equations?
Figure for problem 533203

Hints

- Look at where each line crosses the y-axis. - Use grid points to compare vertical change with horizontal change. - Which part of \(y = mx + b\) determines the y-intercept?

Solution

1. For \(g_1\), the points \((0, 2)\) and \((2, 4)\) give slope \(1\), so \(g_1(x) = x + 2\). 2. For \(g_2\), the points \((0, 2)\) and \((1, 0)\) give slope \(-2\), so \(g_2(x) = -2x + 2\). 3. For \(g_3\), the points \((0, 2)\) and \((4, 4)\) give slope \(\frac{1}{2}\), so \(g_3(x) = \frac{1}{2}x + 2\). 4. For \(g_4\), the points \((0, 2)\) and \((4, 0)\) give slope \(-\frac{1}{2}\), so \(g_4(x) = -\frac{1}{2}x + 2\). 5. All four equations have the same y-intercept, \(b = 2\), so all four lines pass through \((0, 2)\).

Answer

\(g_1(x) = x + 2\), \(g_2(x) = -2x + 2\), \(g_3(x) = \frac{1}{2}x + 2\), and \(g_4(x) = -\frac{1}{2}x + 2\). All four lines have y-intercept \(2\).
5332048
Three lines, \(f\), \(g\), and \(h\), are shown. Write an equation for each line. How can you tell from the equations that the lines are parallel?
Figure for problem 533204

Hints

- A linear equation can be written as \(y = mx + b\). - The value \(b\) tells where the line crosses the y-axis. - Use the grid to determine each line's rise and run. - What must be the same for two nonvertical lines to be parallel?

Solution

1. For \(f\), the y-intercept is \(3\). The line falls \(1\) unit for every \(2\) units to the right, so \(m = -\frac{1}{2}\). Thus \(f(x) = -\frac{1}{2}x + 3\). 2. For \(g\), the y-intercept is \(1\) and the slope is also \(-\frac{1}{2}\). Thus \(g(x) = -\frac{1}{2}x + 1\). 3. For \(h\), the y-intercept is \(-2\) and the slope is again \(-\frac{1}{2}\). Thus \(h(x) = -\frac{1}{2}x - 2\). 4. Nonvertical lines are parallel when they have the same slope. All three equations have slope \(-\frac{1}{2}\).

Answer

\(f(x) = -\frac{1}{2}x + 3\), \(g(x) = -\frac{1}{2}x + 1\), and \(h(x) = -\frac{1}{2}x - 2\). The lines are parallel because all three have slope \(-\frac{1}{2}\).
5332128
The coordinate plane shows three lines, \(f\), \(g\), and \(h\). Find the equation of each line.
Figure for problem 533212

Hints

- A horizontal line has slope \(0\) and an equation of the form \(y = c\). - For the other lines, notice whether each line rises or falls from left to right.

Solution

1. Line \(f\) is horizontal at \(y = 2\), so its slope is \(0\) and its equation is \(f(x) = 2\). 2. Line \(g\) has y-intercept \(1\). Moving \(1\) unit right corresponds to moving \(1\) unit down, so the slope is \(-1\). Thus \(g(x) = -x + 1\). 3. Line \(h\) has y-intercept \(-3\). Moving \(1\) unit right corresponds to moving \(1\) unit up, so the slope is \(1\). Thus \(h(x) = x - 3\).

Answer

\(f(x) = 2\), \(g(x) = -x + 1\), and \(h(x) = x - 3\).
5332578
Consider the line \(g\) shown in the coordinate plane. a) Find the y-intercept \(b\). b) Find the slope \(m\). c) Write the equation in the form \(y = mx + b\).
Figure for problem 533257

Hints

- Find the point where the line crosses the y-axis. - Choose two grid points on the line. - Compare the vertical change with the horizontal change between those points.

Solution

1. The line crosses the y-axis at \((0, 1)\), so \(b = 1\). 2. Using \((0, 1)\) and \((2, 2)\), the slope is \(m = \frac{2 - 1}{2 - 0} = \frac{1}{2}\). 3. Therefore, the equation is \(g(x) = \frac{1}{2}x + 1\).

Answer

a) \(b = 1\) b) \(m = \frac{1}{2}\) c) \(g(x) = \frac{1}{2}x + 1\)
5332588
Find the equation of each line \(f\) and \(h\) shown in the coordinate plane. Use the y-intercept and a slope triangle.
Figure for problem 533258

Hints

- Start by finding where each line crosses the y-axis. - Choose grid points with easy-to-read coordinates for the slope triangle. - Compare the vertical change with the horizontal change. - Use the direction of the line to determine the sign of the slope.

Solution

1. Line \(f\) crosses the y-axis at \((0, 4)\). Using \((0, 4)\) and \((1, 1)\), the slope is \(m = \frac{1 - 4}{1 - 0} = -3\). Therefore, \(f(x) = -3x + 4\). 2. Line \(h\) crosses the y-axis at \((0, -3)\). Using \((0, -3)\) and \((5, -1)\), the slope is \(m = \frac{-1 - (-3)}{5 - 0} = \frac{2}{5}\). Therefore, \(h(x) = \frac{2}{5}x - 3\).

Answer

\(f(x) = -3x + 4\) and \(h(x) = \frac{2}{5}x - 3\).
5332728
The graph shows four lines, \(h_1\) through \(h_4\). Find the equation of each line.
Figure for problem 533272

Hints

- Find a point where each line crosses the y-axis. - What is special about the equation of a line through the origin? - What is the slope of a horizontal line? - Use two points to calculate slope as change in y divided by change in x.

Solution

1. Line \(h_1\) passes through \((0, 0)\) and \((4, 1)\), so its slope is \(\frac{1}{4}\). Thus \(h_1(x) = \frac{1}{4}x\). 2. Line \(h_2\) has y-intercept \(4\) and passes through \((4, 2)\). Its slope is \(\frac{2 - 4}{4 - 0} = -\frac{1}{2}\). Thus \(h_2(x) = -\frac{1}{2}x + 4\). 3. Line \(h_3\) is horizontal at \(y = -3\), so \(h_3(x) = -3\). 4. Line \(h_4\) has y-intercept \(4\) and passes through \((-2, 0)\). Its slope is \(\frac{4 - 0}{0 - (-2)} = 2\). Thus \(h_4(x) = 2x + 4\).

Answer

\(h_1(x) = \frac{1}{4}x\), \(h_2(x) = -\frac{1}{2}x + 4\), \(h_3(x) = -3\), and \(h_4(x) = 2x + 4\).
5332738
A storage tank contains \(10\,\text{gal}\) of water at the start. It is then filled at a constant rate. The graph shows the volume \(V\), in gallons, after \(t\) minutes. The axes use different scales. a) Find the linear function \(V(t)\). b) Find the amount of water in the tank after \(6\) minutes. Then check your result on the graph.
Figure for problem 533273

Hints

- First check the scale used on each axis. - Choose two points on the line that lie exactly on grid intersections. - The point where the line crosses the vertical axis gives the initial value. - Substitute the requested time into your function.

Solution

1. Two convenient points are \((0, 10)\) and \((4, 20)\). 2. The slope is \(m = \frac{20 - 10}{4 - 0} = 2.5\). The y-intercept is \(10\), so \(V(t) = 2.5t + 10\). 3. \(V(6) = 2.5 \cdot 6 + 10 = 25\). The graph also contains the point \((6, 25)\).

Answer

a) \(V(t) = 2.5t + 10\) b) \(25\,\text{gal}\)
5332818
At the start of a \(15\)-minute observation, a rainwater tank already contains \(25\) gallons. A constant inflow adds \(5\) gallons per minute. The graph shows the tank's volume as a function of time. a) Use the graph to determine how many minutes pass before the tank contains \(60\) gallons. b) Write the function, including its domain for the observation interval, and find when the tank contains \(90\) gallons.
Figure for problem 533281

Hints

- The y-intercept is the starting volume. - The slope is the amount added each minute. - Use the stated observation interval for the domain. - Set the function equal to the target volume and solve for time.

Solution

1. The graph reaches \(60\) gallons at \(t=7\) minutes. 2. The initial value is \(25\), and the rate of change is \(5\) gallons per minute, so \(V(t)=5t+25\), with \(0\le t\le15\). Set \(V(t)=90\): \(90=5t+25\), so \(65=5t\) and \(t=13\).

Answer

a) \(7\) minutes b) \(V(t)=5t+25\), with \(0\le t\le15\); \(13\) minutes
5332828
A new candle is \(12\) inches tall. As it burns, its height decreases at a constant rate of \(0.75\) inch per hour. The graph shows candle height over time. a) Use the graph to find the candle's height after \(4\) hours. b) Write the function, including its contextual domain, and find how many hours pass before the candle is \(4.5\) inches tall.
Figure for problem 533282

Hints

- A decreasing height gives a negative slope. - The y-intercept is the initial candle height. - The contextual domain ends when the candle's height reaches \(0\). - Set the function equal to the target height and solve.

Solution

1. At \(t=4\), the graph shows a height of \(9\) inches. 2. The initial height is \(12\), and the rate of change is \(-0.75\) inch per hour, so \(h(t)=12-0.75t\). The candle burns from its initial height to \(0\) inches, so the contextual domain is \(0\le t\le16\). Set \(h(t)=4.5\): \(4.5=12-0.75t\), so \(-7.5=-0.75t\) and \(t=10\).

Answer

a) \(9\) inches b) \(h(t)=12-0.75t\), with \(0\le t\le16\); \(10\) hours
5332888
Two hikers, Alex and Bea, travel along a long straight road. Alex starts at mile marker \(0\) and walks at a constant speed toward higher mile markers. Bea starts at the same time at mile marker \(10\) and walks at a constant speed toward mile marker \(0\). On the graph, curve \(a\) represents Alex and curve \(b\) represents Bea. Write a function of the form \(s(t)=mt+b\) for each hiker on the displayed interval \(0\le t\le4\).
Figure for problem 533288

Hints

- The y-intercept is each hiker's starting position. - A positive slope means moving toward higher mile markers; a negative slope means moving toward lower mile markers. - Use another point on each line to calculate slope.

Solution

1. Alex's graph starts at \(0\) and passes through \((2, 6)\). Its slope is \(m=\frac{6-0}{2-0}=3\), so \(s_A(t)=3t\), with \(0\le t\le4\). 2. Bea's graph starts at \(10\) and passes through \((2, 6)\). Its slope is \(m=\frac{6-10}{2-0}=-2\), so \(s_B(t)=-2t+10\), with \(0\le t\le4\).

Answer

\(s_A(t)=3t\) and \(s_B(t)=-2t+10\), with \(0\le t\le4\) for both functions
5335258
Find the equations of lines \(a\) and \(b\). Pay special attention to the direction of line \(a\).
Figure for problem 533525

Hints

- What is the slope of a line parallel to the x-axis? - For line \(b\), compare how the y-value changes as the x-value changes.

Solution

1. Line \(a\) is horizontal, so its slope is \(0\). It crosses the y-axis at \(-3\), so \(a(x) = -3\). 2. Line \(b\) crosses the y-axis at \(2\). Using \((0, 2)\) and \((-1, 5)\), the slope is \(m = \frac{5 - 2}{-1 - 0} = -3\). Therefore, \(b(x) = -3x + 2\).

Answer

Line \(a\): \(a(x) = -3\); line \(b\): \(b(x) = -3x + 2\)
5335288
Two water tanks are observed at the same time. The graph shows volume \(V\), in gallons, as a function of time \(t\), in minutes. Curve \(a\) represents Tank 1, and curve \(b\) represents Tank 2. Write an equation of the form \(V(t)=mt+b\) for each tank, including the displayed domain.
Figure for problem 533528

Hints

- The y-intercept is the initial volume. - The slope is the change in gallons per minute. - Use two clear points on each graph. - Read each displayed time interval from the graph.

Solution

1. Tank 1 starts with \(10\) gallons, so \(b=10\). After \(5\) minutes it contains \(30\) gallons, so \(m=\frac{30-10}{5-0}=4\). Therefore, \(V_1(t)=4t+10\), with \(0\le t\le12\). 2. Tank 2 starts with \(50\) gallons, so \(b=50\). It is empty after \(10\) minutes, so \(m=\frac{0-50}{10-0}=-5\). Therefore, \(V_2(t)=-5t+50\), with \(0\le t\le10\).

Answer

Tank 1: \(V_1(t)=4t+10\), with \(0\le t\le12\); Tank 2: \(V_2(t)=-5t+50\), with \(0\le t\le10\)
5338778
A taxi company charges a fixed starting fee plus a constant price per mile. The graph shows the total cost \(y\), in dollars, for a trip of \(x\) miles. Find the equation of the line.
Figure for problem 533877

Hints

- Where does the line cross the vertical axis? That value represents the starting fee. - How much does the cost increase for each additional mile? - Use the form \(y = mx + b\).

Solution

1. The line crosses the y-axis at \(4\), so the starting fee is \(\$4\). 2. Using \((0, 4)\) and \((3, 10)\), the slope is \(m = \frac{10 - 4}{3 - 0} = 2\). This means the taxi charges \(\$2\) per mile. 3. Therefore, the equation is \(y = 2x + 4\).

Answer

\(y = 2x + 4\)
5119128
Two runners, Alex and Blake, train on a long, straight trail. Alex starts at mile marker \(0\) and runs at a constant speed of \(8\,\text{mi/h}\). Blake starts at the same time at mile marker \(2\) and runs at a constant speed of \(6\,\text{mi/h}\). Write a linear position function for each runner, then use the models to determine when and where Alex catches Blake.

Hints

- For each runner, identify the starting position and the distance added each hour. - A linear position model has the form starting position plus rate times time. - At the catch-up time, what must be true about the two position values?

Solution

1. Alex starts at \(0\) and gains \(8\) miles each hour, so \(s_A(t)=8t\). 2. Blake starts at mile marker \(2\) and gains \(6\) miles each hour, so \(s_B(t)=2+6t\). 3. At the catch-up time the positions are equal: \(8t=2+6t\). Thus \(2t=2\), so \(t=1\) hour. 4. Substitute \(t=1\): \(s_A(1)=8\) and \(s_B(1)=8\). Alex catches Blake at mile marker \(8\).

Answer

\(s_A(t)=8t\) and \(s_B(t)=2+6t\). Alex catches Blake after \(1\) hour at mile marker \(8\).
5119138
A smartphone battery drains at a constant rate during a certain type of use. It is fully charged at \(10{:}00\) a.m. The table shows the battery level at several times. <table><tr><td>Time</td><td>\(10{:}00\) a.m.</td><td>\(12{:}00\) p.m.</td><td>\(2{:}00\) p.m.</td><td>\(3{:}00\) p.m.</td><td>\(6{:}00\) p.m.</td></tr><tr><td>Battery level (%)</td><td>\(100\)</td><td>\(84\)</td><td>\(68\)</td><td>\(60\)</td><td>\(36\)</td></tr></table> a) Find the average decrease in battery level per hour. Verify that the rate is constant over every interval in the table. b) Estimate the battery level at \(4{:}30\) p.m. Explain your method. c) If the pattern continues, at what time will the battery reach \(0\%\)? d) Write a rule for the battery level \(y\) after \(x\) hours have passed since \(10{:}00\) a.m.

Hints

- For each interval, divide the change in battery level by the elapsed time. - Find how many hours pass from \(10{:}00\) a.m. to \(4{:}30\) p.m. - Set the battery-level expression equal to \(0\) to find when the battery is empty. - Identify the initial value and the change per hour.

Solution

1. From \(10{:}00\) a.m. to \(12{:}00\) p.m., the level decreases by \(16\) percentage points in \(2\) hours, or \(8\) points per hour. The changes over the other intervals are also \(16\div2=8\), \(8\div1=8\), and \(24\div3=8\) percentage points per hour. The rate is constant. 2. By \(4{:}30\) p.m., \(6.5\) hours have passed. The model gives \(100-8(6.5)=48\), so the battery level is \(48\%\). 3. Solve \(100-8x=0\). Then \(x=12.5\), which is \(12\) hours \(30\) minutes after \(10{:}00\) a.m., or \(10{:}30\) p.m. 4. The initial level is \(100\), and it decreases by \(8\) percentage points per hour, so \(y=100-8x\).

Answer

a) A constant decrease of \(8\) percentage points per hour b) \(48\%\) c) \(10{:}30\) p.m. d) \(y=100-8x\)
5119218
A taxi company charges a base fare of \(\$4.50\) plus \(\$2.20\) per mile. a) Write an equation for the total fare \(y\), in dollars, after traveling \(x\) miles. b) Explain what \(4.50\) represents in this situation. c) A passenger pays exactly \(\$26.50\). How many miles did the passenger travel?

Hints

- Separate the fixed charge from the charge that depends on distance. - Identify the rate per mile and the initial value. - For part c), subtract the base fare before dividing by the rate.

Solution

1. The total fare is the mileage charge plus the base fare, so \(y = 2.20x + 4.50\). 2. The value \(4.50\) is the fare charged before any miles are traveled. It is the initial value of the model. 3. Substitute \(y = 26.50\): \(26.50 = 2.20x + 4.50\). 4. Subtract \(4.50\): \(22.00 = 2.20x\). 5. Divide by \(2.20\): \(x = 10\).

