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Graph proportional relationships and compute slope

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5520008
The graph shows a proportional relationship between time and distance. What is the constant of proportionality, and what does it mean in this context?
Figure for problem 552000

Hints

- Use two clear grid points on the line. - Find the change in distance for one unit of time.

Solution

1. Choose two points on the line, such as \((0, 0)\) and \((1, 2)\). 2. The slope is \(\frac{2 - 0}{1 - 0} = 2\). 3. The constant of proportionality is \(2\), meaning the distance increases by \(2\) miles for each hour.

Answer

The constant of proportionality is \(2\). It means a rate of \(2\,\text{miles per hour}\).
5520018
The table shows a proportional relationship. <table><tr><th>\(x\)</th><th>\(y\)</th></tr><tr><td>\(1\)</td><td>\(3\)</td></tr><tr><td>\(2\)</td><td>\(6\)</td></tr><tr><td>\(4\)</td><td>\(12\)</td></tr></table> Find the constant of proportionality and write an equation relating \(y\) to \(x\).

Hints

- Compare \(y\) to \(x\) in more than one row. - A proportional relationship has the form \(y = kx\).

Solution

1. For each row, compute \(\frac{y}{x}\): the ratio is constant at \(3\). 2. Therefore, the constant of proportionality is \(3\). 3. The proportional equation is \(y = 3x\).

Answer

Constant of proportionality: \(3\) Equation: \(y = 3x\)
5120558
At a copy shop, price is proportional to the number of pages. Printing \(40\) pages costs \(\$3.20\). a) Find the price per page. b) Find the total price for \(150\) pages. c) Describe the graph of total price versus number of pages, with number of pages on the x-axis.

Hints

- Find the cost of one page. - What is the price when no pages are printed? - How are page count and price related?

Solution

1. The price per page is \(3.20\div40=\$0.08\). 2. Printing \(150\) pages costs \(150\cdot0.08=\$12.00\). 3. Because the relationship is proportional, the graph is a straight line through the origin. Its slope is \(\$0.08\) per page.

Answer

a) \(\$0.08\) per page. b) \(\$12.00\). c) A line through the origin with slope \(\$0.08\) per page.
5331736
Points \(A\), \(B\), \(C\), and \(D\), along with line \(g\), are shown on the coordinate plane. a) Find the coordinates of all four points. b) Which points lie on line \(g\)?
Figure for problem 533173

Hints

- Read the x-coordinate first and the y-coordinate second. - Trace line \(g\) across the grid. - Check whether the center of each plotted point lies on the displayed line.

Solution

1. Reading the graph gives \(A=(-4, 5)\), \(B=(2, 2)\), \(C=(5, -3)\), and \(D=(-3, -3)\). 2. Inspecting the plotted line shows that points \(B\) and \(D\) lie directly on line \(g\). Points \(A\) and \(C\) do not.

Answer

a) \(A=(-4, 5)\), \(B=(2, 2)\), \(C=(5, -3)\), and \(D=(-3, -3)\) b) Points \(B\) and \(D\)
5332808
A hiker travels at a constant pace. The graph shows distance \(s\), in miles, as a function of time \(t\), in hours. a) Find the hiking speed and write the function. b) How far has the hiker traveled after \(4.5\) hours? c) How long does it take the hiker to travel \(15\) miles?
Figure for problem 533280

Hints

- Use a clear point on the graph to find distance divided by time. - What does the slope represent in this context? - Substitute a time to find distance, or solve the equation for time.

Solution

1. The graph passes through \((5,15)\), so the speed is \(15\div5=3\,\text{mph}\). Because the graph passes through the origin, the function is \(s(t)=3t\). 2. Evaluate \(s(4.5)=3\cdot4.5=13.5\), so the hiker travels \(13.5\) miles. 3. Solve \(15=3t\) to get \(t=5\) hours.

Answer

a) \(3\,\text{mph}\); \(s(t)=3t\) b) \(13.5\) miles c) \(5\) hours
5332908
The graph shows a side view of a planned bicycle ramp. Each grid unit represents \(8\) inches in both directions. The top edge of the ramp begins at \((0, 0)\). Find the slope of the ramp and write an equation for the line.
Figure for problem 533290

Hints

- Count the vertical and horizontal grid changes. - Slope is rise divided by run. - A line beginning at the origin has a simple equation form.

Solution

1. The graph rises \(3\) units while running \(12\) units to the right. The slope is \(m=3 \div 12=0.25\). 2. The line passes through the origin, so its equation is \(y=0.25x\). The physical scale does not change the slope because both axes use the same scale.

Answer

The slope is \(0.25\), and the equation is \(y=0.25x\).
5338758
A conveyor belt at a gravel plant carries material upward. The graph shows a side view of the belt, with horizontal distance on the x-axis and height on the y-axis, both in meters. Find the slope of the conveyor belt.
Figure for problem 533875

Hints

- Choose two points on the line that lie exactly on grid intersections. - Compare the vertical change with the horizontal change. - Use rise over run.

Solution

1. Use two clear points on the line, such as \((0, 0)\) and \((5, 3)\). 2. Calculate the slope: \(m = \frac{3 - 0}{5 - 0} = \frac{3}{5}\). 3. Therefore, the conveyor belt has slope \(\frac{3}{5}\), or \(0.6\).

