Free math worksheets

- The initial value of a function is its output when the input is \(0\). - Read the \(x=0\) row entry for the table and substitute \(0\) into the equation.

1. The initial value is the output when \(x=0\). 2. From the table, \(f(0)=3\). 3. From the equation, \(g(0)=2(0)+1=1\). 4. Because \(3>1\), function \(f\) has the greater initial value.

\(f\) has the greater initial value. The initial values are \(f(0)=3\) and \(g(0)=1\).

- 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.

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.

\(A(v)=45\) 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.

- The y-intercept is the value of \(y\) when \(x = 0\). Where does each graph cross the y-axis? - A negative slope means a line falls from left to right. Compare the steepness of the two decreasing lines. - A linear function with no constant term passes through \((0, 0)\).

1. For \(a(x) = -\frac{2}{3}x + 1\), the y-intercept is \(1\) and the slope is negative with magnitude less than \(1\). Graph p has those features, so p represents \(a\). 2. For \(b(x) = \frac{3}{2}x - 2\), the y-intercept is \(-2\) and the slope is positive. Graph q has those features, so q represents \(b\). 3. For \(c(x) = -2x\), the y-intercept is \(0\), so the line passes through the origin, and its negative slope is steeper than the slope of \(a\). Graph r has those features, so r represents \(c\).

p: \(a(x) = -\frac{2}{3}x + 1\); q: \(b(x) = \frac{3}{2}x - 2\); r: \(c(x) = -2x\)

- Identify the y-intercept in each equation and compare it with the graph. - Check whether each line rises or falls from left to right. - Compare the steepness of each graph with the coefficient of \(x\).

1. Graph \(g_1\) crosses the y-axis at \(3\) and falls \(2\) units for every \(1\) unit to the right, so its slope is \(-2\). Thus \(g_1\) matches equation 1, \(y = -2x + 3\). 2. Graph \(g_2\) crosses the y-axis at \(1\) and rises \(1\) unit for every \(1\) unit to the right, so its slope is \(1\). Thus \(g_2\) matches equation 3, \(y = x + 1\). 3. Graph \(g_3\) crosses the y-axis at \(-2\) and rises \(1\) unit for every \(3\) units to the right, so its slope is \(\frac{1}{3}\). Thus \(g_3\) matches equation 2, \(y = \frac{1}{3}x - 2\).

\(g_1\): 1) \(y = -2x + 3\); \(g_2\): 3) \(y = x + 1\); \(g_3\): 2) \(y = \frac{1}{3}x - 2\)

- Evaluate all three functions at the same input before comparing their outputs. - In the equation \(g(t)=w(t)+s(t)\), what must the solar output be when \(g(t)=w(t)\)? - For the function question, fix one input time and count how many outputs the rule for \(g\) can produce.

1. At noon, the graph gives \(w(12)=2\,\text{MW}\) and \(s(12)=3.6\,\text{MW}\). Therefore, \(g(12)=2+3.6=5.6\,\text{MW}\). 2. Since \(g(t)=w(t)+s(t)\), equality \(g(t)=w(t)\) occurs exactly when \(s(t)=0\). The solar graph is zero from midnight through \(6{:}00\) a.m. and from \(6{:}00\) p.m. through midnight. 3. For each time input \(t\), the functions \(w\) and \(s\) each give exactly one output. Adding those two outputs gives exactly one value of \(g(t)\), so \(g\) is also a function.

a) \(w(12)=2\,\text{MW}\), \(s(12)=3.6\,\text{MW}\), and \(g(12)=5.6\,\text{MW}\) b) Midnight–\(6{:}00\) a.m. and \(6{:}00\) p.m.–midnight, because those are the times when \(s(t)=0\). c) Each input \(t\) produces one wind output and one solar output, so their sum is one unique output \(g(t)\).

- Compare the y-intercept in the equation with where each graph crosses the y-axis. - Does a positive slope make a graph rise or fall from left to right? - Use a slope triangle: if you move one unit to the right, how far should the line move vertically? - You can also substitute an x-value into the equation and check whether the resulting point lies on a candidate graph.