Answer

a) \(y = 2.20x + 4.50\) b) The \(\$4.50\) is the base fare charged regardless of distance. c) The passenger traveled \(10\) miles.
5119248
Two mobile data plans have the following monthly costs. - Plan A: a \(\$10.00\) base fee plus \(\$0.50\) per gigabyte. - Plan B: no base fee and \(\$1.50\) per gigabyte. a) Write an expression for each plan's monthly cost \(C\) as a function of data use \(x\), in gigabytes. b) Find both costs for \(5\,\text{GB}\) and \(15\,\text{GB}\). c) For what data use is Plan A less expensive than Plan B? Explain by comparing the expressions or testing values.

Hints

- A base fee is included even when no data is used. - Make a small table to compare the costs. - Find the input where the two cost expressions are equal.

Solution

1. The cost functions are \(C_A(x)=10+0.50x\) and \(C_B(x)=1.50x\). 2. At \(5\,\text{GB}\), Plan A costs \(\$10.00+\$0.50 \cdot 5=\$12.50\), and Plan B costs \(\$1.50 \cdot 5=\$7.50\). 3. At \(15\,\text{GB}\), Plan A costs \(\$17.50\), and Plan B costs \(\$22.50\). 4. The costs are equal when \(10+0.50x=1.50x\), which gives \(x=10\). Plan A is less expensive when \(x>10\).

Answer

a) \(C_A(x)=10+0.50x\); \(C_B(x)=1.50x\) b) At \(5\,\text{GB}\): Plan A \(\$12.50\), Plan B \(\$7.50\). At \(15\,\text{GB}\): Plan A \(\$17.50\), Plan B \(\$22.50\). c) Plan A is less expensive for more than \(10\,\text{GB}\).
5119548
A candle is \(30\,\text{cm}\) tall when it is lit. After burning at a constant rate for \(4\,\text{hours}\), it is \(22\,\text{cm}\) tall. a) Is the relationship from burn time (in hours) to remaining height (in centimeters) proportional? Briefly explain. b) Write a function \(h(x)\) for the candle's remaining height after \(x\) hours, while the candle is burning. c) Use your function to find the candle's height after \(7.5\,\text{hours}\).

Hints

- Think about the candle's height at time \(0\). Would a proportional relationship start there? - How many centimeters of height does the candle lose each hour? - Start your function with the candle's initial height and account for the amount lost each hour.

Solution

1. A proportional relationship must include \((0, 0)\). Here, the candle has height \(30\,\text{cm}\) at \(0\,\text{hours}\), so the relationship is not proportional. 2. In \(4\,\text{hours}\), the candle loses \(30 - 22 = 8\,\text{cm}\), so its rate of change is \(-8 \div 4 = -2\,\text{cm/h}\). 3. Starting from \(30\,\text{cm}\) and decreasing by \(2\,\text{cm}\) per hour gives \(h(x) = 30 - 2x\), with \(0 \le x \le 15\). 4. Evaluate the function at \(x = 7.5\): \(h(7.5) = 30 - 2 \cdot 7.5 = 15\).

Answer

a) No. The relationship is not proportional because it does not pass through \((0, 0)\); at \(0\,\text{hours}\), the candle is \(30\,\text{cm}\) tall. b) \(h(x) = 30 - 2x\) for \(0 \le x \le 15\). c) \(15\,\text{cm}\)
5120438
Two different candles are lit at the same time. Candle A is initially \(30\,\text{cm}\) tall and is \(18\,\text{cm}\) tall after \(4\) hours. Candle B is initially \(20\,\text{cm}\) tall and burns down at a constant rate of \(1\,\frac{\text{cm}}{\text{h}}\). Let \(t\) be the number of hours after the candles are lit. a) Determine Candle A's constant rate of change and construct linear height functions \(h_A(t)\) and \(h_B(t)\). b) Use your model to find Candle A's height after \(7\) hours and the time when Candle A reaches height \(0\). c) Use the two functions to determine when the candles have the same height and what that height is.

Hints

- Use the two measurements for Candle A to find its change in height per hour. - A linear model needs both an initial value and a signed rate of change. - Once both functions are written, equal heights correspond to equal function outputs at the same input.

Solution

1. Candle A loses \(30-18=12\,\text{cm}\) in \(4\) hours, so its rate of change is \(-3\,\frac{\text{cm}}{\text{h}}\). 2. Therefore, \(h_A(t)=30-3t\) and \(h_B(t)=20-t\). 3. \(h_A(7)=30-3\cdot7=9\), so Candle A is \(9\,\text{cm}\) tall after \(7\) hours. 4. Solve \(30-3t=0\). This gives \(t=10\), so Candle A reaches height \(0\) after \(10\) hours. 5. Set the functions equal: \(30-3t=20-t\). Then \(10=2t\), so \(t=5\). 6. At \(t=5\), \(h_B(5)=20-5=15\). Both candles are \(15\,\text{cm}\) tall.

Answer

a) Candle A's rate is \(-3\,\frac{\text{cm}}{\text{h}}\); \(h_A(t)=30-3t\), \(h_B(t)=20-t\) b) \(9\,\text{cm}\); Candle A reaches height \(0\) after \(10\) hours. c) After \(5\) hours, both candles are \(15\,\text{cm}\) tall.
5120458
Lara watches her smartphone battery during a long train ride while playing a game continuously. At \(12{:}15\) p.m., the battery level is \(88\%\). At \(2{:}45\) p.m., it is \(63\%\). a) Find the decrease in battery level, in percentage points per hour. Under this linear assumption, at what time would the battery reach \(0\%\)? b) According to the model, what would the battery level have been at \(10{:}15\) a.m.? c) Explain why the result from part b shows that the linear model cannot be valid for the entire morning.

Hints

- Convert the elapsed time to hours before finding the rate. - Use the rate to calculate both forward and backward in time. - Consider the greatest possible battery level.

Solution

1. From \(12{:}15\) p.m. to \(2{:}45\) p.m. is \(2.5\) hours. The battery decreases by \(88-63=25\) percentage points, so the rate is \(25\div2.5=10\) percentage points per hour. 2. Starting from \(63\%\), reaching \(0\%\) takes \(63\div10=6.3\) hours, or \(6\) hours \(18\) minutes. That is \(9{:}03\) p.m. 3. At \(10{:}15\) a.m., two hours before the first measurement, the linear model gives \(88+2(10)=108\%\). 4. A battery cannot be charged above \(100\%\). Therefore, the constant-rate model cannot be extended that far backward in time.

Answer

a) \(10\) percentage points per hour; \(9{:}03\) p.m. b) \(108\%\) c) The result exceeds the physical maximum of \(100\%\), so the model is not valid for that earlier time.
5120518
A sports club tracks membership in its tennis and basketball programs. <table> <thead> <tr><th>Year</th><th>2010</th><th>2013</th><th>2016</th><th>2019</th><th>2022</th></tr> </thead> <tbody> <tr><td>Tennis</td><td>\(240\)</td><td>\(230\)</td><td>\(215\)</td><td>\(205\)</td><td>\(190\)</td></tr> <tr><td>Basketball</td><td>\(80\)</td><td>\(110\)</td><td>\(145\)</td><td>\(175\)</td><td>\(210\)</td></tr> </tbody> </table> For parts a)–c), assume each program changed at a constant rate from 2019 onward equal to its average rate from 2019 to 2022. Let \(t\) be the number of years after 2019. a) Construct linear models \(T(t)\) and \(B(t)\) for tennis and basketball membership. b) Use the models to estimate when the memberships were equal. c) Use the basketball model to predict membership in 2025.

Hints

- Use only the 2019 and 2022 values to determine each assumed constant rate for this model. - Each model needs the 2019 membership as its initial value because \(t=0\) represents 2019. - Equal memberships occur when the two model outputs are equal.

Solution

1. Tennis changes from \(205\) to \(190\) in \(3\) years, so its rate is \(\frac{190-205}{3}=-5\) members per year. Thus, \(T(t)=205-5t\). 2. Basketball changes from \(175\) to \(210\) in \(3\) years, so its rate is \(\frac{210-175}{3}=\frac{35}{3}\) members per year. Thus, \(B(t)=175+\frac{35}{3}t\). 3. Set the models equal: \(205-5t=175+\frac{35}{3}t\). 4. Then \(30=\frac{50}{3}t\), so \(t=1.8\). This is about \(1.8\) years after 2019, or approximately 2021. 5. The year 2025 is \(6\) years after 2019. \(B(6)=175+\frac{35}{3}\cdot6=245\).

Answer

a) \(T(t)=205-5t\), \(B(t)=175+\frac{35}{3}t\) b) Approximately 2021 c) \(245\) members
5120578
A pump fills an empty tank at a constant rate. After \(15\) minutes, the tank contains \(450\) gallons. a) Write a function for the volume \(V\), in gallons, after \(t\) minutes. b) A second pump moves twice as much water per minute. Write its function. c) How long would each pump take to fill a \(2700\)-gallon tank? How much time does the faster pump save?

Hints

- First find the gallons pumped per minute. - How does doubling the rate change the equation? - Set the volume equal to the tank capacity and solve for time.

Solution

1. The first pump's rate is \(450\div 15=30\) gallons per minute, so \(V_1(t)=30t\). 2. The second pump's rate is \(2\cdot 30=60\) gallons per minute, so \(V_2(t)=60t\). 3. The first pump takes \(2700\div 30=90\) minutes. 4. The second pump takes \(2700\div 60=45\) minutes. 5. The time saved is \(90-45=45\) minutes.

Answer

a) \(V_1(t)=30t\). b) \(V_2(t)=60t\). c) The first pump takes \(90\) minutes, the second takes \(45\) minutes, and the second saves \(45\) minutes.
5120598
A vehicle uses \(18\) gallons of gasoline to travel \(600\) miles. Assume gasoline used is proportional to distance. a) Write a function of the form \(f(x)=kx\) for the gasoline use \(f(x)\), in gallons, after traveling \(x\) miles. What does \(k\) mean in context? b) While towing a trailer, the vehicle uses an additional \(1.2\) gallons for every \(100\) miles. How many gallons will it use under these conditions on a \(350\)-mile trip?

Hints

- Find the gasoline used for one mile. - Interpret the constant in \(y=kx\) using units. - Convert the added use from per \(100\) miles to per mile before combining rates.

Solution

1. The original rate is \(k=18\div 600=0.03\) gallon per mile. 2. The function is \(f(x)=0.03x\). The constant \(k\) is the gallons used per mile. 3. The added rate is \(1.2\div 100=0.012\) gallon per mile. 4. The towing rate is \(0.03+0.012=0.042\) gallon per mile. 5. For \(350\) miles, the vehicle uses \(350\cdot 0.042=14.7\) gallons.

Answer

a) \(f(x)=0.03x\); \(k=0.03\) is the gasoline used per mile. b) \(14.7\) gallons.
5124528
A class is ordering printed T-shirts. InkWorks charges a one-time setup fee of \(\$24\) plus \(\$7\) per shirt. TeeDirect charges \(\$10\) per shirt with no additional fee. Let \(x\) be the number of shirts ordered. a) Construct cost functions \(I(x)\) and \(T(x)\) for InkWorks and TeeDirect. b) Use the functions to find the cost of \(5\) shirts and \(10\) shirts from each company. c) Starting with how many whole shirts is InkWorks less expensive than TeeDirect?

Hints

- Separate each company's fixed cost from its per-shirt cost when constructing the functions. - Check that the output at \(x=0\) matches each company's setup-fee situation. - After the functions are written, translate “InkWorks is less expensive” into a comparison of their outputs.

Solution

1. InkWorks has a fixed fee of \(24\) dollars and a per-shirt rate of \(7\) dollars, so \(I(x)=24+7x\). TeeDirect has no fixed fee and charges \(10\) dollars per shirt, so \(T(x)=10x\). 2. \(I(5)=24+7\cdot5=59\) and \(T(5)=10\cdot5=50\). 3. \(I(10)=24+7\cdot10=94\) and \(T(10)=100\). 4. InkWorks is less expensive when \(24+7x<10x\). This gives \(24<3x\), so \(x>8\). 5. The least whole-number order greater than \(8\) is \(9\) shirts.

Answer

a) \(I(x)=24+7x\), \(T(x)=10x\) b) For \(5\) shirts: InkWorks \(\$59\), TeeDirect \(\$50\). For \(10\) shirts: InkWorks \(\$94\), TeeDirect \(\$100\). c) \(9\) shirts
5125078
Lucas and Sarah are saving for bicycles. Lucas already has \(\$60\) and adds \(\$4\) each week. Sarah has \(\$10\) and adds \(\$9\) each week. a) Write a linear model for each person's savings after \(w\) weeks. b) Use the models to determine after how many weeks they will have the same amount and state that amount.

Hints

- In each model, identify the starting amount and the amount added each week. - The weekly amount is the coefficient of \(w\), and the starting amount is the constant term. - To find when the balances match, compare the two model outputs for the same value of \(w\).

Solution

1. Lucas starts with \(60\) dollars and adds \(4\) dollars per week, so \(L(w)=60+4w\). 2. Sarah starts with \(10\) dollars and adds \(9\) dollars per week, so \(S(w)=10+9w\). 3. Set the models equal to find when the balances match: \(60+4w=10+9w\). 4. Subtract \(10\) and \(4w\): \(50=5w\), so \(w=10\). 5. Evaluate either model: \(L(10)=60+4\cdot10=100\). Each person has \(\$100\).

Answer

a) Lucas: \(L(w)=60+4w\); Sarah: \(S(w)=10+9w\) b) After \(10\) weeks, each has \(\$100\).
5125708
A large tank contains \(450\,\text{L}\) of water and loses \(15\,\frac{\text{L}}{\text{min}}\) through a leak. At the same time, a second tank, initially empty, is filled at \(25\,\frac{\text{L}}{\text{min}}\). Let \(t\) be the number of minutes after both processes begin. a) Construct volume functions \(V_1(t)\) and \(V_2(t)\) for the two tanks. b) Use the functions to find when the tanks contain the same amount of water and how much each contains then. c) Explain from the two models why the second tank contains more water after that time, while both models remain physically meaningful.

Hints

- A linear volume model combines the starting amount with a signed per-minute change. - Write both functions before setting any amounts equal. - Compare the signs of the two rates to reason about what happens after the intersection.

Solution

1. The first tank starts at \(450\) liters and decreases by \(15\) liters per minute, so \(V_1(t)=450-15t\). 2. The second starts at \(0\) and increases by \(25\) liters per minute, so \(V_2(t)=25t\). 3. Set the functions equal: \(450-15t=25t\). Then \(450=40t\), so \(t=11.25\). 4. \(V_2(11.25)=25\cdot11.25=281.25\). Each tank contains \(281.25\,\text{L}\). 5. After the intersection, \(V_1\) continues to decrease while \(V_2\) continues to increase, so \(V_2(t)>V_1(t)\) after that time within the physical domains.

Answer

a) \(V_1(t)=450-15t\), \(V_2(t)=25t\) b) After \(11.25\) minutes, each tank contains \(281.25\,\text{L}\). c) \(V_1\) decreases while \(V_2\) increases, so after their intersection the second model stays above the first while both apply.
5125848
Two conveyor belts add boxes to two stacks. Stack A starts with \(15\) boxes and receives \(5\) boxes per minute. Stack B starts with \(45\) boxes and receives \(2\) boxes per minute. Let \(t\) be the number of minutes after the belts start. a) Construct functions \(A(t)\) and \(B(t)\) for the number of boxes in each stack. b) Which stack grows faster? Explain using the functions' rates of change. c) Use the functions to find when the stacks contain the same number of boxes and how many boxes each contains then.

Hints

- Use each starting count as the initial value of its model. - The coefficient of time should represent the number of boxes added each minute. - Equal stack sizes mean the two function outputs are equal for the same input time.

Solution

1. Stack A has model \(A(t)=15+5t\), and Stack B has model \(B(t)=45+2t\). 2. Stack A grows faster because its rate of change is \(5\,\frac{\text{boxes}}{\text{min}}\), compared with \(2\,\frac{\text{boxes}}{\text{min}}\) for Stack B. 3. Set \(A(t)=B(t)\): \(15+5t=45+2t\). 4. Then \(3t=30\), so \(t=10\). 5. \(A(10)=15+5\cdot10=65\), so each stack contains \(65\) boxes.

Answer

a) \(A(t)=15+5t\), \(B(t)=45+2t\) b) Stack A c) After \(10\) minutes, each stack contains \(65\) boxes.
5125858
Two candles are lit at the same time. Candle A is \(20\,\text{cm}\) tall and burns down \(1.5\,\frac{\text{cm}}{\text{h}}\). Candle B is \(15\,\text{cm}\) tall and burns down \(0.5\,\frac{\text{cm}}{\text{h}}\). Let \(t\) be the number of hours after the candles are lit. a) Construct height functions \(A(t)\) and \(B(t)\). b) Use the functions to find when the candles have the same height and what that height is. c) Which candle is taller after \(7\) hours? Justify using the functions.

Hints

- Burning gives each height model a negative rate of change. - Build both functions from initial height plus signed rate times time. - Use equal outputs for the same input to model equal heights.