Answer

\(m = \frac{3}{5}\), or \(0.6\)
5338768
A backyard pool is filled at a constant rate. The graph shows the volume of water in gallons as a function of time in minutes. Find the fill rate, or slope, in gallons per minute.
Figure for problem 533876

Hints

- The fill rate is the slope of the line. - Use change in volume divided by change in time between the marked points.

Solution

1. Use the marked points \(P_1(2, 10)\) and \(P_2(6, 30)\). 2. The slope is \(m=\frac{30-10}{6-2}=\frac{20}{4}=5\). Therefore, the pool fills at \(5\,\text{gal/min}\).

Answer

The fill rate is \(5\,\text{gal/min}\).
5546578
Relationship A is given by \(y = 2.5x\). Relationship B is shown in the table. <table><tr><th>\(x\)</th><th>\(y\)</th></tr><tr><td>\(2\)</td><td>\(6\)</td></tr><tr><td>\(4\)</td><td>\(12\)</td></tr><tr><td>\(6\)</td><td>\(18\)</td></tr></table> Which relationship has the greater constant of proportionality, and by how much?

Hints

- In an equation of the form \(y=kx\), where is the constant of proportionality shown? - For the table, compare the ratio \(\frac{y}{x}\) across the rows. - After finding both constants, compare them directly.

Solution

1. In Relationship A, the coefficient of \(x\) is the constant of proportionality, so the constant is \(2.5\). 2. In Relationship B, each ratio \(\frac{y}{x}\) equals \(3\), so its constant of proportionality is \(3\). 3. Relationship B has the greater constant, and the difference is \(3-2.5=0.5\).

Answer

Relationship B has the greater constant of proportionality by \(0.5\).
5546678
Relation A is a proportional relationship shown in the graph. Relation B is shown in the table. <table> <tr><th>\(x\)</th><th>\(y\)</th></tr> <tr><td>\(2\)</td><td>\(5\)</td></tr> <tr><td>\(4\)</td><td>\(10\)</td></tr> </table> Find the constant of proportionality for each relation. Which relation has the greater constant of proportionality, and by how much?
Figure for problem 554667

Hints

- For a proportional relationship, compare \(\frac{y}{x}\). - Read one exact point on Relation A from the graph. - Check whether the table gives the same ratio in both rows before comparing the relations.

Solution

1. From the graph of Relation A, a convenient point is \((4,8)\), so its constant of proportionality is \(\frac{8}{4}=2\). 2. From Relation B's table, \(\frac{5}{2}=2.5\), and the second row confirms \(\frac{10}{4}=2.5\). 3. Relation B has the greater constant of proportionality. 4. The difference is \(2.5-2=0.5\).

Answer

Relation A: \(2\) Relation B: \(2.5\) Relation B is greater by \(0.5\).
5119508
Two tables represent proportional relationships. **Table A:** <table> <tr><td>\(x\)</td><td>\(2\)</td><td>\(4\)</td><td>\(6\)</td></tr> <tr><td>\(y\)</td><td>\(5\)</td><td>\(10\)</td><td>\(15\)</td></tr> </table> **Table B:** <table> <tr><td>\(x\)</td><td>\(3\)</td><td>\(6\)</td><td>\(9\)</td></tr> <tr><td>\(y\)</td><td>\(6\)</td><td>\(12\)</td><td>\(18\)</td></tr> </table> a) Find the constant of proportionality \(k\) for each table and write each equation in the form \(y=kx\). b) Describe one feature the two graphs have in common. c) Which graph is steeper? Justify your answer using the constants of proportionality.

Hints

- Use one ordered pair to calculate \(\frac{y}{x}\). - What feature do all graphs of proportional relationships share? - Compare the change in \(y\) for the same change in \(x\). - How does the coefficient of \(x\) affect a line's steepness?

Solution

1. For Table A, \(k=5\div2=2.5\), so its equation is \(y=2.5x\). For Table B, \(k=6\div3=2\), so its equation is \(y=2x\). 2. Both proportional relationships graph as straight lines through \((0,0)\). 3. Table A's graph is steeper because its slope, \(2.5\), is greater than Table B's slope, \(2\).

Answer

a) Table A: \(k=2.5\), \(y=2.5x\). Table B: \(k=2\), \(y=2x\). b) Both graphs are lines through the origin. c) Table A's graph is steeper because \(2.5>2\).
5119518
Two school copiers work at different constant rates. Copier A prints \(120\) pages in \(3\) minutes. Copier B prints \(150\) pages in \(5\) minutes. For each copier, time \(x\), in minutes, and pages printed \(y\) form a proportional relationship. a) Write an equation for each copier. b) If both relationships are graphed, which graph is steeper? Explain what steepness means in this context. c) Both copiers start at the same time. How many pages has Copier A printed when Copier B finishes \(360\) pages?

Hints

- What does the constant of proportionality represent here? - Find how many pages each copier prints in one minute. - What does a steeper graph mean for pages printed per minute? - First find how long Copier B takes to print \(360\) pages.

Solution

1. Copier A prints \(120\div3=40\) pages per minute, so \(y_A=40x\). Copier B prints \(150\div5=30\) pages per minute, so \(y_B=30x\). 2. Copier A's graph is steeper because \(40>30\). In context, greater steepness means more pages printed per minute. 3. Copier B needs \(360\div30=12\) minutes to print \(360\) pages. 4. In \(12\) minutes, Copier A prints \(40\cdot12=480\) pages.