1. The y-intercept is \(1\), so the correct graph must cross the y-axis at \(1\). This eliminates graph c). 2. The slope is positive, so the line must rise from left to right. This eliminates graph d). 3. A slope of \(0.5 = \frac{1}{2}\) means the line rises \(1\) unit for every \(2\) units to the right. Graph b) has this slope, while graph a) is steeper. 4. Therefore, graph b) represents the function.

b)

- First find where the graph crosses the y-axis. Which parameter is this in \(y = mx + b\)? - Use two points that lie exactly on grid intersections to make a slope triangle. - Check the sign of the slope. Does the line rise or fall from left to right? - Compare the slope you found with the slopes in the answer choices.

1. The line crosses the y-axis at \((0, 3)\), so the y-intercept is \(3\). This eliminates D. 2. The line falls from left to right, so its slope is negative. This eliminates B. 3. Use \((0, 3)\) and \((2, 0)\): \(m = \frac{0 - 3}{2 - 0} = -\frac{3}{2} = -1.5\). 4. The equation is \(g(x) = -1.5x + 3\), which is choice A.

A: \(g(x) = -1.5x + 3\)

- First identify where the line crosses the y-axis. - Use two grid points to calculate rise over run. - Check whether the line rises or falls from left to right to determine the sign of the slope.

1. The line crosses the y-axis at \(2\), so its y-intercept is \(2\). All four choices have this y-intercept. 2. Use \((0, 2)\) and \((4, 0)\): \(m = \frac{0 - 2}{4 - 0} = -\frac{1}{2}\). 3. The equation is \(y = -\frac{1}{2}x + 2\), which is choice D.

D) \(y = -\frac{1}{2}x + 2\)

- On the graph, use two clear points to find change in output divided by change in input. - In the table, compare output changes with the corresponding input changes. - The initial value is the output at \(x=0\) in each representation.

1. From the graph, \(f\) passes through \((0,2)\) and \((4,10)\), so its rate of change is \(\frac{10-2}{4-0}=2\). Its initial value is \(2\). 2. From the table, \(g\) changes by \(4\) when \(x\) changes by \(2\), so its rate of change is \(\frac{4}{2}=2\). Its initial value is \(g(0)=5\). 3. The functions have the same rate of change, but \(g\) has the greater initial value by \(5-2=3\).

The rates are equal at \(2\) output units per input unit. The initial values are \(2\) for \(f\) and \(5\) for \(g\), so \(g\) has the greater initial value by \(3\).

- Test consecutive whole-number movie counts near the flat monthly price. - Compare each result with \(\$9.99\). - At what movie count does Service B first exceed the flat price?

1. For \(4\) movies, Service B costs \(\$1.50 \cdot 4=\$6.00\). For \(8\) movies, it costs \(\$1.50 \cdot 8=\$12.00\). 2. At \(6\) movies, Service B costs \(\$9.00\), which is less than \(\$9.99\). At \(7\) movies, Service B costs \(\$10.50\), which is more than \(\$9.99\). Therefore, Service A is less expensive starting at \(7\) movies.

a) \(4\) movies cost \(\$6.00\); \(8\) movies cost \(\$12.00\). b) Service A is less expensive starting at \(7\) movies per month.

- Evaluate each formula separately. - Find an absolute difference when comparing the two outputs. - Set the expressions equal to locate the boundary value. - Compare their rates of change on either side of that value.

1. At \(30\,^{\circ}\text{C}\), the estimate gives \(2 \cdot 30 + 30 = 90\), while the exact formula gives \(1.8 \cdot 30 + 32 = 86\). 2. At \(25\,^{\circ}\text{C}\), the estimate gives \(2 \cdot 25 + 30 = 80\), and the exact formula gives \(1.8 \cdot 25 + 32 = 77\). The difference is \(3\,^{\circ}\text{F}\). 3. Set the two expressions equal to find their intersection: \(2C + 30 = 1.8C + 32\). 4. Subtract \(1.8C + 30\): \(0.2C = 2\), so \(C = 10\). 5. At \(10\,^{\circ}\text{C}\), both formulas give \(50\,^{\circ}\text{F}\). Since the estimate has the greater rate of change, it is greater only when \(C > 10\). The claim is false because it includes \(C = 10\).

a) Estimate: \(90\,^{\circ}\text{F}\); exact value: \(86\,^{\circ}\text{F}\) b) The estimate is \(3\,^{\circ}\text{F}\) higher. c) The claim is false. The values are equal at \(10\,^{\circ}\text{C}\), and the estimate is higher only for \(C > 10\).