Solution

1. Candle A starts at \(20\) centimeters and decreases by \(1.5\) centimeters per hour, so \(A(t)=20-1.5t\). 2. Candle B starts at \(15\) centimeters and decreases by \(0.5\) centimeter per hour, so \(B(t)=15-0.5t\). 3. Set the functions equal: \(20-1.5t=15-0.5t\). Then \(t=5\). 4. \(A(5)=20-1.5\cdot5=12.5\), so both candles are \(12.5\,\text{cm}\) tall. 5. \(A(7)=9.5\) and \(B(7)=11.5\), so Candle B is taller after \(7\) hours.

Answer

a) \(A(t)=20-1.5t\), \(B(t)=15-0.5t\) b) After \(5\) hours, both are \(12.5\,\text{cm}\) tall. c) Candle B; \(B(7)=11.5\,\text{cm}\) and \(A(7)=9.5\,\text{cm}\).
5125928
Julian compares two monthly cell phone plans. The Basic plan costs \(\$12.00\) plus \(\$0.08\) per minute. The Flat plan costs \(\$22.00\), regardless of the number of minutes used. Let \(m\) be the number of minutes used in a month. a) Construct monthly cost functions \(B(m)\) and \(F(m)\) for the Basic and Flat plans. b) Use the functions to determine the least whole number of minutes for which the Flat plan is less expensive than the Basic plan. Show the comparison as an inequality.

Hints

- A cost function should show both the fixed charge and the amount that depends on minutes. - A constant-price plan is still a linear function with rate of change \(0\). - Compare the two function outputs with a strict inequality, then interpret the whole-number boundary.

Solution

1. The Basic plan has a fixed cost of \(12\) dollars and a rate of \(0.08\) dollar per minute, so \(B(m)=12+0.08m\). 2. The Flat plan has constant cost \(F(m)=22\). 3. Flat is less expensive when \(F(m)<B(m)\): \(22<12+0.08m\). 4. Subtract \(12\): \(10<0.08m\). Divide by \(0.08\): \(125<m\). 5. The least whole number greater than \(125\) is \(126\).

Answer

a) \(B(m)=12+0.08m\), \(F(m)=22\) b) \(126\) minutes
5126108
Luke compares two gym memberships. Power: a \(\$45.00\) sign-up fee plus \(\$25.00\) per month Easy: no sign-up fee and \(\$35.00\) per month Let \(m\) be the number of months of membership. a) Construct total-cost functions \(P(m)\) and \(E(m)\). b) Use the functions to find the total cost of each membership after \(4\) months. c) Starting with which full month is Power less expensive overall than Easy? Justify with a comparison of the two functions.

Hints

- Include the sign-up fee as the initial value of only the Power model. - Check what each model would output at \(m=0\) to verify the fixed-cost structure. - For part c), compare the two function outputs, not just the monthly rates.

Solution

1. Power has model \(P(m)=45+25m\). Easy has model \(E(m)=35m\). 2. \(P(4)=45+25\cdot4=145\), and \(E(4)=35\cdot4=140\). 3. Power is less expensive when \(45+25m<35m\). 4. Then \(45<10m\), so \(m>4.5\). 5. The first full month satisfying the inequality is month \(5\).

Answer

a) \(P(m)=45+25m\), \(E(m)=35m\) b) Power: \(\$145.00\); Easy: \(\$140.00\) c) Month \(5\)
5126118
A family compares two annual cell phone plans. Smart: a \(\$114.00\) annual fee plus \(\$0.05\) per minute Basic: no annual fee and \(\$0.15\) per minute Let \(m\) be the number of minutes used during the year. a) Construct annual cost functions \(S(m)\) and \(B(m)\). b) Compare the annual costs for \(960\) minutes. c) Starting at how many whole minutes per year is Smart less expensive than Basic?

Hints

- Use one consistent annual input variable for both models. - The annual fee is the initial value of the Smart model; Basic has initial value \(0\). - For the threshold, compare the two model outputs with a strict inequality.

Solution

1. Smart has model \(S(m)=114+0.05m\). Basic has model \(B(m)=0.15m\). 2. \(S(960)=114+0.05\cdot960=162\), while \(B(960)=0.15\cdot960=144\). Basic is less expensive at \(960\) minutes. 3. Smart is less expensive when \(114+0.05m<0.15m\). 4. Subtract \(0.05m\): \(114<0.10m\), so \(m>1140\). 5. The least whole-number usage is \(1141\) minutes.

Answer

a) \(S(m)=114+0.05m\), \(B(m)=0.15m\) b) Smart: \(\$162.00\); Basic: \(\$144.00\), so Basic is less expensive. c) \(1141\) minutes
5126128
Two car-sharing companies offer these monthly plans. Company A: a \(\$15.00\) monthly fee plus \(\$0.20\) per mile Company B: no monthly fee and \(\$0.45\) per mile Let \(m\) be the number of miles driven in a month. a) Construct monthly cost functions \(A(m)\) and \(B(m)\). b) Use the functions to find the monthly cost for each company at \(40\) miles. c) For what monthly mileage is Company A less expensive than Company B?

Hints

- Translate each fixed fee and per-mile charge into the corresponding parts of a linear function. - Verify the models by considering what each company would charge at \(0\) miles. - Compare the two outputs with a strict inequality for part c).

Solution

1. Company A has model \(A(m)=15+0.20m\). Company B has model \(B(m)=0.45m\). 2. \(A(40)=15+0.20\cdot40=23\), while \(B(40)=0.45\cdot40=18\). 3. Company A is less expensive when \(15+0.20m<0.45m\). 4. Then \(15<0.25m\), so \(m>60\).

Answer

a) \(A(m)=15+0.20m\), \(B(m)=0.45m\) b) Company A: \(\$23.00\); Company B: \(\$18.00\) c) More than \(60\) miles per month
5128128
A portable solar panel charges a battery. Under constant conditions, the stored energy increases by \(4\) watt-hours, or \(\text{Wh}\), every \(15\) minutes. Assume the battery starts empty and can store at least \(50\,\text{Wh}\). a) How much energy is stored after \(2\) hours? b) After how many hours, minutes, and seconds has the panel delivered exactly \(50\,\text{Wh}\)? c) Write a function in the form \(y=kx\), where \(x\) is time in minutes and \(y\) is energy in watt-hours. Explain what \(k\) means in context.

Hints

- Convert two hours to minutes. - Find the energy added in one minute. - Use the constant rate to work backward from \(50\,\text{Wh}\). - In \(y=kx\), \(k\) represents a rate.

Solution

1. The rate is \(k=\frac{4}{15}\) watt-hour per minute. 2. Two hours is \(120\) minutes, so \(y=120\cdot \frac{4}{15}=32\,\text{Wh}\). 3. For \(50\,\text{Wh}\), \(x=50\cdot \frac{15}{4}=187.5\) minutes. 4. This is \(3\) hours, \(7\) minutes, and \(30\) seconds. 5. The function is \(y=\frac{4}{15}x\). The constant \(k\) is the energy added per minute.

Answer

a) \(32\,\text{Wh}\). b) \(3\) hours, \(7\) minutes, \(30\) seconds. c) \(y=\frac{4}{15}x\); \(k\) is the watt-hours added per minute.
5128568
A vehicle uses gasoline at a constant rate of \(4\) gallons per \(100\) miles. a) Write a function for the volume \(V\), in gallons, used after traveling \(s\) miles. b) Find the gasoline used on a \(450\)-mile trip. c) The tank has \(12\) gallons remaining. How far can the vehicle theoretically travel?

Hints

- Convert the rate per \(100\) miles to a rate per mile. - Identify which quantity depends on the other. - Use the equation forward for part b and solve it backward for part c. - A small value table may help you check the model.

Solution

1. The rate per mile is \(4\div 100=0.04\) gallon per mile. 2. The function is \(V(s)=0.04s\). 3. For \(450\) miles, \(V(450)=0.04\cdot 450=18\) gallons. 4. With \(12\) gallons, solve \(12=0.04s\): \(s=12\div 0.04=300\) miles.

Answer

a) \(V(s)=0.04s\). b) \(18\) gallons. c) \(300\) miles.
5129228
On a sunny day, Anja observes the shadow of a vertical pole in the schoolyard. She marks the shadow's position at \(10{:}00\) a.m. At \(12{:}30\) p.m., the shadow has swept through an angle of \(37.5^\circ\). Assume the shadow angle changes at a constant rate from \(10{:}00\) a.m. through \(4{:}15\) p.m. a) Find the change in angle per hour. b) Write a function for the angle \(\alpha\), in degrees, after \(t\) hours from the first measurement. c) What total angle has the shadow swept through by \(4{:}15\) p.m.?

Hints

- Find the elapsed time between the first two measurements. - Convert minutes to a fraction or decimal part of an hour. - Use the constant change in angle per hour. - What form does a proportional function have?

Solution

1. The elapsed time from \(10{:}00\) a.m. to \(12{:}30\) p.m. is \(2.5\) hours. 2. The rate is \(37.5\div 2.5=15^\circ\) per hour. 3. The function is \(\alpha(t)=15t\). 4. From \(10{:}00\) a.m. to \(4{:}15\) p.m. is \(6.25\) hours. 5. The angle is \(15\cdot 6.25=93.75^\circ\).

Answer

a) \(15^\circ\) per hour. b) \(\alpha(t)=15t\). c) \(93.75^\circ\).
5129628
A linear function \(g\) has graph passing through \(P(-1,7)\) and \(Q(2,-2)\). a) Construct the function in the form \(g(x)=mx+b\). b) Use your function to determine whether \(R(0,5)\) and \(S(1,1)\) are input-output pairs of \(g\).

Hints

- A linear function through two points is determined by its slope and one point. - After finding the slope, use one given point to determine the initial value \(b\). - Only after the function is constructed should you test the proposed ordered pairs.

Solution

1. The slope through \(P\) and \(Q\) is \(m=\frac{-2-7}{2-(-1)}=-3\). 2. Substitute \(P(-1,7)\) into \(g(x)=-3x+b\): \(7=-3(-1)+b\), so \(b=4\). 3. Therefore, \(g(x)=-3x+4\). 4. \(g(0)=4\), not \(5\), so \(R(0,5)\) is not an input-output pair. 5. \(g(1)=1\), so \(S(1,1)\) is an input-output pair.

Answer

a) \(g(x)=-3x+4\) b) \(R(0,5)\) is not an input-output pair; \(S(1,1)\) is.
5129638
An electric-vehicle charging plan has a monthly base fee of \(\$18.50\) plus \(\$0.80\) for each kilowatt-hour (\(\text{kWh}\)) used. Let \(x\) be the number of kilowatt-hours used in a month. a) Construct a monthly cost function \(C(x)\), and explain what its initial value and rate of change represent. b) A customer says a month with \(40\,\text{kWh}\) of charging cost exactly \(\$50.50\). Does the function support the claim? c) How many kilowatt-hours were used if the monthly bill was exactly \(\$74.50\)?

Hints

- Build the function by separating the fixed monthly amount from the usage-dependent amount. - Check what the model gives at input \(0\) and for one additional kilowatt-hour. - Use the completed function for both the forward and reverse questions.

Solution

1. The base fee is the initial value and the per-kilowatt-hour charge is the rate, so \(C(x)=18.50+0.80x\). 2. The initial value \(18.50\) is the monthly cost at zero usage. The rate \(0.80\) means the bill increases by \(\$0.80\) for each additional kilowatt-hour. 3. \(C(40)=18.50+0.80\cdot40=50.50\), so the customer's claim matches the model. 4. Solve \(74.50=18.50+0.80x\). Then \(56=0.80x\), so \(x=70\).

Answer

a) \(C(x)=18.50+0.80x\); the initial value is the \(\$18.50\) base fee, and the rate is \(\$0.80\) per \(\text{kWh}\). b) Yes. c) \(70\,\text{kWh}\)
5129658
A line \(g\) passes through \(P(4, -1)\) and has slope \(m = -0.75\). A student wants to find more points on the line using slope triangles. a) Starting at \(P\), the student moves 8 units left. How many units must the student move up or down to land on the line again? b) What is the vertical change when the student moves exactly 1 unit right? c) Find an equation for line \(g\).

Hints

- How are slope, horizontal change, and vertical change related? - What does a negative slope tell you about the direction of the vertical change? - How can you use the given point and slope to find the y-intercept?

Solution

1. For part a, \(\Delta x = -8\). Since \(\Delta y = m\Delta x\), \(\Delta y = -0.75 \cdot (-8) = 6\). The student moves 6 units up. 2. For part b, \(\Delta x = 1\), so \(\Delta y = -0.75 \cdot 1 = -0.75\). The student moves \(0.75\) unit down. 3. Write \(g(x) = -0.75x + b\). Using \(P(4, -1)\), \(-1 = -0.75(4) + b\), so \(b = 2\). Therefore, \(g(x) = -0.75x + 2\).

Answer

a) 6 units up b) \(\Delta y = -0.75\), so \(0.75\) unit down c) \(g(x) = -0.75x + 2\)
5129708
A small backyard pool already contains \(30\) gallons of water. A garden hose adds water at a constant rate of \(4\) gallons per minute. a) Write a function \(V(t)\) for the volume of water, in gallons, after \(t\) minutes. b) How much water is in the pool after \(20\) minutes? c) How long will it take for the pool to contain \(150\) gallons?

Hints

- Which value represents the amount of water at time \(0\)? - How much does the volume change each minute? - Decide whether you need to substitute a time into the function or solve an equation for time.

Solution

1. The initial volume is \(30\) gallons and the rate of change is \(4\) gallons per minute, so \(V(t) = 4t + 30\). 2. Evaluate at \(t = 20\): \(V(20) = 4 \cdot 20 + 30 = 110\). The pool contains \(110\) gallons. 3. Set the volume equal to \(150\): \(150 = 4t + 30\). Then \(120 = 4t\), so \(t = 30\) minutes.

Answer

a) \(V(t) = 4t + 30\) b) \(110\) gallons c) \(30\) minutes
5129718
Two candles are lit at the same time. Candle A starts at \(24\,\text{cm}\) and burns down at \(2\,\text{cm/h}\). Candle B starts at \(18\,\text{cm}\) and burns down at \(1.2\,\text{cm/h}\). a) Write a function for the height \(h\), in centimeters, of each candle after \(t\) hours. b) Which candle is taller after \(5\) hours? Show your calculation. c) After how many hours will the two candles have the same height?

Hints

- How does a constant burn rate affect the sign of the rate in the equation? - What must be true about the two function values when the candles have the same height? - Write the two height expressions before comparing them.

Solution

1. Candle A is modeled by \(h_A(t) = 24 - 2t\). Candle B is modeled by \(h_B(t) = 18 - 1.2t\). 2. After \(5\) hours, \(h_A(5) = 24 - 2 \cdot 5 = 14\,\text{cm}\) and \(h_B(5) = 18 - 1.2 \cdot 5 = 12\,\text{cm}\). Candle A is taller. 3. Set the heights equal: \(24 - 2t = 18 - 1.2t\). Then \(6 = 0.8t\), so \(t = 7.5\).

Answer

a) \(h_A(t) = 24 - 2t\) and \(h_B(t) = 18 - 1.2t\) b) Candle A; after \(5\) hours it is \(14\,\text{cm}\) tall and Candle B is \(12\,\text{cm}\) tall. c) \(7.5\) hours
5129728
A delivery service charges a fixed fee plus a constant amount per mile. A \(10\)-mile delivery costs \(\$17.00\), and a \(25\)-mile delivery costs \(\$32.00\). a) Write a function \(C(d)\) for the total cost, in dollars, of a delivery that travels \(d\) miles. b) What is the fixed fee? c) Find the cost of a \(40\)-mile delivery.

Hints

- Treat distance and total cost as ordered pairs. - How much does the cost change when the distance changes? - After finding the per-mile charge, subtract the distance-based part from a total cost to find the fixed fee.

Solution

1. Use the two data points \((10, 17)\) and \((25, 32)\). The rate per mile is \(m = \frac{32 - 17}{25 - 10} = 1\). 2. Substitute \((10, 17)\) into \(C(d) = md + b\): \(17 = 10 + b\), so \(b = 7\). 3. The cost function is \(C(d) = d + 7\), so the fixed fee is \(\$7.00\). 4. Evaluate at \(d = 40\): \(C(40) = 40 + 7 = 47\).

Answer

a) \(C(d) = d + 7\) b) \(\$7.00\) c) \(\$47.00\)
5130018
A candle burns at a constant rate. After \(3\) hours, it is \(7\) inches tall. Its height decreases by \(0.75\) inch each hour. a) Write a function \(h(t)\) for the candle's height, in inches, after \(t\) hours. b) How tall was the candle when it was first lit?

Hints

- Should the rate of change be positive or negative when the candle gets shorter? - What time represents the moment the candle is first lit? - Which information gives you a point on the graph of the height function?

Solution

1. Because the candle gets shorter, the rate of change is \(m = -0.75\) inch per hour. 2. The information after \(3\) hours gives the point \((3, 7)\). 3. Substitute into \(h(t) = mt + b\): \(7 = -0.75 \cdot 3 + b\), so \(b = 9.25\). 4. Therefore, \(h(t) = -0.75t + 9.25\). The initial height is \(h(0) = 9.25\) inches.