Answer

a) Copier A: \(y=40x\). Copier B: \(y=30x\). b) Copier A's graph is steeper; slope represents pages per minute. c) Copier A has printed \(480\) pages.
5119528
The graph of a proportional relationship passes through \(P(4, 14)\). a) Find the constant of proportionality \(k\) and write the equation of the relationship. b) Use calculation to determine whether \(Q(7, 24.5)\) also lies on the graph. c) A student claims, “If I double the \(x\)-value, the \(y\)-value always doubles.” Is the student correct? Justify your answer generally using \(y=kx\).

Hints

- Substitute the known point into \(y=kx\). - Test a point by checking whether its coordinates satisfy the equation. - Consider what happens to a product when one factor is doubled. - Try a numerical example before writing a general algebraic explanation.

Solution

1. Substitute \((4,14)\) into \(y=kx\): \(14=4k\), so \(k=3.5\) and the equation is \(y=3.5x\). 2. For \(x=7\), the equation gives \(y=3.5\cdot7=24.5\), so \(Q\) lies on the graph. 3. Replacing \(x\) by \(2x\) gives \(y_{\text{new}}=k(2x)=2(kx)=2y\). Therefore, doubling \(x\) doubles \(y\) in any proportional relationship of the form \(y=kx\).

Answer

a) \(k=3.5\), so \(y=3.5x\). b) Yes, because \(3.5\cdot7=24.5\). c) Yes. Since \(k(2x)=2(kx)\), doubling the input doubles the output.
5119668
A graph represents a cyclist riding at a constant speed. The x-axis shows time in minutes, and the y-axis shows distance in miles. The point \(P(15, 3)\) lies on the graph. a) Explain what \(P(15, 3)\) means in context. b) Find the constant of proportionality \(k\) and explain what it tells you about the cyclist. c) How far does the cyclist travel in \(40\) minutes? d) How long does the cyclist take to travel \(18\) miles?

Hints

- What does each coordinate of an ordered pair represent on this graph? - Divide distance by time to find the rate per minute. - Use the distance per minute for a longer time. - Divide the target distance by the distance traveled per minute.

Solution

1. The point \((15,3)\) means the cyclist travels \(3\) miles in \(15\) minutes. 2. The constant of proportionality is \(k=3\div15=0.2\) mile per minute, which is the cyclist's speed. 3. In \(40\) minutes, the distance is \(0.2\cdot40=8\) miles. 4. Traveling \(18\) miles takes \(18\div0.2=90\) minutes, or \(1\) hour \(30\) minutes.

Answer

a) The cyclist travels \(3\) miles in \(15\) minutes. b) \(k=0.2\) mile per minute. c) \(8\) miles. d) \(90\) minutes, or \(1\) hour \(30\) minutes.
5120568
Two mobile-data plans charge in proportion to the amount of data used. Plan A: \(\$1.50\) for \(500\,\text{MB}\). Plan B: \(\$2.40\) for \(1000\,\text{MB}\). a) Which plan has the lower cost per megabyte? Justify your answer with calculations. b) If cost is graphed against data used, with data on the x-axis, which plan's graph is steeper? Explain the connection between cost per megabyte and slope.

Hints

- Divide each price by the matching number of megabytes. - What does a greater slope mean in a cost graph? - Compare the costs for the same amount of data.

Solution

1. Plan A costs \(1.50\div500=\$0.003\) per megabyte. Plan B costs \(2.40\div1000=\$0.0024\) per megabyte. 2. Since \(0.0024<0.003\), Plan B has the lower cost per megabyte. 3. In a cost-versus-data graph, slope equals cost per megabyte. Plan A's graph is steeper because \(0.003>0.0024\).

Answer

a) Plan B is cheaper at \(\$0.0024\) per megabyte; Plan A costs \(\$0.003\) per megabyte. b) Plan A's graph is steeper because its cost per megabyte is greater.
5120608
Two pumps fill a water tank. Pump A moves \(1200\) gallons in \(4\) hours. Pump B moves \(1500\) gallons in \(6\) hours. For each pump, time and volume form a proportional relationship. a) Find each pump's rate in gallons per hour. b) When both pumps run together, their rates add. How much water enters the tank in \(5\) hours? c) Describe the graphs for Pump A, Pump B, and their combined rate, with time on the x-axis and volume on the y-axis. Compare the steepness of the three lines and explain.

Hints

- Find how much each pump moves in one hour. - Add the rates when both pumps work at the same time. - How does a rate in gallons per hour relate to graph slope?

Solution

1. Pump A's rate is \(1200\div4=300\) gallons per hour. Pump B's rate is \(1500\div6=250\) gallons per hour. 2. Their combined rate is \(300+250=550\) gallons per hour. In \(5\) hours, they move \(5\cdot550=2750\) gallons. 3. The three graphs are lines through \((0,0)\): \(V_A=300t\), \(V_B=250t\), and \(V_C=550t\). 4. Slope represents pumping rate, so the combined graph is steepest, followed by Pump A and then Pump B because \(550>300>250\).

Answer

a) Pump A: \(300\) gallons per hour. Pump B: \(250\) gallons per hour. b) \(2750\) gallons. c) The graphs are the lines \(V_A=300t\), \(V_B=250t\), and \(V_C=550t\), all through the origin. The combined graph is steepest, Pump A is next, and Pump B is least steep.
5131138
Three points have coordinates \(P_1(1.2, 3.6)\), \(P_2(2.5, 7.5)\), and \(P_3(5, 15.5)\). Determine whether all three points can lie on the same line through the origin. Justify your answer by testing proportionality.