- Interpret the coefficient in Paint B's equation. - Find Paint A's coverage for one gallon. - Use the better unit rate to find the gallons needed for the target area.

1. Paint A covers \(425\div 5=85\) square feet per gallon. 2. In \(y=90x\), the coefficient means Paint B covers \(90\) square feet per gallon. 3. Since \(90>85\), Paint B has greater coverage. 4. The amount needed is \(630\div 90=7\) gallons.

a) Paint B; it covers \(90\,\text{ft}^2\) per gallon, compared with \(85\,\text{ft}^2\) for Paint A. b) \(7\) gallons.

- Simplify each expression separately. - In \(f\), check what happens to the two \(x^2\)-terms. - Equivalent expressions have the same simplified form for every value of the variable.

1. Distribute in \(f\): \(\frac{1}{2}x(4-2x)=2x-x^2\), so \(f(x)=2x-x^2+x^2+3=2x+3\). 2. Distribute in \(g\): \(g(x)=2x+3\). 3. Both expressions simplify to \(2x+3\), so they are equivalent for every value of \(x\).

\(f(x)=2x+3\) and \(g(x)=2x+3\). Yes, they are equivalent for every value of \(x\).

- How can you find the vertical change when the slope and horizontal change are known? - To compare a fraction and a decimal, rewrite them in the same form. - Which value tells you how steep a line is: the height of one chosen slope triangle or the slope itself?

1. For line \(k\), \(\Delta y = \frac{2}{3} \cdot 3 = 2\). 2. For line \(l\), \(\Delta y = 0.6 \cdot 5 = 3\). 3. The slopes are \(\frac{2}{3}\) and \(0.6\). Since \(\frac{2}{3} \approx 0.667 > 0.6\), line \(k\) is steeper. The larger height of the second slope triangle does not make \(l\) steeper because the two triangles have different horizontal widths.

a) Line \(k\): 2 units; line \(l\): 3 units b) Line \(k\) is steeper because \(\frac{2}{3} > 0.6\).

- How do you find slope from two points in a table? - What does slope tell you when comparing two linear functions? - At an intersection point, what must be true about the two function values?

1. From the table, the slope of \(f\) is \(m_f = \frac{8 - 2}{3 - 1} = \frac{6}{2} = 3\). 2. Use \((1, 2)\) in \(f(x) = 3x + b\): \(2 = 3 \cdot 1 + b\), so \(b = -1\). Thus \(f(x) = 3x - 1\). 3. The slopes are \(3\) for \(f\) and \(4\) for \(g\), so \(g\) has the greater slope. 4. Set the functions equal: \(3x - 1 = 4x - 1\). Then \(x = 0\). 5. Substituting gives \(y = -1\), so the intersection point is \((0, -1)\).

a) \(f(x) = 3x - 1\) b) \(g\) has the greater slope. c) \((0, -1)\)

- How are speed, time, and distance related? - What does the coefficient in \(y=kx\) represent? - Find Tim's rate from one distance-time pair.

1. Tim's speed is \(12.5\div 5=2.5\) meters per second; the second pair confirms \(30\div 12=2.5\). 2. Sarah's speed is the coefficient \(2.4\) meters per second. Since \(2.5>2.4\), Tim is faster. 3. Tim runs \(2.5\cdot 20=50\) meters. 4. For Sarah, solve \(60=2.4t\): \(t=60\div 2.4=25\) seconds.

a) Tim is faster: \(2.5\,\text{m/s}\) compared with Sarah's \(2.4\,\text{m/s}\). b) \(50\) meters. c) \(25\) seconds.