Answer

a) \(h(t) = -0.75t + 9.25\) b) \(9.25\) inches
5130218
A candle burns at a constant rate. Two hours after it is lit, it is \(7\) inches tall. After \(5\) hours, it is \(4\) inches tall. 1. Write a function \(f(x)\) for the candle's height, in inches, after \(x\) hours. 2. How tall was the candle when it was lit? 3. After how many hours will the candle be completely burned down? 4. What domain for \(x\) makes sense in this context?

Hints

- Represent the two time-and-height measurements as points. - Find the slope from those two points. - What does the function value at time \(0\) represent? - What is the candle's height when it is completely burned down?

Solution

1. Use the points \((2, 7)\) and \((5, 4)\): \(m = \frac{4 - 7}{5 - 2} = -1\) inch per hour. 2. Substitute \((2, 7)\) into \(f(x) = -x + b\): \(7 = -2 + b\), so \(b = 9\). Thus \(f(x) = -x + 9\). 3. The initial height is \(f(0) = 9\) inches. 4. The candle is burned down when \(f(x) = 0\): \(0 = -x + 9\), so \(x = 9\). 5. In context, time begins at \(0\) and ends when the height reaches \(0\), so \(0 \le x \le 9\).

Answer

1. \(f(x) = -x + 9\) 2. \(9\) inches 3. \(9\) hours 4. \([0, 9]\)
5130228
Water flows into a large tank at a constant rate. After \(10\) minutes, the tank contains \(110\) gallons. After \(25\) minutes, it contains \(200\) gallons. 1. Write a function \(V(x)\) for the volume of water, in gallons, after \(x\) minutes. 2. How much water was in the tank when the timing began? 3. The tank holds \(290\) gallons. After how many minutes will it be full?

Hints

- What does the slope represent in this situation? - Use the two time-volume pairs to find the rate per minute. - Once you know the rate, how can you find the amount at time \(0\)? - To find when the tank is full, set the function equal to its capacity.

Solution

1. From \((10, 110)\) and \((25, 200)\), the rate is \(m = \frac{200 - 110}{25 - 10} = 6\) gallons per minute. 2. Substitute \((10, 110)\) into \(V(x) = 6x + b\): \(110 = 6 \cdot 10 + b\), so \(b = 50\). Thus \(V(x) = 6x + 50\). 3. The initial volume is \(V(0) = 50\) gallons. 4. Set the volume equal to capacity: \(290 = 6x + 50\). Then \(240 = 6x\), so \(x = 40\).

Answer

1. \(V(x) = 6x + 50\) 2. \(50\) gallons 3. \(40\) minutes
5130308
A rain barrel contains \(80\) gallons of water at time \(t = 0\). Water drains from an open spigot at a constant rate. After \(40\) minutes, \(60\) gallons remain. a) Write a linear function \(V(t)\) for the amount of water, in gallons, after \(t\) minutes. b) After how many minutes will the barrel be empty?

Hints

- What value represents the amount of water at time \(0\)? - How much water is lost per minute? - What value of \(V(t)\) represents an empty barrel?

Solution

1. The initial value is \(V(0) = 80\). 2. The rate of change is \(m = \frac{60 - 80}{40 - 0} = -0.5\) gallon per minute. Thus \(V(t) = -0.5t + 80\). 3. The barrel is empty when \(V(t) = 0\): \(0 = -0.5t + 80\), so \(t = 160\).

Answer

a) \(V(t) = -0.5t + 80\) b) \(160\) minutes
5130318
Two water tanks are observed at the same time. Tank A starts with \(500\,\text{L}\) and is drained at a constant rate of \(12\,\text{L/min}\). Tank B starts with \(200\,\text{L}\) and is filled at a constant rate of \(8\,\text{L/min}\). Write a function for the volume of each tank after \(t\) minutes. Then find when the tanks contain the same amount of water and what that common volume is.

Hints

- For each tank, decide whether the volume is increasing or decreasing. - What equation represents the moment when the two volumes are equal? - After finding the time, substitute it into either model to find the common volume.

Solution

1. Tank A is modeled by \(V_A(t) = 500 - 12t\). Tank B is modeled by \(V_B(t) = 200 + 8t\). 2. Set the volumes equal: \(500 - 12t = 200 + 8t\). 3. Then \(300 = 20t\), so \(t = 15\). 4. Substitute \(t = 15\): \(V_A(15) = 500 - 12 \cdot 15 = 320\). Therefore, after \(15\) minutes both tanks contain \(320\,\text{L}\).

Answer

\(V_A(t) = 500 - 12t\) and \(V_B(t) = 200 + 8t\). After \(15\) minutes, both tanks contain \(320\,\text{L}\).
5130328
A candle is \(10\) inches tall when it is lit. After \(3\) hours, it is \(8.5\) inches tall. a) Write a function \(h(t)\) for the candle's height, in inches, after \(t\) hours. b) Explain what the slope \(m\) and y-intercept \(b\) mean in this situation. c) Find the candle's height after \(5\) hours.

Hints

- Is the candle's height increasing or decreasing as it burns? - Use the two height measurements to find the change per hour. - Think about what the function value at time \(0\) represents.

Solution

1. The initial height is the y-intercept, so \(b = 10\). 2. The slope is \(m = \frac{8.5 - 10}{3 - 0} = -0.5\) inch per hour. Thus \(h(t) = -0.5t + 10\). 3. The y-intercept \(10\) is the candle's initial height. The slope \(-0.5\) means its height decreases by \(0.5\) inch each hour. 4. Evaluate at \(t = 5\): \(h(5) = -0.5 \cdot 5 + 10 = 7.5\) inches.

Answer

a) \(h(t) = -0.5t + 10\) b) \(b = 10\) is the initial height, and \(m = -0.5\) means the candle loses \(0.5\) inch of height per hour. c) \(7.5\) inches
5130428
A water tank fills at a constant rate. After \(10\) minutes, it contains \(350\) gallons. After \(25\) minutes, it contains \(800\) gallons. a) Write a linear function \(V(t)\) for the volume, in gallons, after \(t\) minutes. b) How much water was in the tank at \(t = 0\)? c) The tank holds \(1700\) gallons. After how many minutes will it be full?

Hints

- Treat time as the input and volume as the output. - Use the two measurements to find how many gallons are added each minute. - What does the function value at \(t = 0\) represent? - To find the fill time, set the function equal to the tank's capacity.

Solution

1. Use \((10, 350)\) and \((25, 800)\): \(m = \frac{800 - 350}{25 - 10} = 30\) gallons per minute. 2. Substitute \((10, 350)\) into \(V(t) = 30t + b\): \(350 = 30 \cdot 10 + b\), so \(b = 50\). 3. Therefore, \(V(t) = 30t + 50\), and the initial amount is \(50\) gallons. 4. Set the volume equal to capacity: \(1700 = 30t + 50\). Then \(1650 = 30t\), so \(t = 55\) minutes.

Answer

a) \(V(t) = 30t + 50\) b) \(50\) gallons c) \(55\) minutes
5130438
A taxi company uses a linear fare. A \(5\)-mile ride costs \(\$14.50\), and a \(12\)-mile ride costs \(\$28.50\). a) Write a function \(C(x)\) for the fare, in dollars, for a ride of \(x\) miles. b) Explain what the slope \(m\) and y-intercept \(b\) mean in this situation. c) Find the fare for a \(20\)-mile ride.

Hints

- How much does the fare change when the distance increases? - What cost would remain when the distance is \(0\) miles? - Substitute the requested distance into your function.

Solution

1. From \((5, 14.50)\) and \((12, 28.50)\), the slope is \(m = \frac{28.50 - 14.50}{12 - 5} = 2\) dollars per mile. 2. Substitute \((5, 14.50)\) into \(C(x) = 2x + b\): \(14.50 = 2 \cdot 5 + b\), so \(b = 4.50\). 3. Therefore, \(C(x) = 2x + 4.50\). 4. The slope means the fare increases by \(\$2.00\) per mile, and the y-intercept is the \(\$4.50\) starting fare. 5. Evaluate at \(x = 20\): \(C(20) = 2 \cdot 20 + 4.50 = 44.50\).

Answer

a) \(C(x) = 2x + 4.50\) b) The slope is a charge of \(\$2.00\) per mile, and the y-intercept is the \(\$4.50\) starting fare. c) \(\$44.50\)
5130448
Air temperature in the atmosphere can be modeled as decreasing linearly with altitude. At \(2000\) feet above sea level, a meteorologist measures \(50^\circ\text{F}\). At \(10{,}000\) feet, the temperature is \(18^\circ\text{F}\). a) Write a linear function \(T(h)\) for the temperature, in degrees Fahrenheit, at altitude \(h\) feet. b) What temperature does the model predict at sea level? c) At what altitude does the model predict the freezing temperature of \(32^\circ\text{F}\)?

Hints

- The temperature decreases as altitude increases, so what sign should the slope have? - What altitude represents sea level? - For part c, the temperature is known. Substitute it into the model and solve for altitude.

Solution

1. Use \((2000, 50)\) and \((10{,}000, 18)\): \(m = \frac{18 - 50}{10{,}000 - 2000} = -0.004\) degree Fahrenheit per foot. 2. Substitute \((2000, 50)\) into \(T(h) = -0.004h + b\): \(50 = -0.004 \cdot 2000 + b\), so \(b = 58\). Thus \(T(h) = -0.004h + 58\). 3. At sea level, \(T(0) = 58^\circ\text{F}\). 4. Set the temperature equal to freezing: \(32 = -0.004h + 58\). Then \(-26 = -0.004h\), so \(h = 6500\) feet.

Answer

a) \(T(h) = -0.004h + 58\) b) \(58^\circ\text{F}\) c) \(6500\) feet
5130938
Two monthly phone plans are available. The Basic plan charges a monthly fee of \(\$8.00\) plus \(\$0.15\) per calling minute. The Smart plan charges a monthly fee of \(\$14.00\) plus \(\$0.05\) per calling minute. a) Write a function for the monthly cost \(y\) of each plan in terms of the number of calling minutes \(x\). b) Find the number of calling minutes for which the two plans have the same monthly cost. c) Explain which plan is less expensive for lighter use and which is less expensive for heavier use.

Hints

- What part of each cost is fixed each month? - How much does each total increase for one additional calling minute? - What equation represents the point where the two plans cost the same? - After finding the break-even point, compare the per-minute rates.

Solution

1. The Basic plan is modeled by \(B(x) = 0.15x + 8.00\). The Smart plan is modeled by \(S(x) = 0.05x + 14.00\). 2. Set the costs equal: \(0.15x + 8.00 = 0.05x + 14.00\). 3. Then \(0.10x = 6.00\), so \(x = 60\). 4. At \(60\) minutes, both plans cost \(\$17.00\). For fewer than \(60\) minutes, Basic is less expensive; for more than \(60\) minutes, Smart is less expensive.

Answer

a) \(B(x) = 0.15x + 8.00\) and \(S(x) = 0.05x + 14.00\) b) \(60\) minutes c) Basic is less expensive below \(60\) minutes; Smart is less expensive above \(60\) minutes.
5130958
Two hiking groups are on the same trail to a mountain summit. Group A starts at a parking lot and hikes at a constant speed of \(3\,\text{mi/h}\). Group B starts at the same time from a lodge \(2\) miles farther along the trail and hikes at \(2\,\text{mi/h}\). a) Write a function for each group's distance \(s\), in miles, from the parking lot after \(t\) hours. b) Find when and where Group A catches Group B. c) The summit is \(7\) miles from the parking lot. Determine mathematically whether Group A catches Group B before either group reaches the summit.

Hints

- How far from the parking lot is Group B at \(t = 0\)? - What equation represents the moment when the groups are at the same location? - Compare the meeting distance with the total distance to the summit.

Solution

1. Group A is modeled by \(s_A(t) = 3t\). Group B starts \(2\) miles ahead, so \(s_B(t) = 2 + 2t\). 2. Set the distances equal: \(3t = 2 + 2t\). 3. Solving gives \(t = 2\), so Group A catches Group B after \(2\) hours. 4. The meeting point is \(s_A(2) = 3 \cdot 2 = 6\) miles from the parking lot. 5. Since \(6 < 7\), the catch-up happens before the summit, exactly \(1\) mile before it.

Answer

a) \(s_A(t) = 3t\) and \(s_B(t) = 2 + 2t\) b) After \(2\) hours, at \(6\) miles from the parking lot. c) Yes. The catch-up point is \(1\) mile before the \(7\)-mile summit.
5130968
A tablet battery is charging. At the start, the battery level is \(10\%\). After \(24\) minutes, it has increased at a constant rate to \(40\%\). a) Write a linear function \(y\) for the battery level, in percent, after \(x\) minutes. b) Assuming the charging rate stays constant, after how many minutes will the battery reach \(100\%\)? c) Describe the meaningful graph of the model by stating its two endpoints and whether it is increasing, decreasing, or constant. You may sketch it on paper to help.

Hints

- Use the battery level at time \(0\) as the initial value. - Find the constant rate from the change in battery percentage over \(24\) minutes. - A full battery corresponds to an output of \(100\%\). - For part c, use the starting point and the full-charge point to describe the meaningful graph segment.

Solution

1. At \(x=0\), the battery level is \(10\%\), so the y-intercept is \(10\). 2. Over \(24\) minutes, the battery level increases by \(30\) percentage points, so \(m=\frac{30}{24}=1.25\) percentage points per minute. Thus \(y=1.25x+10\). 3. For a full battery, set \(y=100\): \(100=1.25x+10\). Solving gives \(x=72\). 4. In the meaningful model interval \(0\le x\le72\), the graph is an increasing line segment from \((0, 10)\) to \((72, 100)\). The point \((24, 40)\) lies on that segment.

Answer

a) \(y=1.25x+10\) b) \(72\) minutes c) An increasing line segment from \((0, 10)\) to \((72, 100)\), for \(0\le x\le72\).
5130988
A rectangular backyard pool is filled at a constant rate. At \(1{:}00\) p.m., the water is \(16\,\text{in.}\) deep. At \(3{:}00\) p.m., it is \(40\,\text{in.}\) deep. Let \(t\) be the number of hours after \(1{:}00\) p.m., and let \(d(t)\) be the water depth in inches. a) Construct a linear depth function \(d(t)\) from the two measurements. State its rate of change. b) Extending the same model backward, at what clock time does \(d(t)=0\)? c) The pool is full at a depth of \(64\,\text{in.}\). At what clock time does the model reach that depth?

Hints

- Use the two depth measurements to determine the signed change per hour. - Because \(t=0\) is fixed at \(1{:}00\) p.m., the measured depth at that time becomes the initial value of the function. - After constructing the function, use target depths as output values and solve for \(t\).

Solution

1. From \(1{:}00\) p.m. to \(3{:}00\) p.m., the depth increases by \(40-16=24\) inches in \(2\) hours, so the rate is \(12\,\frac{\text{in.}}{\text{h}}\). 2. Since \(t=0\) represents \(1{:}00\) p.m. and the depth is then \(16\) inches, \(d(t)=16+12t\). 3. Solve \(0=16+12t\). Then \(t=-\frac{4}{3}\) hour, which is \(1\) hour \(20\) minutes before \(1{:}00\) p.m., or \(11{:}40\) a.m. 4. Solve \(64=16+12t\). Then \(48=12t\), so \(t=4\), which is \(5{:}00\) p.m.

Answer

a) \(d(t)=16+12t\); rate of change \(12\,\frac{\text{in.}}{\text{h}}\) b) \(11{:}40\) a.m. c) \(5{:}00\) p.m.
5131158
A vehicle travels at a constant speed during a highway test. The table shows time \(t\), in hours, and distance \(s\), in miles. <table> <tr><td>Time \(t\), in hours</td><td>\(0.5\)</td><td>\(1.5\)</td><td>\(2\)</td></tr> <tr><td>Distance \(s\), in miles</td><td>\(30\)</td><td>\(90\)</td><td>\(120\)</td></tr> </table> a) Use calculations to determine whether the table could represent a proportional relationship. b) State the physical condition that must hold throughout the trip for this proportionality to be exact. c) Write the function for the relationship.

Hints

- How are time, distance, and speed related? - What must be constant in a proportional table? - Use the table to find the coefficient in the equation.

Solution

1. Calculate \(s\div t\): \(30\div 0.5=60\), \(90\div 1.5=60\), and \(120\div 2=60\). 2. Since the ratios are equal, the relationship is proportional. 3. The vehicle's speed must remain constant at \(60\) miles per hour. 4. The function is \(s(t)=60t\).

Answer

a) Yes. Every ratio \(s\div t\) equals \(60\). b) The speed must remain constant. c) \(s(t)=60t\).
5131318
Two electricity providers offer different monthly plans. Provider A: no monthly base charge and \(\$0.40\) per kilowatt-hour, or \(\text{kWh}\). Provider B: a \(\$10.00\) monthly base charge and \(\$0.30\) per kilowatt-hour. a) Write a function for monthly cost \(y\), in dollars, as a function of usage \(x\), in kilowatt-hours, for each provider. b) Which plan represents a proportional relationship? Justify your answer mathematically. c) How does Provider B's base charge change its graph compared with a line through the origin?

Hints

- Separate each cost into a fixed part and a usage-dependent part. - What should the cost be at zero usage in a proportional relationship? - How does a fixed starting value affect the y-intercept?