Hints

- What distinguishes a line through the origin from other lines? - What ratio is constant for points satisfying \(y=kx\)? - Test the ratio for every point.

Solution

1. Points on a line through the origin satisfy \(y=kx\), so \(y\div x\) must be constant. 2. For \(P_1\), \(3.6\div 1.2=3\). 3. For \(P_2\), \(7.5\div 2.5=3\). 4. For \(P_3\), \(15.5\div 5=3.1\). 5. Since the third ratio differs, the points do not all lie on one line through the origin.

Answer

No. The first two ratios \(y\div x\) equal \(3\), but the third equals \(3.1\).
5131198
A line through the origin passes through \(P(4.5, 1.8)\). a) Find the equation of the line. b) Determine whether \(Q(7, 2.8)\) also lies on the line. c) A point \(R(a, 4.4)\) lies on the line. Find \(a\).

Hints

- What form does the equation of a line through the origin have? - Use the known point to find the slope. - Test a point by substituting its coordinates. - Solve the equation backward when the y-coordinate is known.

Solution

1. A line through the origin has equation \(y=kx\). Using \(P\), \(k=1.8\div4.5=0.4\), so the equation is \(y=0.4x\). 2. For \(x=7\), the equation gives \(y=0.4\cdot7=2.8\), so \(Q\) lies on the line. 3. For \(R(a,4.4)\), solve \(4.4=0.4a\). Thus \(a=4.4\div0.4=11\).

Answer

a) \(y=0.4x\). b) Yes, because \(0.4\cdot7=2.8\). c) \(a=11\).
5131408
Two runners, Anna and Ben, train for a charity run at constant speeds. - Anna runs \(200\) meters in \(40\) seconds. - Ben runs \(300\) meters in \(75\) seconds. a) Find each runner's speed in meters per second. Interpret each speed as the constant of proportionality for time \(\to\) distance. b) Who runs faster? c) If both motions are graphed with time on the x-axis and distance on the y-axis, whose graph lies above the other for \(t>0\)? Justify your answer without drawing.

Hints

- Find how far each runner travels in one second. - What does slope represent on a distance-time graph? - For two lines through the origin, which is higher when its slope is greater?

Solution

1. Anna's speed is \(200\div40=5\,\text{m/s}\). Ben's speed is \(300\div75=4\,\text{m/s}\). Their equations are \(s_A(t)=5t\) and \(s_B(t)=4t\). 2. Anna is faster because \(5>4\). 3. Anna's graph lies above Ben's for \(t>0\) because it has the greater slope and therefore gives a greater distance at every positive time.

Answer

a) Anna: \(5\,\text{m/s}\). Ben: \(4\,\text{m/s}\). b) Anna. c) Anna's graph lies above Ben's because its slope is greater.
5227008
Two liquids, A and B, have proportional relationships between volume \(V\), in \(\text{cm}^3\), and mass \(m\), in grams. For Liquid A, \(20\,\text{cm}^3\) has a mass of \(16\,\text{g}\). For Liquid B, \(15\,\text{cm}^3\) has a mass of \(13.5\,\text{g}\). 1) Find the density of each liquid in \(\text{g/cm}^3\). 2) If both relationships are graphed with volume on the x-axis and mass on the y-axis, which line is steeper? Explain using the densities. 3) Find the mass of \(50\,\text{cm}^3\) of Liquid B. 4) Find the volume of Liquid A that has a mass of \(64\,\text{g}\).

Hints

- What does mass divided by volume represent? - How is the constant of proportionality related to the slope? - Use the density to find mass from volume. - Rearrange the mass equation to solve for volume.

Solution

1. For Liquid A, \(\rho_A=16\div20=0.8\,\text{g/cm}^3\). For Liquid B, \(\rho_B=13.5\div15=0.9\,\text{g/cm}^3\). 2. Density is the slope of each mass-versus-volume line. Since \(0.9>0.8\), the line for Liquid B is steeper. 3. For Liquid B, \(m=0.9\cdot50=45\,\text{g}\). 4. For Liquid A, solve \(64=0.8V\). Thus \(V=64\div0.8=80\,\text{cm}^3\).

Answer

1) \(\rho_A=0.8\,\text{g/cm}^3\); \(\rho_B=0.9\,\text{g/cm}^3\) 2) Liquid B has the steeper line because \(0.9>0.8\). 3) \(45\,\text{g}\) 4) \(80\,\text{cm}^3\)
5239598
At a farmers market, the price \(P\), in dollars, of organic apples is proportional to the weight \(w\), in pounds. A \(2.5\)-pound bag costs \(\$6.25\). 1. Find the price per pound and write an equation for price as a function of weight. 2. Make a value table for \(1\), \(2\), \(3\), and \(4\) pounds. 3. The graph below shows the proportional relationship using \(1\) grid unit per pound on the x-axis and \(2\) dollars per grid unit on the y-axis. Name at least three points on the line. 4. Use the graph's linear pattern to estimate the cost of \(3.5\) pounds and the weight that costs \(\$5.00\). Then verify both values by calculation.
Figure for problem 523959

Hints

- First find the price of one pound. - A proportional relationship graphs as a line through the origin. - Multiply each weight by the unit price to complete the table. - Use horizontal and vertical guide lines when reading values from the graph.