- How do you find slope from two points? - Translate the verbal description of Liquid 2 into a rate and an initial value. - Which number in \(mt + b\) represents the rate, and which represents the starting value? - Substitute \(t = 10\) into all three functions for the final comparison.

1. Liquid 1 is already given: \(T_1(t) = 4t + 68\). 2. Liquid 2 has initial value \(65\) and rate \(5\), so \(T_2(t) = 5t + 65\). 3. For Liquid 3, the slope is \(m = \frac{83 - 74}{5 - 2} = 3\). Using \((2, 74)\), \(74 = 3 \cdot 2 + b\), so \(b = 68\). Thus \(T_3(t) = 3t + 68\). 4. The initial temperatures are \(68^\circ\text{F}\), \(65^\circ\text{F}\), and \(68^\circ\text{F}\), so Liquids 1 and 3 tie for the highest initial temperature. 5. The rates are \(4\), \(5\), and \(3\) degrees Fahrenheit per minute, so Liquid 2 warms fastest. 6. At \(t = 10\), the temperatures are \(T_1(10) = 108^\circ\text{F}\), \(T_2(10) = 115^\circ\text{F}\), and \(T_3(10) = 98^\circ\text{F}\).

a) \(T_1(t) = 4t + 68\); \(T_2(t) = 5t + 65\); \(T_3(t) = 3t + 68\) b) Liquids 1 and 3, at \(68^\circ\text{F}\) c) Liquid 2, at \(5^\circ\text{F}\) per minute d) After \(10\) minutes: Liquid 1 is \(108^\circ\text{F}\), Liquid 2 is \(115^\circ\text{F}\), and Liquid 3 is \(98^\circ\text{F}\).

- Rewrite the slopes in the same form before comparing them. - Where is the y-intercept visible in slope-intercept form? - What happens to the slope when it is doubled?

1. The slopes are \(\frac{2}{3}\) and \(0.6\). Since \(\frac{2}{3} \approx 0.667 > 0.6\), \(f\) is steeper. 2. Both functions have y-intercept \(4\), so both graphs pass through \((0, 4)\). 3. Twice the slope of \(f\) is \(2 \cdot \frac{2}{3} = \frac{4}{3}\). Keeping the y-intercept \(4\) gives \(h(x) = \frac{4}{3}x + 4\).

a) \(f\) is steeper because \(\frac{2}{3} > 0.6\). b) Both y-intercepts are \((0, 4)\). c) \(h(x) = \frac{4}{3}x + 4\)

- What does the coefficient of \(x\) represent in \(y=4.80x\)? - Find Variety A's price for one pound. - You can compare the total prices or first find the difference in price per pound.

1. The unit price for Variety A is \(13.50\div 3=\$4.50\) per pound. 2. From \(y=4.80x\), the unit price for Variety B is \(\$4.80\) per pound. 3. Variety B is more expensive because \(4.80>4.50\). 4. The difference per pound is \(\$4.80-\$4.50=\$0.30\). For \(2.5\,\text{lb}\), the difference is \(0.30\cdot 2.5=\$0.75\).

a) Variety B costs more: \(\$4.80\) per pound compared with \(\$4.50\) per pound for Variety A. b) Variety B costs \(\$0.75\) more for \(2.5\,\text{lb}\).

- Use slope-intercept form to write \(g\). - Compare the absolute values of the slopes to compare steepness. - At an intersection, both functions have the same output for the same input.

1. From the given slope and y-intercept, \(g(x) = 0.5x - 1\). 2. Steepness is determined by the absolute value of the slope. Since \(|-2| = 2\) and \(|0.5| = 0.5\), the graph of \(f\) is steeper. 3. At the intersection, \(-2x + 4 = 0.5x - 1\). Thus, \(5 = 2.5x\), so \(x = 2\). Then \(y = 0\), giving \(S(2, 0)\).

a) \(g(x) = 0.5x - 1\) b) \(f\) is steeper because \(|-2| > |0.5|\). c) \(S(2, 0)\)

- Use the stated constant-rate condition to determine Panel A's energy production per hour. - Compare the two panels using energy produced per hour. - For part c, first determine how much energy Panel B produces in \(6\) hours.