Solution

1. Provider A is \(y_A=0.40x\). Provider B is \(y_B=0.30x+10\). 2. Provider A is proportional because it has the form \(y=kx\) and \(y_A(0)=0\). 3. Provider B is not proportional because \(y_B(0)=10\), so the output-to-input ratio is not constant. 4. The base charge shifts Provider B's graph upward; its y-intercept is \(10\) instead of \(0\).

Answer

a) Provider A: \(y=0.40x\). Provider B: \(y=0.30x+10\). b) Only Provider A is proportional. c) The base charge shifts Provider B's graph upward so it crosses the y-axis at \(10\), not at the origin.
5131738
Two water tanks are being adjusted at the same time. Tank 1 starts with \(2500\,\text{L}\) of water and loses \(150\,\text{L}\) per hour. Tank 2 starts with \(500\,\text{L}\) of water and gains \(250\,\text{L}\) per hour. a) Write a function for the volume \(V\), in liters, of each tank after \(t\) hours. b) Find when the two tanks contain the same amount of water and determine that common volume. c) Tank 2 has a maximum capacity of \(3000\,\text{L}\). How much water remains in Tank 1 when Tank 2 becomes full?

Hints

- Decide whether each tank's volume is increasing or decreasing. - Part b asks when the two volume functions have the same value. - For part c, first find when Tank 2 reaches its maximum capacity, then use that time in the model for Tank 1.

Solution

1. The volume functions are \(V_1(t) = 2500 - 150t\) and \(V_2(t) = 500 + 250t\). 2. Set the volumes equal: \(2500 - 150t = 500 + 250t\). 3. Then \(2000 = 400t\), so \(t = 5\) hours. 4. The common volume is \(V_2(5) = 500 + 250 \cdot 5 = 1750\,\text{L}\). 5. Tank 2 is full when \(3000 = 500 + 250t\). Thus \(2500 = 250t\), so \(t = 10\) hours. 6. At that time, \(V_1(10) = 2500 - 150 \cdot 10 = 1000\,\text{L}\).

Answer

a) \(V_1(t) = 2500 - 150t\) and \(V_2(t) = 500 + 250t\) b) After \(5\) hours, both tanks contain \(1750\,\text{L}\). c) \(1000\,\text{L}\) remains in Tank 1.
5131928
A candle is \(9.5\) inches tall when it is lit. After \(4\) hours, it is \(5.5\) inches tall. Its height decreases at a constant rate. a) Write a function \(h(t)\) for the candle's height, in inches, after \(t\) hours. b) After how many hours and minutes will the candle be completely burned down?

Hints

- What does the candle's starting height tell you about the y-intercept? - How much height does the candle lose each hour? - What height represents a completely burned-down candle? - How do you convert a fraction of an hour to minutes?

Solution

1. The initial height is \(9.5\) inches, so the y-intercept is \(9.5\). 2. The rate of change is \(m = \frac{5.5 - 9.5}{4 - 0} = -1\) inch per hour. Thus \(h(t) = -t + 9.5\). 3. The candle is burned down when \(h(t) = 0\): \(0 = -t + 9.5\), so \(t = 9.5\) hours. 4. Since \(0.5\) hour is \(30\) minutes, the total time is \(9\) hours and \(30\) minutes.

Answer

a) \(h(t) = -t + 9.5\) b) \(9\) hours and \(30\) minutes
5131938
A liquid cools at a constant rate. After \(2\) minutes, its temperature is \(170^\circ\text{F}\). After \(10\) minutes, its temperature is \(130^\circ\text{F}\). a) Write a linear function \(T(t)\) for the temperature, in degrees Fahrenheit, after \(t\) minutes. b) Explain what the slope and y-intercept mean in this situation.

Hints

- How do you find the slope from two time-temperature points? - Which quantity changes as time passes? - What does the function value at time \(0\) represent? - What does a negative slope mean in this context?

Solution

1. Use \((2, 170)\) and \((10, 130)\): \(m = \frac{130 - 170}{10 - 2} = -5\) degrees Fahrenheit per minute. 2. Substitute \((2, 170)\) into \(T(t) = -5t + b\): \(170 = -5 \cdot 2 + b\), so \(b = 180\). Thus \(T(t) = -5t + 180\). 3. The slope \(-5\) means the temperature decreases by \(5^\circ\text{F}\) each minute. The y-intercept \(180\) is the temperature at time \(0\).

Answer

a) \(T(t) = -5t + 180\) b) The slope means the temperature decreases by \(5^\circ\text{F}\) per minute, and the y-intercept means the initial temperature was \(180^\circ\text{F}\).
5131948
A water container fills at a constant rate. After \(5\) minutes, it contains \(5.5\) gallons. After \(12\) minutes, it contains \(9\) gallons. a) Write a function \(V(t) = mt + b\) for the volume, in gallons, after \(t\) minutes. b) How much water was in the container at time \(0\)? c) How much water will be in the container after half an hour?

Hints

- First find how many gallons are added per minute. - How can you use a known time-volume pair to find the amount at time \(0\)? - How many minutes are in half an hour? - Keep the time units consistent when you evaluate the function.

Solution

1. The rate is \(m = \frac{9 - 5.5}{12 - 5} = 0.5\) gallon per minute. 2. Substitute \((5, 5.5)\) into \(V(t) = 0.5t + b\): \(5.5 = 0.5 \cdot 5 + b\), so \(b = 3\). 3. Thus \(V(t) = 0.5t + 3\), and the initial amount is \(3\) gallons. 4. Half an hour is \(30\) minutes. Evaluate \(V(30) = 0.5 \cdot 30 + 3 = 18\) gallons.

Answer

a) \(V(t) = 0.5t + 3\) b) \(3\) gallons c) \(18\) gallons
5132108
A prepaid phone account starts with a balance of \(\$20.00\). Each minute of calling costs \(\$0.12\). a) Write a function \(G(t)\) for the remaining balance, in dollars, after \(t\) minutes of calls. b) How much money remains after \(45\) minutes of calls? c) After how many minutes will exactly \(\$5.00\) remain? d) Is this relationship proportional? Explain.

Hints

- Identify the fixed starting amount and the amount that changes each minute. - What must be true of the graph or equation of a proportional relationship? - For parts b and c, substitute the given value into your function and solve for the unknown.

Solution

1. The starting balance is \(\$20.00\), and the balance decreases by \(\$0.12\) per minute, so \(G(t) = 20 - 0.12t\). 2. Evaluate at \(t = 45\): \(G(45) = 20 - 0.12 \cdot 45 = 14.60\), so \(\$14.60\) remains. 3. Set the balance equal to \(\$5.00\): \(5 = 20 - 0.12t\). Then \(0.12t = 15\), so \(t = 125\) minutes. 4. A proportional relationship has the form \(y = kx\) and passes through \((0, 0)\). Here \(G(0) = 20\), so the relationship is not proportional.

Answer

a) \(G(t) = 20 - 0.12t\) b) \(\$14.60\) c) \(125\) minutes d) No. Since \(G(0) = 20\), the graph does not pass through the origin.
5135178
Two trains travel in the same direction on parallel tracks. A local train travels at a constant speed of \(60\,\text{mi/h}\). An express train travels at \(80\,\text{mi/h}\). At the start of the observation, the local train is \(2\) miles ahead. a) Write a function for each train's distance \(s\), in miles, after \(t\) minutes. b) After how many minutes does the express train catch the local train? c) How far has the express train traveled from the start of the observation by that time?

Hints

- Convert each speed to miles per minute because time is measured in minutes. - How does the initial lead appear in the local train model? - At the catch-up time, what must be true about the two distances?

Solution

1. Convert the speeds to miles per minute: \(60 \div 60 = 1\,\text{mi/min}\) and \(80 \div 60 = \frac{4}{3}\,\text{mi/min}\). 2. The local train is modeled by \(s_L(t) = t + 2\), and the express train by \(s_E(t) = \frac{4}{3}t\). 3. Set the distances equal: \(t + 2 = \frac{4}{3}t\). 4. Then \(2 = \frac{1}{3}t\), so \(t = 6\) minutes. 5. The express train has traveled \(s_E(6) = \frac{4}{3} \cdot 6 = 8\) miles.

Answer

a) \(s_L(t) = t + 2\) and \(s_E(t) = \frac{4}{3}t\), with \(t\) in minutes and \(s\) in miles. b) \(6\) minutes c) \(8\) miles
5137588
A school is comparing two online photo-printing offers. PhotoNow charges \(\$0.12\) per photo plus a flat \(\$3.50\) shipping fee. PrintPlus charges \(\$0.17\) per photo plus a flat \(\$1.50\) shipping fee. Write a function for the total cost \(y\) of each offer for \(x\) photos. Then find the number of photos for which the two offers cost exactly the same amount.

Hints

- Which part of each cost is fixed, and which part depends on the number of photos? - What is the general form of a linear cost function? - Set the two cost expressions equal to find the break-even quantity.

Solution

1. The cost functions are \(P(x) = 0.12x + 3.50\) and \(Q(x) = 0.17x + 1.50\). 2. Set the costs equal: \(0.12x + 3.50 = 0.17x + 1.50\). 3. Then \(2.00 = 0.05x\). 4. Solving gives \(x = 40\). 5. Therefore, the two offers cost the same amount for exactly \(40\) photos.

Answer

\(P(x) = 0.12x + 3.50\) and \(Q(x) = 0.17x + 1.50\). The costs are equal at \(x = 40\) photos.
5137818
A candle burns at a constant rate. Two hours after it is lit, it is \(7.5\) inches tall. After \(5\) hours, it is \(6\) inches tall. a) Write a function \(h(t)\) for the candle's height, in inches, after \(t\) hours. b) Explain what the y-intercept and slope mean in this situation.

Hints

- How much does the candle's height change over the three-hour interval? - What sign should the rate have when the candle gets shorter? - What does time \(t = 0\) represent?

Solution

1. Use the points \((2, 7.5)\) and \((5, 6)\): \(m = \frac{6 - 7.5}{5 - 2} = -0.5\) inch per hour. 2. Substitute \((2, 7.5)\) into \(h(t) = -0.5t + b\): \(7.5 = -0.5 \cdot 2 + b\), so \(b = 8.5\). 3. Therefore, \(h(t) = -0.5t + 8.5\). 4. The y-intercept \(8.5\) is the candle's starting height. The slope \(-0.5\) means the candle loses \(0.5\) inch of height each hour.

Answer

a) \(h(t) = -0.5t + 8.5\) b) The y-intercept \(8.5\) is the initial height in inches, and the slope \(-0.5\) means the height decreases by \(0.5\) inch per hour.
5137938
A point \(P(x, y)\) lies on a line described by this rule: “The y-value is \(4\) more than twice the x-value.” a) Write a linear equation that represents the rule. b) Determine algebraically whether \(A(3, 10)\) and \(B(-2, 1)\) lie on the line. c) A point \(C\) on the line has y-coordinate \(15\). Find its x-coordinate.

Hints

- Translate “twice the x-value” as \(2x\), then account for the additional \(4\). - To test a point, substitute both coordinates into the equation. - For part c, substitute the known y-value and solve for \(x\).

Solution

1. “Twice the x-value” is \(2x\), and “4 more” gives \(y = 2x + 4\). 2. For \(A\), \(2 \cdot 3 + 4 = 10\), so \(A(3, 10)\) lies on the line. 3. For \(B\), \(2 \cdot (-2) + 4 = 0\), not \(1\), so \(B(-2, 1)\) does not lie on the line. 4. For \(C\), solve \(15 = 2x + 4\). Then \(11 = 2x\), so \(x = 5.5\).

Answer

a) \(y = 2x + 4\) b) \(A(3, 10)\) lies on the line; \(B(-2, 1)\) does not. c) \(x = 5.5\)
5138188
Two climbing gyms offer different monthly pricing plans. Summit Climbing Gym charges a \(\$10\) monthly fee plus \(\$5\) per visit. Peak Climbing Gym has no monthly fee and charges \(\$7.50\) per visit. Let \(v\) be the number of visits in a month. a) Construct monthly cost functions \(S(v)\) and \(P(v)\). b) Use the functions to find the number of visits for which the plans cost exactly the same amount. c) For \(5\) visits, which plan costs less and by how much? d) Starting at how many whole visits per month is Summit less expensive overall?

Hints

- Translate the monthly fee and per-visit charge into the initial value and rate of each cost function. - Equal costs mean equal function outputs for the same number of visits. - Use the constructed functions for both the sample comparison and the whole-number threshold.

Solution

1. Summit has model \(S(v)=10+5v\). Peak has model \(P(v)=7.50v\). 2. Set the functions equal: \(10+5v=7.50v\). Then \(2.50v=10\), so \(v=4\). Both plans cost \(\$30\). 3. \(S(5)=10+5\cdot5=35\), and \(P(5)=7.50\cdot5=37.50\). Summit is cheaper by \(\$2.50\). 4. The plans are equal at \(4\) visits, and Summit has the smaller per-visit rate after the fixed fee is covered. Therefore, Summit is less expensive for every whole-number visit count greater than \(4\), starting at \(5\) visits.

Answer

a) \(S(v)=10+5v\), \(P(v)=7.50v\) b) \(4\) visits; both cost \(\$30\). c) Summit; it saves \(\$2.50\). d) \(5\) visits
5152328
Two water tanks are being adjusted at the same time. Tank A starts with \(100\) gallons of water, and \(15\) gallons are added each minute. Tank B starts with \(400\) gallons of water, and \(10\) gallons are drained each minute. a) Write a function for the volume \(y\), in gallons, in each tank after \(x\) minutes. b) Explain what the slope and y-intercept of the function for Tank B mean in this situation. c) Find when the tanks contain the same amount of water and how much water each tank contains at that time.

Hints

- Identify the starting amount and the amount that changes each minute for each tank. - Does the amount in Tank B increase or decrease? What sign should its slope have? - What equation represents the moment when the two tanks have the same volume?

Solution

1. Tank A starts at \(100\) gallons and gains \(15\) gallons per minute, so \(A(x) = 15x + 100\). Tank B starts at \(400\) gallons and loses \(10\) gallons per minute, so \(B(x) = -10x + 400\). 2. For Tank B, the y-intercept \(400\) is the initial volume in gallons. The slope \(-10\) means the volume decreases by \(10\) gallons each minute. 3. Set the functions equal: \(15x + 100 = -10x + 400\). 4. Then \(25x = 300\), so \(x = 12\). 5. Substituting gives \(A(12) = 15 \cdot 12 + 100 = 280\). Therefore, after \(12\) minutes, each tank contains \(280\) gallons.

Answer

a) \(A(x) = 15x + 100\) and \(B(x) = -10x + 400\). b) For Tank B, the y-intercept \(400\) is the starting volume in gallons, and the slope \(-10\) means \(10\) gallons are drained each minute. c) After \(12\) minutes, both tanks contain \(280\) gallons.
5154828
Two siblings, Liam and Sarah, are saving money. Liam starts with \(\$20\) and saves another \(\$5\) each week. Sarah starts with \(\$50\) but saves only \(\$2\) each week. a) Write a function for each person's savings \(y\) after \(x\) weeks. b) After how many weeks will they have the same amount of money? c) How much money will each person have at that time? d) Interpret the slope of each function in this situation.

Hints

- In a linear function, which value represents the amount at week \(0\)? - What equation represents the time when both savings amounts are equal? - How does a one-week increase in \(x\) change each person's savings?

Solution

1. Liam's savings are modeled by \(L(x) = 5x + 20\). Sarah's savings are modeled by \(S(x) = 2x + 50\). 2. Set the functions equal: \(5x + 20 = 2x + 50\). 3. Then \(3x = 30\), so \(x = 10\). They have the same amount after \(10\) weeks. 4. Substituting gives \(L(10) = 5 \cdot 10 + 20 = 70\), so each person has \(\$70\). 5. The slopes, \(5\) and \(2\), are the weekly savings rates in dollars per week.

Answer

a) \(L(x) = 5x + 20\) and \(S(x) = 2x + 50\). b) \(10\) weeks. c) Each person has \(\$70\). d) The slopes are the weekly savings rates: \(\$5\) per week for Liam and \(\$2\) per week for Sarah.
5223688
An electric scooter rental offers two pricing plans: Plan A: a \(\$1.20\) activation fee plus \(\$0.20\) per minute Plan B: no activation fee and \(\$0.30\) per minute a) Write a cost rule for each plan for a ride lasting \(x\) minutes. b) Find the cost of a \(15\)-minute ride under each plan. Which plan is less expensive? c) Find the ride length for which the two plans cost the same.

Hints

- Separate each fixed fee from its per-minute rate. - Substitute \(15\) into both cost rules. - Equal costs are represented by an equation.

Solution

1. The cost rules are \(C_A = 0.20x + 1.20\) and \(C_B = 0.30x\). 2. For \(x = 15\), Plan A costs \(0.20 \cdot 15 + 1.20 = 4.20\) dollars. 3. Plan B costs \(0.30 \cdot 15 = 4.50\) dollars, so Plan A is less expensive for a \(15\)-minute ride. 4. Set the costs equal: \(0.20x + 1.20 = 0.30x\). 5. Subtract \(0.20x\): \(1.20 = 0.10x\). 6. Divide by \(0.10\): \(x = 12\).