Solution

1. The unit price is \(6.25\div2.5=\$2.50\) per pound, so \(P(w)=2.5w\). 2. The prices for \(1\), \(2\), \(3\), and \(4\) pounds are \(\$2.50\), \(\$5.00\), \(\$7.50\), and \(\$10.00\), respectively. 3. Three points on the displayed line are \((0,0)\), \((2,5)\), and \((4,10)\). 4. From the graph, \(3.5\) pounds costs about \(\$8.75\), and \(\$5.00\) buys about \(2\) pounds. Calculation verifies these values: \(2.5\cdot3.5=8.75\) and \(5.00\div2.5=2\).

Answer

1. \(\$2.50\) per pound; \(P(w)=2.5w\). 2. <table> <tr><th>Weight</th><td>\(1\,\text{lb}\)</td><td>\(2\,\text{lb}\)</td><td>\(3\,\text{lb}\)</td><td>\(4\,\text{lb}\)</td></tr> <tr><th>Price</th><td>\(\$2.50\)</td><td>\(\$5.00\)</td><td>\(\$7.50\)</td><td>\(\$10.00\)</td></tr> </table> 3. One valid set is \((0,0)\), \((2,5)\), and \((4,10)\). 4. The graph gives approximately \(\$8.75\) for \(3.5\) pounds and approximately \(2\) pounds for \(\$5.00\); calculation verifies both values exactly.
5239608
Two candles burn at constant rates. The height lost \(h\), in centimeters, is proportional to burning time \(t\), in hours. - Candle A loses \(3\,\text{cm}\) in \(2\) hours. - Candle B loses \(3\,\text{cm}\) in \(3\) hours. 1. Write equations \(h_A(t)\) and \(h_B(t)\) for the height lost by each candle. 2. Which candle burns down faster? Explain using the slopes. 3. Find the height lost by each candle after \(4.5\) hours. 4. After how many hours is the combined height loss \(10\,\text{cm}\)?

Hints

- A slope gives the height lost in one hour. - The candle with the greater height loss per hour burns down faster. - Add the two hourly rates to model their combined height loss.

Solution

1. Candle A's rate is \(3\div2=1.5\,\text{cm/h}\), and Candle B's rate is \(3\div3=1\,\text{cm/h}\). Thus \(h_A(t)=1.5t\) and \(h_B(t)=t\). 2. Candle A burns down faster because its slope is greater: \(1.5>1\). 3. After \(4.5\) hours, \(h_A(4.5)=1.5\cdot4.5=6.75\,\text{cm}\), and \(h_B(4.5)=4.5\,\text{cm}\). 4. The combined height loss is \(1.5t+t=2.5t\). Solving \(2.5t=10\) gives \(t=4\) hours.

Answer

1. \(h_A(t)=1.5t\); \(h_B(t)=t\) 2. Candle A burns down faster because \(1.5>1\). 3. Candle A: \(6.75\,\text{cm}\); Candle B: \(4.5\,\text{cm}\) 4. \(4\) hours
5241698
A farm stand sells blueberries for \(\$1.20\) per \(0.1\) pound. 1) Write an equation for the price \(y\), in dollars, of \(x\) pounds of blueberries. 2) A second stand charges \(\$10.00\) per pound. Describe how the graphs of the two offers would differ. 3) Explain mathematically why the first relationship is proportional. 4) Find the price of \(0.35\) pound of blueberries at the first stand.

Hints

- First find the price of one full pound. - What does slope represent in this context? - What happens to price when weight is multiplied by a factor? - Use the weight unit in the equation consistently.

Solution

1. The price per pound is \(\$1.20\cdot10=\$12.00\), so \(y=12x\). 2. The first graph has slope \(12\), and the second has slope \(10\). The second graph is less steep because its unit price is lower. 3. The ratio \(\frac{y}{x}\) is always \(12\), so the first relationship is proportional. 4. For \(0.35\) pound, \(y=12\cdot0.35=\$4.20\).

Answer

1) \(y=12x\) 2) The second stand's graph is less steep because its slope is \(10\) rather than \(12\). 3) The ratio of price to weight is constant. 4) \(\$4.20\)
5241708
A cyclist travels at a constant speed. After \(15\) minutes, the cyclist has traveled exactly \(4\) miles. 1) Find the constant of proportionality for time \(t\), in hours, and distance \(d\), in miles. What does it mean? 2) How would the graph change if the cyclist doubled the speed? 3) Find the distance traveled in \(1.5\) hours. 4) Find the number of minutes needed to travel \(10\) miles.

Hints

- Convert minutes to hours before finding miles per hour. - What happens to slope when a rate is doubled? - Rearrange the proportional equation to solve for the unknown. - Convert a decimal number of hours back to minutes.

Solution

1. Convert \(15\) minutes to \(0.25\) hour. The constant is \(k=4\div0.25=16\,\text{mi/h}\), which is the cyclist's speed. 2. Doubling the speed doubles the slope, so the graph would be steeper. 3. In \(1.5\) hours, \(d=16\cdot1.5=24\) miles. 4. Traveling \(10\) miles takes \(10\div16=0.625\) hour. Converting gives \(0.625\cdot60=37.5\) minutes.