1. Panel A's rate is \(1.2\div5=0.24\,\text{kWh/h}\), so \(E_A(t)=0.24t\). 2. In \(8\) hours, Panel A produces \(0.24\cdot8=1.92\,\text{kWh}\). 3. Panel B produces more per hour because \(0.25>0.24\). 4. Panel B produces \(0.25\cdot6=1.5\,\text{kWh}\) in \(6\) hours. Solve \(1.5=0.24t\): \(t=1.5\div0.24=6.25\) hours, or \(6\) hours \(15\) minutes.

a) \(1.92\,\text{kWh}\) b) Panel B, because \(0.25\,\text{kWh/h}>0.24\,\text{kWh/h}\) c) \(6.25\) hours, or \(6\) hours \(15\) minutes

- What does the coefficient of \(t\) tell you about Beta's rate? - How much does Alpha print in one hour? - Find each printer's time for the same volume. - Convert the difference in hours to minutes.

1. Alpha's rate is \(50 \div 2=25\,\text{cm}^3/\text{h}\). Beta's rate is \(22.5\,\text{cm}^3/\text{h}\). Since \(25>22.5\), Alpha is faster. 2. Alpha prints \(25\), \(50\), \(75\), and \(100\,\text{cm}^3\) in \(1\), \(2\), \(3\), and \(4\) hours. 3. Alpha needs \(180 \div 25=7.2\) hours. Beta needs \(180 \div 22.5=8\) hours. The difference is \(0.8\) hour, or \(0.8 \cdot 60=48\) minutes.

a) Alpha is faster: \(25\,\text{cm}^3/\text{h}\) compared with \(22.5\,\text{cm}^3/\text{h}\). b) <table> <tr><th>Time</th><td>\(1\,\text{h}\)</td><td>\(2\,\text{h}\)</td><td>\(3\,\text{h}\)</td><td>\(4\,\text{h}\)</td></tr> <tr><th>Volume</th><td>\(25\,\text{cm}^3\)</td><td>\(50\,\text{cm}^3\)</td><td>\(75\,\text{cm}^3\)</td><td>\(100\,\text{cm}^3\)</td></tr> </table> c) \(48\) minutes.

- Look at where each graph crosses the y-axis. - How does the y-intercept relate to the constant term in each equation? - What does the slope represent in this pricing context? - Substitute a rental time into each equation to compare costs. - Look at where the graphs intersect and which graph is lowest before and after that point.

1. Match the graphs by their y-intercepts. Graph a passes through \((0, 0)\), so it represents \(f_1(x) = 4x\). Graph b has y-intercept \(6\), so it represents \(f_2(x) = 2x + 6\). Graph c has y-intercept \(9\), so it represents \(f_3(x) = x + 9\). 2. At \(x = 2\), \(f_1\) costs \(4 \cdot 2 = 8\), \(f_2\) costs \(2 \cdot 2 + 6 = 10\), and \(f_3\) costs \(2 + 9 = 11\). Plan \(f_1\) is least expensive at \(\$8\), and Plan \(f_3\) is most expensive at \(\$11\), so the savings are \(\$3\). 3. The three plans have the same cost at \(x = 3\), where each costs \(\$12\). For \(0 \le x < 3\), Plan \(f_1\) has the lowest cost. At \(x = 3\), all three plans tie. For \(x > 3\), Plan \(f_3\) has the lowest cost.

a) a: \(f_1(x) = 4x\); b: \(f_2(x) = 2x + 6\); c: \(f_3(x) = x + 9\) b) Plan \(f_1\) costs \(\$8\) and is least expensive. The savings compared with Plan \(f_3\) at \(\$11\) are \(\$3\). c) Plan \(f_1\) for \(0 \le x < 3\); all three plans at \(x = 3\); Plan \(f_3\) for \(x > 3\).

- Apply the vertical line test to each graph. - A proportional linear function must pass through the origin. - For Equation (II), think about what y-values can correspond to one allowed x-value and what shape that creates.