Answer

a) \(C_A = 0.20x + 1.20\); \(C_B = 0.30x\) b) Plan A costs \(\$4.20\), and Plan B costs \(\$4.50\). Plan A is less expensive. c) The plans cost the same for a \(12\)-minute ride.
5227128
A technician records the temperature of a cooling liquid: <table> <tr> <td>Time \(t\) in \(\text{min}\)</td> <td>\(0\)</td><td>\(8\)</td><td>\(20\)</td> </tr> <tr> <td>Temperature \(T\) in \(^\circ\text{C}\)</td> <td>\(90\)</td><td>\(74\)</td><td>\(50\)</td> </tr> </table> a) Show using calculations that the temperature can be modeled by a linear function over this time interval. b) Find a linear model of the form \(T(t) = mt + b\). c) Find the temperature after exactly \(15\) minutes. d) Assuming the linear pattern continues, when will the liquid reach \(26\,^\circ\text{C}\)?

Hints

- How can you compare rates of change between consecutive pairs of data points? - Which table value gives the initial value of the model? - Substitute the given time into your model. - For the last part, set the model equal to the target temperature and solve for time.

Solution

1. From \(0\) to \(8\) minutes, the rate of change is \(\frac{74 - 90}{8 - 0} = -2\,^\circ\text{C}/\text{min}\). From \(8\) to \(20\) minutes, it is \(\frac{50 - 74}{20 - 8} = -2\,^\circ\text{C}/\text{min}\). The rates are equal, so the data are linear. 2. At \(t = 0\), the temperature is \(90\,^\circ\text{C}\), so \(b = 90\). The model is \(T(t) = -2t + 90\). 3. \(T(15) = -2 \cdot 15 + 90 = 60\), so the temperature is \(60\,^\circ\text{C}\). 4. Set \(T(t) = 26\): \(26 = -2t + 90\). Solving gives \(t = 32\).

Answer

a) Both intervals have rate of change \(-2\,^\circ\text{C}/\text{min}\), so the data are linear. b) \(T(t) = -2t + 90\) c) \(60\,^\circ\text{C}\) d) \(32\,\text{min}\)
5231218
A cylindrical container has an inside diameter of \(12\,\text{cm}\) and a height of \(30\,\text{cm}\). Water is poured into the container. a) Write a function \(V\) that gives the water volume \(V(h)\), in \(\text{cm}^3\), for a water height \(h\), in centimeters. b) State the domain \(D\) and range \(R\) for this situation. Round the upper endpoint of the range to the nearest hundredth.

Hints

- How can you find the radius from the diameter? - Which formula gives the volume of a cylinder? - What water heights are physically possible, from empty to full? - How are the smallest and largest function values related to the container height?

Solution

1. The radius is half the diameter, so \(r=6\,\text{cm}\). 2. The area of the circular base is \(B=\pi r^2=\pi\cdot6^2=36\pi\,\text{cm}^2\). 3. Since cylinder volume is base area times height, \(V(h)=Bh=36\pi h\). 4. The water height can range from empty to full, so \(D=[0, 30]\). 5. The minimum volume is \(V(0)=0\). The maximum is \(V(30)=36\pi\cdot30=1080\pi\approx3{,}392.92\,\text{cm}^3\). Therefore, \(R=[0, 1080\pi]\), or approximately \([0, 3{,}392.92]\), in cubic centimeters.

Answer

a) \(V(h)=36\pi h\) b) \(D=[0, 30]\); \(R=[0, 1080\pi]\approx[0, 3{,}392.92]\). Here, \(h\) is measured in centimeters and \(V(h)\) is measured in cubic centimeters.
5231228
A right triangular prism has a base with legs \(6\,\text{cm}\) and \(8\,\text{cm}\). The prism is \(25\,\text{cm}\) tall and stands on its triangular base while it is filled with liquid. a) Write a function \(f\) that gives the liquid volume \(f(x)\), in \(\text{cm}^3\), for a fill height \(x\), in centimeters. b) Find the fill height when the prism contains exactly \(0.36\,\text{L}\) of liquid. c) Briefly explain why \(f\) is a linear function.

Hints

- Recall the area formula for a right triangle. - Convert liters to cubic centimeters before solving. - Think about whether the horizontal cross-sectional area changes as the liquid rises. - A linear function has a constant rate of change.

Solution

1. The area of the right triangular base is \(B=\frac{1}{2}\cdot6\cdot8=24\,\text{cm}^2\). 2. Prism volume is base area times height, so \(f(x)=Bx=24x\). 3. Convert the given volume: \(0.36\,\text{L}=360\,\text{cm}^3\). 4. Set \(24x=360\) and solve: \(x=360\div24=15\). The fill height is \(15\,\text{cm}\). 5. The horizontal cross-sectional area remains constant as the liquid rises, so the volume increases at a constant rate of \(24\,\text{cm}^3\) per centimeter. Thus, \(f(x)=24x\) is linear.

Answer

a) \(f(x)=24x\) b) \(x=15\,\text{cm}\) c) The cross-sectional area is constant, so the volume increases by a constant amount for each centimeter of fill height.
5237868
A rain barrel initially contains \(10\) gallons of water and has a capacity of \(50\) gallons. During a storm, water flows into the barrel at a constant rate of \(1\) gallon per minute. 1) Write a function for the amount of water \(y\), in gallons, after \(x\) minutes, including the physical domain. 2) Make a table for \(x=0\), \(x=10\), \(x=15\), and \(x=30\). 3) Describe the graph of the function on its physical domain. 4) Determine algebraically when the barrel becomes full.

Hints

- Identify the amount in the barrel at time \(0\). - Identify how much water is added each minute. - Use the form \(y=mx+b\) and restrict the domain to the time before the barrel is full. - Set \(y=50\) and solve for \(x\).

Solution

1. The initial value is \(10\) and the rate of change is \(1\), so \(y=x+10\). The barrel is modeled from the start of the storm until it reaches \(50\) gallons, so the physical domain is \(0\le x\le40\). 2. Evaluating the function gives \(y(0)=10\), \(y(10)=20\), \(y(15)=25\), and \(y(30)=40\). 3. The graph is the line segment with y-intercept \(10\) and slope \(1\), from \((0, 10)\) to \((40, 50)\). 4. Solve \(50=x+10\). Subtracting \(10\) gives \(x=40\), so the barrel becomes full after \(40\) minutes.

Answer

1) \(y=x+10\), for \(0\le x\le40\) 2) <table><tr><td>Time \(x\) in minutes</td><td>\(0\)</td><td>\(10\)</td><td>\(15\)</td><td>\(30\)</td></tr><tr><td>Water \(y\) in gallons</td><td>\(10\)</td><td>\(20\)</td><td>\(25\)</td><td>\(40\)</td></tr></table> 3) The line segment from \((0, 10)\) to \((40, 50)\) 4) \(40\) minutes
5239968
Two water tanks begin draining at the same time. Tank 1 starts with \(500\,\text{gal}\) and loses \(10\,\frac{\text{gal}}{\text{min}}\). Tank 2 starts with \(800\,\text{gal}\) and loses \(30\,\frac{\text{gal}}{\text{min}}\). Let \(t\) be the number of minutes after draining begins. a) Construct volume functions \(V_1(t)\) and \(V_2(t)\) for the two tanks while each model is physically meaningful. b) Use the functions to determine when Tank 1 contains exactly twice as much water as Tank 2. Verify that both modeled volumes are nonnegative at that time.

Hints

- Each draining model needs its starting volume and a negative rate of change. - Express the “twice as much” condition using the two named functions rather than a one-off verbal equation. - Check the resulting time against the physical requirement that neither volume be negative.

Solution

1. Tank 1 has model \(V_1(t)=500-10t\). Tank 2 has model \(V_2(t)=800-30t\). 2. The condition is \(V_1(t)=2V_2(t)\), so \(500-10t=2(800-30t)\). 3. Distribute and solve: \(500-10t=1600-60t\), so \(50t=1100\) and \(t=22\). 4. \(V_1(22)=500-220=280\) and \(V_2(22)=800-660=140\). Both volumes are nonnegative, and \(280=2\cdot140\).

Answer

a) \(V_1(t)=500-10t\), \(V_2(t)=800-30t\) b) \(22\) minutes; the volumes are \(280\,\text{gal}\) and \(140\,\text{gal}\), respectively.
5241368
Two monthly cell phone plans are available. Plan A: a \(\$10\) monthly fee plus \(\$0.15\) per minute Plan B: no monthly fee and \(\$0.25\) per minute Let \(m\) be the number of minutes used in a month. a) Construct monthly cost functions \(A(m)\) and \(B(m)\). b) Use the functions to find the number of minutes for which the plans have the same monthly cost. c) Which plan costs less for \(150\) minutes? Justify using your functions.

Hints

- Build each cost function from its fixed fee and per-minute rate. - The no-fee plan should have output \(0\) at input \(0\). - After constructing the functions, equal cost means equal outputs at the same input.

Solution

1. Plan A has model \(A(m)=10+0.15m\). Plan B has model \(B(m)=0.25m\). 2. Set the functions equal: \(10+0.15m=0.25m\). Then \(10=0.10m\), so \(m=100\). 3. \(A(150)=10+0.15\cdot150=32.50\), while \(B(150)=0.25\cdot150=37.50\). Plan A is less expensive.

Answer

a) \(A(m)=10+0.15m\), \(B(m)=0.25m\) b) \(100\) minutes c) Plan A; \(\$32.50\) compared with \(\$37.50\).
5241818
Lena leaves for a hike at \(9{:}00\) a.m. and walks at a constant speed of \(3\) miles per hour. At \(10{:}30\) a.m., her brother Max realizes that she forgot her water bottle. He immediately rides his bike after her along the same route at \(12\) miles per hour. a) Write a function for each person's distance \(y\), in miles, from the starting point after \(x\) hours since Lena left. b) Identify two points on each person's graph and explain what an intersection of the graphs represents. c) Determine what time Max catches Lena and how far they are from the starting point. d) Explain what the slope of each line means in this situation.

Hints

- How can you account for Max starting \(1.5\) hours after Lena? - Which input values give convenient points on each function? - What does it mean when the two distance values are equal? - How is each rate related to the slope of its function?

Solution

1. Let \(x\) be the number of hours since \(9{:}00\) a.m., and let \(y\) be the distance from the starting point in miles. 2. Lena's distance is \(L(x) = 3x\) for \(x \ge 0\). Max starts \(1.5\) hours later, so his distance is \(M(x) = 12(x - 1.5)\) for \(x \ge 1.5\). 3. Lena's graph includes \((0, 0)\) and \((2, 6)\). Max's graph begins at \((1.5, 0)\) and also includes \((2, 6)\). An intersection represents a time when Lena and Max are the same distance from the starting point. 4. Set the distances equal: \(3x = 12(x - 1.5)\). This gives \(3x = 12x - 18\), so \(x = 2\). The distance is \(3 \cdot 2 = 6\) miles. 5. Two hours after \(9{:}00\) a.m. is \(11{:}00\) a.m. The slopes \(3\) and \(12\) represent the walking and biking speeds in miles per hour.

Answer

a) \(L(x) = 3x\) for \(x \ge 0\), and \(M(x) = 12(x - 1.5)\) for \(x \ge 1.5\). b) Lena's graph includes \((0, 0)\) and \((2, 6)\). Max's graph includes \((1.5, 0)\) and \((2, 6)\). An intersection means they are the same distance from the starting point at the same time. c) Max catches Lena at \(11{:}00\) a.m., \(6\) miles from the starting point. d) Each slope represents that person's speed in miles per hour.
5241828
Towns \(A\) and \(B\) are \(300\) miles apart along a straight rail line. A freight train leaves Town \(A\) traveling toward Town \(B\) at a constant speed of \(60\) miles per hour. At the same time, a passenger train leaves Town \(B\) traveling toward Town \(A\) at \(90\) miles per hour. a) Write a function for each train's distance \(y\), in miles, from Town \(A\) after \(x\) hours. Identify two points on each function. b) Compare the trains' distances at whole-number times to estimate when and where they meet. c) Calculate the exact meeting time and distance to check your estimate.

Hints

- If a train is moving toward Town \(A\), what sign should its rate of change have when distance is measured from Town \(A\)? - What is the passenger train's starting distance from Town \(A\)? - At what whole-number time do the two functions have the same value? - Set the two distance expressions equal to check the estimate algebraically.

Solution

1. Let \(x\) be time in hours and \(y\) be distance from Town \(A\) in miles. The freight train is modeled by \(F(x) = 60x\). The passenger train starts \(300\) miles from Town \(A\) and moves toward it, so \(P(x) = 300 - 90x\). 2. The freight-train function includes \((0, 0)\) and \((1, 60)\). The passenger-train function includes \((0, 300)\) and \((1, 210)\). 3. At \(x=1\), the trains are at distances \(60\) miles and \(210\) miles from Town \(A\). At \(x=2\), both distances are \(120\) miles, so the comparison indicates that they meet after about \(2\) hours, \(120\) miles from Town \(A\). 4. To find the exact intersection, set the functions equal: \(60x = 300 - 90x\). 5. Then \(150x = 300\), so \(x = 2\) hours. The distance from Town \(A\) is \(F(2) = 60 \cdot 2 = 120\) miles.

Answer

a) \(F(x) = 60x\) and \(P(x) = 300 - 90x\). Two points on \(F\) are \((0, 0)\) and \((1, 60)\); two points on \(P\) are \((0, 300)\) and \((1, 210)\). b) Comparing whole-number times gives an estimated meeting point of \((2, 120)\). c) The exact values are \(2\) hours and \(120\) miles from Town \(A\).
5241938
A smartphone battery is at \(15\%\) when charging begins. After \(20\) minutes, it is at \(45\%\). Assume the battery percentage increases linearly with time. a) Write a linear function for the battery percentage \(y\) after \(x\) minutes. b) How many percentage points are added per minute? Interpret this value as the slope. c) According to the model, how many minutes after charging begins will the battery reach \(100\%\)?

Hints

- First find the total change in battery percentage over the 20-minute interval. - Divide that change by the time interval to find the rate per minute. - What value represents the battery percentage at time \(0\)? - For a full battery, set the output equal to \(100\).

Solution

1. The slope is \(m = \frac{45 - 15}{20} = 1.5\) percentage points per minute. 2. The initial value is \(15\), so the model is \(y = 1.5x + 15\). 3. The slope \(1.5\) means the battery percentage increases by \(1.5\) percentage points each minute. 4. Set \(y = 100\): \(100 = 1.5x + 15\). Then \(85 = 1.5x\), so \(x = \frac{170}{3} \approx 56.7\) minutes.

Answer

a) \(y = 1.5x + 15\) b) \(1.5\) percentage points per minute c) \(\frac{170}{3}\) minutes, or about \(56.7\) minutes
5241988
A smartphone battery is at \(12\%\) when it is connected to a charger that increases the battery level at a constant rate of \(0.8\) percentage point per minute. a) Write a linear function for the battery level \(L\), in percent, after \(t\) minutes. b) A fast charger starts at the same battery level and reaches \(100\%\) in exactly \(80\) minutes. Find its charging rate in percentage points per minute. c) How many minutes does the fast charger save compared with the first charger when charging from \(12\%\) to \(100\%\)?

Hints

- Identify the starting battery level and the constant rate of increase. - How many percentage points must the battery gain to reach \(100\%\)? - To find a constant rate, divide the total increase by the elapsed time. - Compare the two times needed to reach full charge.

Solution

1. The initial value is \(12\), and the charging rate is \(0.8\) percentage point per minute. Thus \(L(t) = 0.8t + 12\). 2. The fast charger must increase the battery level by \(100 - 12 = 88\) percentage points in \(80\) minutes. Its rate is \(\frac{88}{80} = 1.1\) percentage points per minute. 3. For the first charger, solve \(100 = 0.8t + 12\). This gives \(t = 110\) minutes. The time saved is \(110 - 80 = 30\) minutes.

Answer

a) \(L(t) = 0.8t + 12\) b) \(1.1\) percentage points per minute c) \(30\) minutes
5242028
A hot-air balloon is \(2800\,\text{ft}\) above the ground when it begins to descend at a constant rate of \(5\,\text{ft/s}\). 1) Write a linear function for the balloon's height \(h\), in feet, after \(t\) seconds. 2) How high is the balloon after \(4\) minutes of descending? 3) When will the balloon reach the ground? Give the time in minutes and seconds.

Hints

- Because the balloon is descending, what sign should the rate of change have? - Make sure the time units match before substituting into the function. - What height represents the moment the balloon reaches the ground? - How can you convert a total number of seconds into minutes and remaining seconds?

Solution

1. The initial height is \(2800\) feet, and the rate of change is \(-5\) feet per second. Thus \(h(t) = 2800 - 5t\). 2. Four minutes is \(4 \cdot 60 = 240\) seconds. Then \(h(240) = 2800 - 5 \cdot 240 = 1600\). The balloon is \(1600\,\text{ft}\) above the ground. 3. At ground level, \(h = 0\). Solve \(0 = 2800 - 5t\), which gives \(t = 560\) seconds. Since \(560 = 9 \cdot 60 + 20\), this is \(9\) minutes \(20\) seconds.