Answer

1) \(k=16\); it represents a speed of \(16\,\text{mi/h}\). 2) The graph would have twice the slope and would be steeper. 3) \(24\) miles 4) \(37.5\) minutes
5241718
Three proportional functions \(f\), \(g\), and \(h\) are represented by lines through the origin. The equations are \(f(x)=\frac{3}{4}x\) and \(g(x)=1.2x\). The graph of \(h\) passes through \((5, 2)\). a) Write the equation of \(h\). b) Order the functions from least steep to most steep. c) Determine whether \((8, 6)\) lies on the graph of \(f\).

Hints

- How can a point on a line through the origin be used to find its slope? - What does the coefficient of \(x\) tell you about steepness? - Substitute the point's x-coordinate into the equation.

Solution

1. Since \(h\) is proportional, substitute \((5,2)\) into \(y=kx\): \(2=5k\), so \(k=\frac{2}{5}=0.4\). Thus \(h(x)=0.4x\). 2. The slopes are \(0.4\), \(0.75\), and \(1.2\). Since \(0.4<0.75<1.2\), the order from least steep to most steep is \(h\), \(f\), \(g\). 3. \(f(8)=\frac{3}{4}\cdot8=6\), so \((8,6)\) lies on the graph of \(f\).

Answer

a) \(h(x)=0.4x\), or \(h(x)=\frac{2}{5}x\) b) \(h\), \(f\), \(g\) c) Yes, because \(f(8)=6\).
5241728
The graph shows a line \(p\) through the origin with a slope triangle. a) Write the equation of \(p\). b) Without calculating, explain what happens to \(y\) when \(x\) is tripled. c) The value \(y=10\) is outside the displayed graph. Use your equation from part a) to find the x-value that gives \(y=10\).
Figure for problem 524172

Hints

- Compare the vertical change with the horizontal change shown by the slope triangle. - What does proportional mean when an input is multiplied by a factor? - For part c), use the equation rather than extending the displayed graph.

Solution

1. The slope is rise divided by run: \(m=\frac{2}{3}\). Therefore, \(p(x)=\frac{2}{3}x\). 2. In a proportional relationship, multiplying \(x\) by \(3\) multiplies \(y\) by \(3\), so \(y\) also triples. 3. Use the equation to solve \(10=\frac{2}{3}x\). Then \(x=10\div\frac{2}{3}=10\cdot\frac{3}{2}=15\).

Answer

a) \(p(x)=\frac{2}{3}x\) b) The y-value triples. c) \(x=15\)
5241738
A line through the origin also passes through \((5, -2)\). 1. Write its equation in the form \(y=mx\). 2. Without graphing, identify the quadrants through which the line passes. Explain. 3. Give the coordinates of another point on the line, other than the origin.

Hints

- What is the equation form of a line through the origin? - Substitute the known point to determine the slope. - How does the sign of the slope affect the quadrants through which the line passes? - Use the equation to generate another point.

Solution

1. Substitute the point into \(y=mx\): \(-2=5m\), so \(m=-0.4\). Thus the equation is \(y=-0.4x\). 2. A negative slope gives positive \(y\) for negative \(x\) and negative \(y\) for positive \(x\), so the line passes through Quadrants II and IV. 3. For example, if \(x=10\), then \(y=-0.4\cdot10=-4\), so \((10, -4)\) is another point on the line.

Answer

1. \(y=-0.4x\) 2. Quadrants II and IV because the slope is negative. 3. For example, \((10, -4)\)
5241748
Three lines through the origin have equations \(f(x)=-1.5x\), \(g(x)=-\frac{1}{3}x\), and \(h(x)=-4x\). 1. Order the graphs from least steep to most steep. Justify your order using absolute values of the slopes. 2. A point \(R\) lies on \(f\) and has y-coordinate \(6\). Find its x-coordinate. 3. The constant of proportionality for \(h\) is multiplied by \(-1\). Describe how the new line compares with the original line.

Hints

- Steepness depends on the absolute value of the slope. - A point on a graph must satisfy its equation. - What changes when only the sign of a slope changes? - Which quadrants contain points whose coordinates have the same sign?

Solution

1. The slope magnitudes are \(\frac{1}{3}\), \(1.5\), and \(4\). Therefore, the order from least steep to most steep is \(g\), \(f\), \(h\). 2. Solve \(6=-1.5x\): \(x=6\div(-1.5)=-4\). 3. The new slope is \((-4)(-1)=4\). The line changes from decreasing to increasing and passes through Quadrants I and III instead of II and IV. Its steepness is unchanged.

Answer

1. \(g\), \(f\), \(h\), because \(\left|-\frac{1}{3}\right|<|-1.5|<|-4|\) 2. \(x=-4\) 3. The new line is \(y=4x\). It increases instead of decreases, but has the same steepness.
5244168
A pump fills an empty tank at a constant rate. After \(4\) minutes, the tank contains \(140\) gallons; after \(10\) minutes, it contains \(350\) gallons. a) Find an equation \(f(x)=kx\), where \(x\) is time in minutes and \(f(x)\) is volume in gallons. Explain the meaning of \(k\). b) A technician records another point, \((15, 520)\). Determine whether it lies on the graph and explain what the result means. c) In general, if \((x_Q, y_Q)\), with \(x_Q\ne0\), lies on this graph, what must \(\frac{y_Q}{x_Q}\) equal? What is true for a point with nonzero x-coordinate that does not lie on the graph?

Hints

- Find the slope of a line through the origin from a known point. - What does slope mean for a filling process? - Substitute the recorded x-value into the equation and compare outputs. - What ratio is constant in a proportional relationship?