Answer

1) \(h(t) = 2800 - 5t\) 2) \(1600\,\text{ft}\) 3) \(9\) minutes \(20\) seconds
5244118
Two water tanks begin filling at the same time. Tank A starts with \(150\) gallons of water and receives another \(20\) gallons each minute. Tank B starts with only \(50\) gallons but fills at a rate of \(32.5\) gallons per minute. a) Write a function for the volume \(y\), in gallons, in each tank after \(x\) minutes. b) Find how many minutes it takes for the tanks to contain the same amount of water. c) How much water is in each tank at that time?

Hints

- Identify the starting volume and filling rate for each tank in the form \(y = mx + b\). - What equation represents the time when the two volumes are equal? - How do you solve an equation when the variable appears on both sides? - Check your value of \(x\) in both original functions.

Solution

1. Tank A is modeled by \(A(x) = 20x + 150\). Tank B is modeled by \(B(x) = 32.5x + 50\). 2. Set the functions equal: \(20x + 150 = 32.5x + 50\). 3. Subtract \(20x\) and \(50\) to get \(100 = 12.5x\). Therefore, \(x = 8\). 4. Substitute \(x = 8\): \(A(8) = 20 \cdot 8 + 150 = 310\). Both tanks contain \(310\) gallons after \(8\) minutes.

Answer

a) \(A(x) = 20x + 150\) and \(B(x) = 32.5x + 50\). b) \(8\) minutes. c) \(310\) gallons.
5262188
In a proportional relationship between \(x\) and \(y\), every increase of \(4\) in \(x\) produces an increase of \(14\) in \(y\). a) Write the equation in the form \(y=kx\). b) What x-value gives \(y=49\)?

Hints

- What does the ratio \(\frac{\Delta y}{\Delta x}\) represent? - How is that ratio related to \(k\)? - Rearrange the equation when the output is known.

Solution

1. The constant of proportionality is the ratio of the changes: \(k=14 \div 4=3.5\). 2. The equation is \(y=3.5x\). 3. Solve \(49=3.5x\): \(x=49 \div 3.5=14\).

Answer

a) \(y=3.5x\) b) \(x=14\)
5268508
A smartphone battery drains at a constant rate while the phone is in use. At \(10{:}00\) a.m., the battery level is \(80\%\). At \(11{:}20\) a.m., it is \(60\%\). a) Find the battery loss in percentage points per hour. b) Assuming the phone has been used continuously at this rate, at what time was it unplugged at \(100\%\)? c) If the same rate continues, at what time will the battery reach \(0\%\)?

Hints

- How many minutes pass between the two battery readings? - How many percentage points does the battery lose during that interval? - Find the loss per minute first, then convert it to a per-hour rate. - How long would a \(20\)-percentage-point drop take at that rate? - How long would a \(100\)-percentage-point drop take at the same rate?

Solution

1. From \(10{:}00\) a.m. to \(11{:}20\) a.m. is \(80\) minutes. The battery level decreases by \(20\) percentage points, so the rate is \(20 \div 80 = 0.25\) percentage point per minute, or \(0.25 \cdot 60 = 15\) percentage points per hour. 2. Going from \(100\%\) to \(80\%\) is a decrease of \(20\) percentage points, which takes \(20 \div 0.25 = 80\) minutes. Eighty minutes before \(10{:}00\) a.m. is \(8{:}40\) a.m. 3. Going from \(100\%\) to \(0\%\) takes \(100 \div 0.25 = 400\) minutes, or \(6\) hours \(40\) minutes. Starting at \(8{:}40\) a.m., the battery reaches \(0\%\) at \(3{:}20\) p.m.

Answer

a) \(15\) percentage points per hour b) \(8{:}40\) a.m. c) \(3{:}20\) p.m.
5268688
Towns A and B are \(180\) miles apart along a highway. At \(12{:}00\) p.m., a truck leaves Town A and drives toward Town B at a constant speed of \(60\) miles per hour. At \(12{:}45\) p.m., a car leaves Town B and drives toward Town A at \(90\) miles per hour. Let \(t\) be the number of hours after \(12{:}00\) p.m., and measure position as distance from Town A. a) Construct the truck's position function \(T(t)\). For \(t\ge0.75\), construct the car's position function \(C(t)\). b) Use the two position functions to determine when the vehicles meet and the distance of the meeting point from Town A.

Hints

- Fix the coordinate system first: both functions must measure position from the same town. - The car's delayed start changes the input inside its model; convert \(45\) minutes to hours before constructing it. - The vehicles meet when the two position functions have the same output.

Solution

1. The truck starts at Town A and moves \(60\) miles per hour away from it, so \(T(t)=60t\). 2. At \(t=0.75\), the car starts \(180\) miles from Town A and then its distance from Town A decreases by \(90\) miles per hour. Thus, for \(t\ge0.75\), \(C(t)=180-90(t-0.75)\). 3. At the meeting time, \(T(t)=C(t)\): \(60t=180-90(t-0.75)\). 4. Simplifying gives \(60t=247.5-90t\), so \(150t=247.5\) and \(t=1.65\). 5. This is \(1\) hour \(39\) minutes after noon, or \(1{:}39\) p.m. 6. The meeting position is \(T(1.65)=60\cdot1.65=99\) miles from Town A.

Answer

a) \(T(t)=60t\); for \(t\ge0.75\), \(C(t)=180-90(t-0.75)\) b) \(1{:}39\) p.m., \(99\) miles from Town A
5321908
A large water tank is being drained at a constant rate. The graph shows the remaining volume \(V\), in gallons, after \(t\) minutes. a) Choose two convenient points from the graph and use them to find the slope \(m\). Explain what the slope means in this situation. b) Find the y-intercept \(b\) from the graph and write the linear function in the form \(V(t) = mt + b\). c) After how many minutes will the tank be empty?
Figure for problem 532190

Hints

- Look for points on the line that lie exactly on grid intersections. - Use change in volume divided by change in time to find the slope. - What volume does the graph show when \(t = 0\)? - How do the slope and initial value fit into \(V(t) = mt + b\)? - What value of \(V\) represents an empty tank?

Solution

1. Two convenient points are \((0, 250)\) and \((5, 150)\). The slope is \(m = \frac{150 - 250}{5 - 0} = -20\,\text{gal/min}\). This means the amount of water decreases by \(20\) gallons each minute. 2. The graph crosses the y-axis at \((0, 250)\), so \(b = 250\). Therefore, \(V(t) = -20t + 250\). 3. The tank is empty when \(V(t) = 0\). Solve \(0 = -20t + 250\), giving \(t = 12.5\) minutes.

Answer

a) \(m = -20\,\text{gal/min}\); the tank loses \(20\) gallons of water per minute. b) \(V(t) = -20t + 250\) c) \(12.5\) minutes
5322138
The coordinate plane uses different scales on the x-axis and y-axis. a) Find the equation of the linear function \(g\) shown in the graph. b) Calculate \(g(6)\), then check your result by reading the corresponding point from the graph.
Figure for problem 532213

Hints

- Describe what the graph shows before calculating. - Pay close attention to the scale on each axis. - What are the coordinates of the axis intercepts? - How can two points be used to calculate the slope? - Which value can you read directly as the y-intercept?

Solution

1. The line crosses the y-axis at \((0, 10)\) and the x-axis at \((-2, 0)\). 2. The slope is \(m = \frac{10 - 0}{0 - (-2)} = 5\), and the y-intercept is \(10\). Therefore, \(g(x) = 5x + 10\). 3. \(g(6) = 5 \cdot 6 + 10 = 40\). The graph also shows the point \((6, 40)\), confirming the calculation.

Answer

a) \(g(x) = 5x + 10\) b) \(g(6) = 40\)
5322158
Maya and Noah ride bicycles along the same route. Maya leaves the starting point at \(10{:}00\) a.m. Noah leaves the same point later but rides faster. The graph labels Maya's distance function \(f\) and Noah's distance function \(g\). a) Who rides faster? Find each rider's speed in miles per hour. b) Use the graph to determine when Noah catches Maya and how far they have traveled by then. c) Write an equation for each rider's distance \(y\), in miles, as a function of clock time \(x\), measured in hours. Solve the system to find the intersection, and compare it with your answer to part b).
Figure for problem 532215

Hints

- Find each speed by dividing distance traveled by elapsed time. - The intersection represents the time and distance at which the riders are together. - Use each rider's starting time and constant speed to write a linear equation. - Set the two distance expressions equal to solve for the meeting time.

Solution

1. Maya travels \(8\) miles from \(10{:}00\) a.m. to \(11{:}00\) a.m., so her speed is \(8\,\text{mph}\). Noah travels \(12\) miles from \(10{:}30\) a.m. to \(11{:}30\) a.m., so his speed is \(12\,\text{mph}\). Noah rides faster. 2. The graphs intersect at \((11.5, 12)\). Therefore, Noah catches Maya at \(11{:}30\) a.m., after each rider has traveled \(12\) miles. 3. Maya's equation is \(f(x)=8(x-10)=8x-80\) for \(x\ge10\). Noah's equation is \(g(x)=12(x-10.5)=12x-126\) for \(x\ge10.5\). Set the expressions equal: \(8x-80=12x-126\), so \(4x=46\) and \(x=11.5\). Then \(y=8(11.5-10)=12\). This matches the graph.

Answer

a) Noah; Maya rides at \(8\,\text{mph}\), and Noah rides at \(12\,\text{mph}\). b) \(11{:}30\) a.m.; \(12\) miles c) Maya: \(f(x)=8x-80\), \(x\ge10\); Noah: \(g(x)=12x-126\), \(x\ge10.5\); the intersection is \((11.5, 12)\).
5332028
A water tank is filled at a constant rate. The function \(V(t)\) gives the volume in gallons after \(t\) minutes. Two measurements, \(P_1\) and \(P_2\), are shown on the graph. a) Find the inflow rate, or slope, in gallons per minute. b) Find the volume already in the tank at \(t=0\). c) Write the function \(V(t)\). d) Find the volume after \(15\) minutes.
Figure for problem 533202

Hints

- Use change in volume divided by change in time to find the slope. - Substitute one point into \(V(t)=mt+b\) to find the initial value. - Evaluate the completed function at \(t=15\).

Solution

1. The points are \(P_1(2, 100)\) and \(P_2(8, 250)\). The slope is \(m=\frac{250-100}{8-2}=\frac{150}{6}=25\,\text{gal/min}\). 2. Substitute \(P_1\) into \(V(t)=25t+b\): \(100=25\cdot2+b\), so \(b=50\) gallons. 3. The function is \(V(t)=25t+50\). 4. \(V(15)=25\cdot15+50=425\), so the tank contains \(425\) gallons.

Answer

a) \(25\,\text{gal/min}\) b) \(50\) gallons c) \(V(t)=25t+50\) d) \(425\) gallons
5332138
Two cell phone plans have different monthly fees and different per-minute charges. Let \(x\) be the number of minutes used in a month. On the graph, curve \(a\) represents Plan A and curve \(b\) represents Plan B. Write a cost function for each plan, including the contextual domain. What does the intersection of the two lines mean in context?
Figure for problem 533213

Hints

- The y-intercept represents the monthly fee. - The slope represents the cost per minute. - The number of minutes used cannot be negative. - At an intersection, both plans have the same cost for the same number of minutes.

Solution

1. Plan A has a y-intercept of \(5\), representing a \(\$5\) monthly fee. Using the point \((100, 15)\), its slope is \(m=\frac{15-5}{100-0}=0.10\). Thus, \(a(x)=0.10x+5\), with \(x\ge0\). 2. Plan B has a y-intercept of \(10\), representing a \(\$10\) monthly fee. Using \((100, 15)\), its slope is \(m=\frac{15-10}{100-0}=0.05\). Thus, \(b(x)=0.05x+10\), with \(x\ge0\). 3. The lines intersect at \((100, 15)\). This means both plans cost \(\$15\) when \(100\) minutes are used.

Answer

Plan A: \(a(x)=0.10x+5\), \(x\ge0\); Plan B: \(b(x)=0.05x+10\), \(x\ge0\); at \(100\) minutes, both plans cost \(\$15\).
5332148
Use the graph to write an equation for the line \(f\). Then read the zero, or x-intercept, from the graph and verify it algebraically.
Figure for problem 533214

Hints

- At a zero, the y-value is \(0\). - To find the slope exactly, use two points where the line passes through grid intersections.

Solution

1. The graph crosses the y-axis at \(-4\), so \(b = -4\). Using \((0, -4)\) and \((5, 0)\), the slope is \(m = \frac{0 - (-4)}{5 - 0} = \frac{4}{5} = 0.8\). Thus \(f(x) = 0.8x - 4\). 2. From the graph, the zero is \(x = 5\). 3. Verify algebraically: \(0.8x - 4 = 0\). Then \(0.8x = 4\), so \(x = 5\).

Answer

Equation: \(f(x) = 0.8x - 4\). Zero: \(x = 5\).
5332288
Liam leaves a trailhead at \(9{:}00\) a.m. and walks at a constant speed of \(4\) miles per hour. At \(10{:}30\) a.m., Marie rides after him along the same route at \(12\) miles per hour. The graph shows each person's distance from the trailhead as a function of time \(t\), in hours since \(9{:}00\) a.m. The graph uses \(l\) for Liam and \(m\) for Marie. a) Construct the two distance functions \(l(t)\) and \(m(t)\) from the situation. State the time restriction on Marie's function. b) Use the graph to estimate when and how far from the trailhead Marie catches Liam. c) Use the functions to calculate the meeting time and distance exactly, and give the clock time.
Figure for problem 533228

Hints

- Build each distance function from its starting time, starting distance, and speed before using the intersection. - Marie's delayed start should appear in her input expression, not as an added distance. - The graph provides an estimate; equality of the two constructed functions provides the exact meeting event.

Solution

1. Liam starts at distance \(0\) at \(t=0\) and walks \(4\) miles per hour, so \(l(t)=4t\). 2. Marie starts \(1.5\) hours after \(9{:}00\) a.m., so for \(t\ge1.5\), \(m(t)=12(t-1.5)\). 3. The graph intersection is about \((2.25,9)\), suggesting a meeting \(2.25\) hours after \(9{:}00\) a.m. and \(9\) miles from the trailhead. 4. Solve \(4t=12(t-1.5)\). Then \(4t=12t-18\), so \(8t=18\) and \(t=2.25\). 5. The distance is \(l(2.25)=4\cdot2.25=9\) miles. Since \(0.25\) hour is \(15\) minutes, the clock time is \(11{:}15\) a.m.

Answer

a) \(l(t)=4t\); \(m(t)=12(t-1.5)\) for \(t\ge1.5\) b) About \(2.25\) hours after \(9{:}00\) a.m. and \(9\) miles from the trailhead c) Exactly \(t=2.25\), so \(11{:}15\) a.m. at \(9\) miles
5332298
Two cargo boats, the Albatross and the Osprey, travel along a \(100\)-mile route between two harbors. The Albatross leaves Harbor A toward Harbor B at \(20\) miles per hour. At the same time, the Osprey leaves Harbor B toward Harbor A at \(30\) miles per hour. The graph shows each boat's distance from Harbor A as a function of travel time \(t\). The graph uses \(a\) for the Albatross and \(o\) for the Osprey. a) Construct position functions \(a(t)\) and \(o(t)\), measuring distance from Harbor A. b) Use the graph to estimate the meeting time and meeting distance from Harbor A. c) Use the two functions to confirm the meeting time and distance exactly.
Figure for problem 533229

Hints

- Use the same reference point, Harbor A, for both position functions. - One boat's distance from Harbor A increases while the other's decreases; the signs of the rates should reflect that. - Use the graph only for an estimate, then equate the constructed functions for the exact result.

Solution

1. The Albatross starts at Harbor A, so \(a(t)=20t\). 2. The Osprey starts \(100\) miles from Harbor A and its distance from Harbor A decreases by \(30\) miles per hour, so \(o(t)=100-30t\). 3. The graph intersection is about \((2,40)\), suggesting a meeting after \(2\) hours, \(40\) miles from Harbor A. 4. Set the functions equal: \(20t=100-30t\). Then \(50t=100\), so \(t=2\). 5. \(a(2)=20\cdot2=40\), so the exact meeting point is \(40\) miles from Harbor A.

Answer

a) \(a(t)=20t\), \(o(t)=100-30t\) b) About \(2\) hours and \(40\) miles from Harbor A c) Exactly \(2\) hours and \(40\) miles from Harbor A
5332658
Four lines in the coordinate plane all pass through the same point \(P\). a) Find the coordinates of \(P\). b) Find the equation of each of the four lines. c) A fifth line \(h_5\) also passes through \(P\) and is parallel to the x-axis. Find its equation.
Figure for problem 533265

Hints

- Find the point where all four lines meet. - A point on the y-axis has x-coordinate \(0\). - Since all four lines share the same y-intercept, focus on finding each slope. - What is the slope of a horizontal line?

Solution

1. All four lines meet on the y-axis at \(-2\), so \(P(0, -2)\). 2. Because every line passes through \(P(0, -2)\), each has y-intercept \(-2\). For \(h_1\), using \((1, 0)\) gives slope \(2\), so \(h_1(x) = 2x - 2\). For \(h_2\), using \((2, -4)\) gives slope \(-1\), so \(h_2(x) = -x - 2\). For \(h_3\), using \((2, -1)\) gives slope \(\frac{1}{2}\), so \(h_3(x) = \frac{1}{2}x - 2\). For \(h_4\), using \((1, -5)\) gives slope \(-3\), so \(h_4(x) = -3x - 2\). 3. A line parallel to the x-axis has slope \(0\). Passing through \(P(0, -2)\), its equation is \(h_5(x) = -2\).