Solution

1. Compute the constant ratios: \(140\div4=35\) and \(350\div10=35\). Thus \(f(x)=35x\), where \(35\) is the filling rate in gallons per minute. 2. Evaluate \(f(15)=35\cdot15=525\). Since \(525\ne520\), the recorded point is not on the graph; the measurement is inconsistent with the constant-rate model. 3. Every point on the graph with nonzero x-coordinate satisfies \(\frac{y_Q}{x_Q}=35\). A point with nonzero x-coordinate that is not on the graph has a different ratio.

Answer

a) \(f(x)=35x\); \(k=35\) is the rate in gallons per minute. b) No. \(f(15)=525\ne520\), so the measurement is inconsistent with the model. c) On the graph, \(\frac{y_Q}{x_Q}=35\). Off the graph, the ratio is not \(35\).
5245108
Two proportional linear functions \(f\) and \(g\) have graphs through the origin. The slope of \(f\) is \(-2\). The graph of \(g\) passes through \((4, -6)\). 1. Write equations for \(f\) and \(g\). 2. Which line decreases more steeply? Explain. 3. Determine whether \((-1.5, 3)\) lies on \(f\), on \(g\), or on neither graph.

Hints

- How do slope and direction relate for a line? - What does the absolute value of slope tell you about steepness? - Substitute the point's coordinates to test whether it satisfies each equation. - How can one point determine the slope of a line through the origin?

Solution

1. Since \(f\) has slope \(-2\), \(f(x)=-2x\). For \(g\), \(-6=4m\), so \(m=-1.5\) and \(g(x)=-1.5x\). 2. Since \(|-2|>|-1.5|\), \(f\) decreases more steeply. 3. Evaluate \(f(-1.5)=-2(-1.5)=3\), so the point lies on \(f\). Also, \(g(-1.5)=(-1.5)(-1.5)=2.25\ne3\), so it does not lie on \(g\).

Answer

1. \(f(x)=-2x\); \(g(x)=-1.5x\) 2. \(f\) decreases more steeply because \(|-2|>|-1.5|\). 3. The point lies on \(f\) only.
5245158
The proportional functions are \(f(x)=2.5x\), \(g(x)=-0.4x\), and \(h(x)=\frac{1}{4}x\). a) Which graph is steepest? Justify your answer using the slopes. b) Through which quadrants does the graph of \(g\) pass? Explain how the sign of the slope determines this. c) For each function, give one point \((x, y)\) with \(x>0\) that lies on its graph.

Hints

- What does \(|m|\) tell you about a line? - How do y-values behave for positive x-values when the slope is negative? - Choose an x-value and use each equation to find y.

Solution

1. Steepness is determined by the absolute value of slope. Since \(|2.5|>|-0.4|\) and \(|2.5|>\left|\frac{1}{4}\right|\), \(f\) is steepest. 2. The slope of \(g\) is negative, so for positive \(x\), \(y\) is negative, and for negative \(x\), \(y\) is positive. Thus the line passes through Quadrants II and IV. 3. Examples are \((2, 5)\) for \(f\), \((5, -2)\) for \(g\), and \((4, 1)\) for \(h\).

Answer

a) \(f\) is steepest because \(|2.5|\) is the greatest slope magnitude. b) Quadrants II and IV because the slope is negative. c) For example, \(f\): \((2, 5)\); \(g\): \((5, -2)\); \(h\): \((4, 1)\).
5280768
Two electricity plans have no monthly base fee, so cost \(y\), in dollars, is proportional to energy use \(x\), in kilowatt-hours. Plan A is modeled by \(y=0.40x\). For Plan B, \(100\,\text{kWh}\) costs \(\$35.00\). a) Write the equation for Plan B. b) Compare the costs for \(500\,\text{kWh}\). Which plan is less expensive, and by how much? c) Which graph is steeper? Explain using the constants of proportionality. d) Determine whether \((120, 48)\) lies on the graph of Plan A.

Hints

- Find the price of one kilowatt-hour for Plan B. - What does the coefficient of \(x\) tell you about the graph? - Substitute a point's x-coordinate into the equation to test it.

Solution

1. Plan B's unit rate is \(\$35.00\div100=\$0.35\) per kilowatt-hour, so its equation is \(y=0.35x\). 2. Plan A costs \(\$0.40\cdot500=\$200.00\), while Plan B costs \(\$0.35\cdot500=\$175.00\). Plan B is \(\$25.00\) less expensive. 3. Plan A has the steeper graph because its constant of proportionality, \(0.40\), is greater than Plan B's \(0.35\). 4. For Plan A, \(0.40\cdot120=48\), so \((120,48)\) lies on the graph.

Answer

a) \(y=0.35x\) b) Plan B is \(\$25.00\) less expensive: \(\$175.00\) compared with \(\$200.00\). c) Plan A is steeper because \(0.40>0.35\). d) Yes, because \(0.40\cdot120=48\).
5322178
Water is poured at a constant rate into a cylindrical container. The graph shows water height \(h\), in centimeters, as a function of time \(t\), in seconds. a) Use the graph to explain why the relationship is proportional. Find the constant of proportionality with units and explain its meaning. b) Read the water height after \(5\) seconds from the graph. c) Calculate when the water height reaches \(20\,\text{cm}\).
Figure for problem 532217

Hints

- Describe the graph's shape and starting point. - Use a clearly readable point to find height divided by time. - What physical quantity is measured in centimeters per second? - Use the equation based on the graph to solve for the final time.

Solution

1. The graph is a straight line through \((0,0)\), so the relationship is proportional. It contains \((5,8)\), so \(k=8\div5=1.6\,\text{cm/s}\). This means the water height rises \(1.6\,\text{cm}\) each second. 2. At \(t=5\), the graph shows \(h=8\,\text{cm}\). 3. Using \(h(t)=1.6t\), solve \(20=1.6t\): \(t=20\div1.6=12.5\) seconds.

Answer

a) The graph is a line through the origin. The constant is \(1.6\,\text{cm/s}\), meaning the water height rises \(1.6\,\text{cm}\) per second. b) \(8\,\text{cm}\) c) \(12.5\) seconds
5332688
Four graphs of functions in the form \(y = mx\) are shown. a) Match graphs \(p_1\) through \(p_4\) with the slopes \(2\), \(0.5\), \(-1\), and \(-0.25\). b) What happens to a graph as \(|m|\) gets closer to zero? c) What point do all graphs of this form share? What are these relationships called?
Figure for problem 533268

Hints

- Use rise over run for each labeled line. - Compare how sharply the lines rise or fall. - Inspect the graph for any point that appears on every line. - Think about a line whose slope is very close to zero.

Solution

1. Graph \(p_1\) passes through \((1, 2)\), so its slope is \(2\). Graph \(p_2\) passes through \((2, 1)\), so its slope is \(0.5\). Graph \(p_3\) passes through \((1, -1)\), so its slope is \(-1\). Graph \(p_4\) passes through \((4, -1)\), so its slope is \(-0.25\). 2. As \(|m|\) approaches zero, the line becomes less steep and gets closer to the x-axis. 3. Every graph passes through \((0, 0)\). These are proportional relationships.

Answer

a) \(p_1\): \(2\); \(p_2\): \(0.5\); \(p_3\): \(-1\); \(p_4\): \(-0.25\) b) The graph becomes less steep and closer to the x-axis. c) They all contain \((0, 0)\); they are proportional relationships.
5349408
Line \(g\) passes through the origin and the labeled point \(A\). 1. State the coordinates of \(A\). 2. Write the equation of \(g\) in the form \(y=mx\). 3. The y-value \(-7.5\) is outside the displayed graph. Use your equation from part 2 to find the x-coordinate of a point \((x, -7.5)\) on the line.
Figure for problem 534940

Hints

- Pay attention to signs when reading coordinates. - For a line through the origin, slope can be found using \(\frac{y}{x}\). - For part 3, use the equation rather than extending the displayed graph.

Solution

1. The graph shows \(A=(-2,3)\). 2. The slope is \(m=3\div(-2)=-1.5\), so the equation is \(y=-1.5x\). 3. Use the equation to solve \(-7.5=-1.5x\). Then \(x=(-7.5)\div(-1.5)=5\).

Answer

1. \((-2, 3)\) 2. \(y=-1.5x\) 3. \(x=5\)
5546588
The graph shows the amount of water Tank A receives over time. Tank B receives \(21\,\text{gal}\) in \(7\) minutes at a constant rate. Which tank has the greater unit rate of water added? State each rate and explain how the graph represents Tank A's rate.
Figure for problem 554658

Hints

- On a proportional graph, what does the slope represent? - Read a convenient point on Tank A's line and compare rise to run from the origin. - Convert Tank B's information to a per-minute rate before comparing.

Solution

1. From the graph, Tank A's line passes through the origin and points such as \((1,4)\), so its unit rate is \(4\,\text{gal/min}\). 2. Tank B's unit rate is \(\frac{21}{7}=3\,\text{gal/min}\). 3. Tank A has the greater unit rate. 4. On Tank A's graph, the slope is the unit rate: the water amount rises \(4\) gallons for each \(1\)-minute increase.

Answer

Tank A: \(4\,\text{gal/min}\); Tank B: \(3\,\text{gal/min}\). Tank A has the greater unit rate. The slope of Tank A's graph represents its unit rate.
5245168
Consider all lines through the origin in the form \(y=mx\). a) How does a line with \(0<m<1\) compare with the line \(y=x\)? b) The line \(p\) has equation \(y=4x\). Reflecting it across the y-axis creates a line \(q\). Write the equation of \(q\) and explain. c) Give an equation of a line that passes through Quadrants I and III and is less steep than \(y=0.5x\).

Hints

- Compare the slope with the slope of \(y=x\). - What happens to x-coordinates under reflection across the y-axis? - What sign must the slope have to pass through Quadrants I and III? - Less steep means the slope magnitude is closer to zero.

Solution

1. The line \(y=x\) has slope \(1\). A slope between \(0\) and \(1\) is positive but smaller, so the line increases more gradually and lies closer to the x-axis. 2. Reflection across the y-axis replaces \(x\) with \(-x\): \(y=4(-x)=-4x\). Thus \(q\) has equation \(y=-4x\). 3. To pass through Quadrants I and III, the slope must be positive. To be less steep than \(y=0.5x\), it must satisfy \(0<m<0.5\). One possible equation is \(y=0.2x\).

Answer

a) It is less steep than \(y=x\) and lies closer to the x-axis. b) \(y=-4x\) c) For example, \(y=0.2x\)

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