Answer

a) \(P(0, -2)\) b) \(h_1(x) = 2x - 2\); \(h_2(x) = -x - 2\); \(h_3(x) = \frac{1}{2}x - 2\); \(h_4(x) = -3x - 2\) c) \(h_5(x) = -2\)
5332668
The point \(A(0, 1.5)\) is shown in the coordinate plane. a) Find the equations of the three lines \(g_1\), \(g_2\), and \(g_3\), which all pass through \(A\). b) What parameter do all three equations have in common when written in the form \(y = mx + b\)? c) Give an equation for another line \(g_4\) that also passes through \(A\) and has a negative slope whose absolute value is less than the absolute value of the slope of \(g_1\).
Figure for problem 533266

Hints

- Since all the lines pass through the same point on the y-axis, what does that tell you about their y-intercepts? - From point \(A\), use another grid point on each line to calculate its slope. - A negative slope with smaller absolute value is closer to \(0\) and represents a less steep decreasing line.

Solution

1. Since every line passes through \(A(0, 1.5)\), each has y-intercept \(b = 1.5\). 2. For \(g_1\), using \((1, -0.5)\), the slope is \(m = \frac{-0.5 - 1.5}{1 - 0} = -2\). Thus \(g_1(x) = -2x + 1.5\). 3. For \(g_2\), using \((1, 2)\), the slope is \(m = \frac{2 - 1.5}{1 - 0} = 0.5\). Thus \(g_2(x) = 0.5x + 1.5\). 4. Line \(g_3\) is horizontal, so its slope is \(0\). Thus \(g_3(x) = 1.5\). 5. For \(g_4\), any slope satisfying \(-2 < m < 0\) works. For example, with \(m = -1\), \(g_4(x) = -x + 1.5\).

Answer

a) \(g_1(x) = -2x + 1.5\); \(g_2(x) = 0.5x + 1.5\); \(g_3(x) = 1.5\) b) They all have the same y-intercept, \(b = 1.5\). c) For example, \(g_4(x) = -x + 1.5\). Any slope \(m\) with \(-2 < m < 0\) works.
5332678
The graph shows a family of parallel lines. a) Find the slope \(m\) that all the lines have in common. b) Write equations for lines \(k\), \(l\), and \(n\). c) Describe how the equation changes when a line is shifted vertically while remaining parallel. Which parameter changes?
Figure for problem 533267

Hints

- What must be true about the slopes of parallel nonvertical lines? - Choose a line with two easy-to-read grid points and calculate rise over run. - The y-intercept is the value \(b\) in \(y = mx + b\).

Solution

1. Use two points on one line. On line \(l\), for example, \((0, 1)\) and \((4, -2)\) give \(m = \frac{-2 - 1}{4 - 0} = -\frac{3}{4}\). Since the lines are parallel, they all have this slope. 2. Read the y-intercepts from the graph. Line \(k\) has y-intercept \(3\), line \(l\) has y-intercept \(1\), and line \(n\) has y-intercept \(-1\). Therefore, \(k: y = -\frac{3}{4}x + 3\), \(l: y = -\frac{3}{4}x + 1\), and \(n: y = -\frac{3}{4}x - 1\). 3. A vertical shift changes only the y-intercept \(b\). Shifting up increases \(b\); shifting down decreases \(b\). The slope remains unchanged.

Answer

a) \(m = -\frac{3}{4}\) b) \(k: y = -\frac{3}{4}x + 3\); \(l: y = -\frac{3}{4}x + 1\); \(n: y = -\frac{3}{4}x - 1\) c) The y-intercept \(b\) changes, while the slope \(m\) stays the same.
5332748
A candle is \(8\,\text{in.}\) tall when it is lit. Its height \(h\), in inches, decreases at a constant rate as it burns for \(t\) hours. a) Use the graph to write the linear function \(h(t)\). Pay attention to the scale on each axis. b) Find the candle's height after \(12\) hours and check the value on the graph.
Figure for problem 533274

Hints

- Since the candle gets shorter, what sign should the slope have? - Check how much one grid step represents on each axis. - Use two easy-to-read points to find the change in height per hour. - Substitute the requested time into your function.

Solution

1. Two convenient points are \((0, 8)\) and \((8, 4)\). 2. The slope is \(m = \frac{4 - 8}{8 - 0} = -\frac{1}{2}\). The initial height is \(8\), so \(h(t) = -\frac{1}{2}t + 8\). 3. \(h(12) = -\frac{1}{2} \cdot 12 + 8 = 2\). The graph also shows \((12, 2)\).

Answer

a) \(h(t) = -\frac{1}{2}t + 8\) b) \(2\,\text{in.}\)
5332918
A line \(g\) passes through \(P(-1, 4)\) and \(Q(2, -2)\), as shown in the graph. a) Find the equation of line \(g\). b) Use substitution to determine whether \(R(4, -6)\) also lies on the line.
Figure for problem 533291

Hints

- Use the coordinates of \(P\) and \(Q\) to find the slope. - Once you know the slope, substitute one point into \(y = mx + b\) to find \(b\). - To test \(R\), substitute its x-coordinate into your equation and compare the result with its y-coordinate.

Solution

1. The slope is \(m = \frac{-2 - 4}{2 - (-1)} = -2\). Using \(Q(2, -2)\), \(-2 = -2 \cdot 2 + b\), so \(b = 2\). Therefore, \(g(x) = -2x + 2\). 2. Substitute \(x = 4\): \(g(4) = -2 \cdot 4 + 2 = -6\). This matches the y-coordinate of \(R\), so \(R\) lies on the line.

Answer

a) \(g(x) = -2x + 2\) b) Yes, \(R(4, -6)\) lies on \(g\).
5349468
Write equations for lines \(g_1\), \(g_2\), and \(g_3\). Then read the intersection point of \(g_1\) and \(g_2\) directly from the graph.
Figure for problem 534946

Hints

- For \(g_1\) and \(g_2\), first read the y-intercept. - Use another grid point to calculate the slope. - The intersection is where the two lines cross.

Solution

1. Line \(g_1\) has y-intercept \(-2\) and passes through \((4, 0)\). Its slope is \(m = \frac{0 - (-2)}{4 - 0} = \frac{1}{2}\), so \(g_1: y = \frac{1}{2}x - 2\). 2. Line \(g_2\) has y-intercept \(4\) and passes through \((4, 0)\). Its slope is \(m = \frac{0 - 4}{4 - 0} = -1\), so \(g_2: y = -x + 4\). 3. Line \(g_3\) is horizontal at \(y = 1\), so \(g_3: y = 1\). 4. The graph shows \(g_1\) and \(g_2\) crossing at \((4, 0)\).

Answer

\(g_1: y = \frac{1}{2}x - 2\); \(g_2: y = -x + 4\); \(g_3: y = 1\) Intersection of \(g_1\) and \(g_2\): \((4, 0)\)
5120448
A climber assumes that air temperature decreases at a constant rate as elevation increases. The climber measures \(61\,\text{°F}\) at \(2000\,\text{ft}\) and \(47\,\text{°F}\) at \(6000\,\text{ft}\). a) Use a linear model to predict the temperature at a \(10{,}000\)-foot summit. b) A plane flies at \(39{,}000\,\text{ft}\). What temperature does the model predict at that elevation? c) Why is it problematic to use this simple linear model at extreme elevations?

Hints

- Find the temperature change for each \(1000\)-foot increase in elevation. - Compare each target elevation with the lower measurement. - Consider whether one local rate should remain valid through every layer of the atmosphere.

Solution

1. The temperature changes by \(47-61=-14\,\text{°F}\) over an elevation increase of \(6000-2000=4000\,\text{ft}\). The rate is \(\frac{-14}{4000}=-0.0035\,\text{°F}\) per foot, or \(-3.5\,\text{°F}\) per \(1000\) feet. 2. The summit is \(8000\) feet above the first measurement. The predicted change is \(8(-3.5)=-28\,\text{°F}\), so the predicted temperature is \(61-28=33\,\text{°F}\). 3. The plane is \(37{,}000\) feet above the first measurement. The predicted change is \(37(-3.5)=-129.5\,\text{°F}\), so the model predicts \(61-129.5=-68.5\,\text{°F}\). 4. Atmospheric conditions and the rate of temperature change vary by altitude and weather. A rate estimated from two mountain measurements should not be assumed valid throughout the atmosphere.

Answer

a) \(33\,\text{°F}\) b) \(-68.5\,\text{°F}\) c) The constant rate is reasonable only over a limited elevation range; atmospheric conditions change at much greater heights.
5120668
A cylindrical container starts empty and has a base area of \(200\,\text{cm}^2\). It is filled at a constant rate of \(500\,\text{mL}\) per minute. a) How many centimeters per minute does the water height increase? b) Write a function for the water height \(h\), in centimeters, after \(t\) minutes. c) How would the graph change for a container with a base area of \(100\,\text{cm}^2\) filled at the same rate? Compare the slopes. d) Suppose a container consists of two stacked cylinders. The lower cylinder has base area \(200\,\text{cm}^2\) and height \(10\,\text{cm}\). The upper cylinder has base area \(100\,\text{cm}^2\) and height \(10\,\text{cm}\). Describe the water-height graph and what happens after \(4\) minutes.

Hints

- Use the cylinder volume relationship \(V=Bh\). - A constant inflow gives a constant increase in volume. - Compare how the same volume changes height when the base area is smaller.

Solution

1. Since \(500\,\text{mL}=500\,\text{cm}^3\), use \(V=Bh\): \(500=200h\). Thus, the height increases by \(2.5\,\text{cm}\) per minute. 2. Starting empty, the height function is \(h(t)=2.5t\). 3. For base area \(100\,\text{cm}^2\), \(500=100h\), so the height increases by \(5\,\text{cm}\) per minute. The slope doubles. 4. In the lower section, the graph has slope \(2.5\). The water reaches \(10\,\text{cm}\) after \(10\div2.5=4\) minutes. In the upper section, the slope is \(5\). Therefore, the graph has a corner at \((4, 10)\) and becomes twice as steep after \(4\) minutes.

Answer

a) \(2.5\,\text{cm}\) per minute b) \(h(t)=2.5t\) c) The slope doubles to \(5\,\text{cm}\) per minute. d) The graph has slope \(2.5\) until \((4, 10)\), then slope \(5\); the water rises twice as fast in the narrower upper section.
5129788
A cylindrical candle is initially \(12.5\,\text{in.}\) tall. After it is lit, its height decreases at a constant rate of \(0.6\,\text{in.}\) per hour. a) Write a function for the candle's height \(h\), in inches, after \(t\) hours. b) How long will it take the candle to reach half its original height? Give the answer in hours and minutes. c) Suppose the candle were shaped like a pyramid, becoming narrower toward the top. Explain why its height would not decrease linearly if the same volume of wax burned each hour.

Hints

- A decreasing quantity has a negative rate of change. - Set the height equal to half of \(12.5\). - Convert the decimal part of an hour to minutes by multiplying by \(60\). - Compare the height removed by the same wax volume in narrow and wide sections.

Solution

1. The initial height is \(12.5\), and the rate of change is \(-0.6\), so \(h(t)=12.5-0.6t\). 2. Half the original height is \(6.25\) inches. Solve \(6.25=12.5-0.6t\). Then \(0.6t=6.25\), so \(t=\frac{6.25}{0.6}\approx10.4167\) hours. The decimal part is \(0.4167(60)\approx25\) minutes, so the time is about \(10\) hours \(25\) minutes. 3. A pyramid has a smaller cross section near the top and a larger cross section near the bottom. Burning the same volume removes more height in a narrow section than in a wide section, so the height-change rate is not constant.

Answer

a) \(h(t)=12.5-0.6t\) b) About \(10\) hours \(25\) minutes c) The changing cross-sectional area makes the height decrease at a changing rate.
5129938
A rectangular water tank holds at most \(375\) gallons. It already contains \(50\) gallons when filling begins. An inlet adds \(30\) gallons in \(8\) minutes at a constant rate. a) Find the inflow rate in gallons per minute and write a function \(V(t)\) for the volume, in gallons, after \(t\) minutes. b) How long will it take to fill the tank completely? Give the time in hours, minutes, and seconds. c) Suppose a blockage cuts the inflow rate in half. Describe how the slope of the graph and the time needed to fill the tank would change.

Hints

- Divide the amount added by the elapsed time to find the rate per minute. - How many more gallons are needed to reach capacity? - In this context, what does the slope represent? - For the same amount of water, what happens to the required time when the rate is cut in half?

Solution

1. The inflow rate is \(30 \div 8 = 3.75\) gallons per minute. With an initial volume of \(50\) gallons, \(V(t) = 3.75t + 50\). 2. The tank is full when \(V(t) = 375\): \(375 = 3.75t + 50\). Thus \(325 = 3.75t\), so \(t = 86\frac{2}{3}\) minutes. 3. Since \(86\frac{2}{3}\) minutes is \(1\) hour plus \(26\frac{2}{3}\) minutes, the fill time is \(1\) hour, \(26\) minutes, and \(40\) seconds. 4. Halving the inflow rate halves the slope to \(1.875\) gallons per minute. The amount still needed is unchanged, so the fill time doubles to \(173\frac{1}{3}\) minutes.

Answer

a) The inflow rate is \(3.75\) gallons per minute, and \(V(t) = 3.75t + 50\). b) \(1\) hour, \(26\) minutes, and \(40\) seconds c) The slope is halved, and the fill time doubles.
5137798
Two candles are lit at the same time. Candle A is \(5\,\text{cm}\) tall and burns down at a constant rate of \(10\,\text{mm}\) per hour. Candle B is \(12\,\text{cm}\) tall and, each hour, loses an amount equal to \(10\%\) of its original height. Use linear models to determine when the candles would have the same height. Then decide whether that actually happens while the candles are burning.

Hints

- First find how many centimeters Candle B loses each hour. - Write a linear equation for the height of each candle after \(t\) hours. - After finding a time when the equations are equal, determine how long each candle can actually burn. - Can a candle have a negative height?

Solution

1. Candle A loses \(10\,\text{mm} = 1\,\text{cm}\) per hour. Candle B loses \(10\%\) of \(12\,\text{cm}\), so its rate is \(0.10 \cdot 12 = 1.2\,\text{cm}\) per hour. 2. Let \(t\) be the number of hours after the candles are lit. The height models are \(A(t) = 5 - t\) and \(B(t) = 12 - 1.2t\). 3. Set the heights equal: \(5 - t = 12 - 1.2t\). This gives \(0.2t = 7\), so \(t = 35\). 4. At \(t = 35\), the model gives a height of \(-30\,\text{cm}\), which is not physically possible. Candle A burns out after \(5\) hours, and Candle B burns out after \(10\) hours. 5. Therefore, the linear models are equal only after both candles would already be gone, so the candles never have the same height while they are burning.

Answer

The linear models are equal after \(35\) hours, at the theoretical height \(-30\,\text{cm}\). Candle A burns out after \(5\) hours and Candle B after \(10\) hours, so the candles never have the same height while they are actually burning.
5142088
At the beginning of an observation, a smartphone battery is at \(10\%\). It is connected to a fast charger and reaches \(90\%\) after \(40\) minutes. The charger is then removed, and a \(20\)-minute video call lowers the battery level to \(60\%\). a) Write a function for battery level \(y\), in percent, as a function of elapsed time \(x\), in minutes, for each phase. b) Compare the charging rate with the magnitude of the discharging rate. By how many percentage points per minute do they differ? c) At what elapsed times was the battery level exactly \(75\%\)? d) If the phone continued to discharge at the same rate after the call, how many additional minutes would it take to reach \(0\%\)?

Hints

- Find the slope of each segment from its two endpoints. - Keep the sign of each slope separate from its magnitude. - Check both phases for a battery level of \(75\%\). - Divide the remaining battery level by the discharge rate.

Solution

1. The key points are \((0, 10)\), \((40, 90)\), and \((60, 60)\). 2. During charging, the slope is \(\frac{90-10}{40}=2\), so \(y=2x+10\) for \(0\le x\le40\). 3. During use, the slope is \(\frac{60-90}{60-40}=-1.5\), so \(y=90-1.5(x-40)=-1.5x+150\) for \(40\le x\le60\). 4. The charging rate is \(2\) percentage points per minute, and the discharging-rate magnitude is \(1.5\). They differ by \(0.5\) percentage point per minute. 5. During charging, solve \(75=2x+10\) to get \(x=32.5\). During use, solve \(75=-1.5x+150\) to get \(x=50\). 6. From \(60\%\), discharging at \(1.5\) points per minute takes \(60\div1.5=40\) additional minutes.

Answer

a) Charging: \(y=2x+10\), \(0\le x\le40\); use: \(y=-1.5x+150\), \(40\le x\le60\) b) \(0.5\) percentage point per minute c) At \(x=32.5\) minutes and \(x=50\) minutes d) \(40\) additional minutes

All problems may be used, copied and printed free of charge for school and tutoring, including paid tutoring. Commercial adaptations as well as publication or redistribution on the internet are not permitted.