1. Graph a) passes the vertical line test, so it is a function. Graph b) fails the vertical line test because many x-values correspond to two y-values. 2. Graph a) is linear but not proportional because it is a line that does not pass through the origin. 3. Graph a) matches Equation (I), \(y=\frac{1}{2}x+1\). Graph b) matches Equation (II), \(x=y^2-2\), which is a sideways parabola.

1) Graph a) is a function; Graph b) is not. 2) Graph a) is linear but not proportional. 3) Graph a) matches (I); Graph b) matches (II).

- First decide which displayed graph is a straight line through the origin, which is a straight line with a nonzero y-intercept, and which is curved. - A proportional linear function has a straight graph through the origin; another straight graph can still be linear without being proportional. - Test one clearly readable point from each graph in the candidate equations to confirm the match.

1. Graph a) is Equation (3), \(g(x)=-1.5x\). The point \((2, -3)\) lies on the graph because \(g(2)=-1.5\cdot2=-3\). The graph is a line through the origin, so it is proportional. 2. Graph b) is Equation (2), \(f(x)=2x+1\). The point \((0, 1)\) lies on the graph because \(f(0)=1\). The graph is a line that does not pass through the origin, so it is linear but not proportional. 3. Graph c) is Equation (1), \(h(x)=x^2\). The point \((2, 4)\) lies on the graph because \(h(2)=4\). The graph is curved, so it is nonlinear.

a) Equation (3); proportional; example point \((2, -3)\) b) Equation (2); linear but not proportional; example point \((0, 1)\) c) Equation (1); nonlinear; example point \((2, 4)\)

- Put the equation into slope-intercept form before looking at the choices. - In \(y=mx+b\), identify separately what \(m\) and \(b\) tell you about the graph. - A correct choice must match both the direction/steepness and the y-axis crossing.

1. Solve for \(y\): \(x-2y=4\) gives \(-2y=4-x\), so \(y=\frac{1}{2}x-2\). 2. The rate of change is \(m=\frac{1}{2}\), and the initial value is \(b=-2\). 3. Graph a) rises from left to right with slope \(\frac{1}{2}\) and crosses the y-axis at \(-2\). The other graphs have either the wrong slope sign or the wrong initial value.

a) \(y=\frac{1}{2}x-2\); \(m=\frac{1}{2}\), \(b=-2\) b) a)

- Read the graph's y-intercept and use two graph points to find the rate of change of \(f\). - The verbal description of \(g\) states both its starting output and its change per input unit. - For part b, evaluate both functions at the same input before comparing their outputs.

1. From the graph, \(f\) has initial value \(12\). Using \((0,12)\) and \((6,0)\), its rate of change is \(\frac{0-12}{6-0}=-2\). 2. The verbal description gives \(g\) an initial value of \(9\) and a rate of change of \(-1.5\). 3. Thus \(f\) starts \(3\) units higher and decreases faster because \(-2<-1.5\). 4. At \(x=4\), the graph gives \(f(4)=4\). For \(g\), \(g(4)=9-1.5\cdot4=3\). Therefore, \(f(4)>g(4)\).

a) \(f\): initial value \(12\), rate \(-2\); \(g\): initial value \(9\), rate \(-1.5\). Function \(f\) starts higher and decreases faster. b) \(f\), because \(f(4)=4\) and \(g(4)=3\).

- One matching input-output pair does not determine an entire function. - Find the constant rate in the table and compare it with the coefficient of \(x\) in the equation. - Test a different input that appears in the table.

1. From the table, \(f\) increases by \(6\) when \(x\) increases by \(2\), so its rate of change is \(3\). 2. From the equation, \(g\) has rate of change \(2\). 3. The functions happen to have the same output at \(x=2\), but equal output at one input does not make two functions identical. Their different rates show that they cannot agree for every input. 4. For example, at \(x=0\), \(f(0)=2\) while \(g(0)=4\).

Jordan's conclusion is incorrect. Function \(f\) has rate \(3\), while \(g\) has rate \(2\). They agree at \(x=2\) only at that input; for example, \(f(0)=2\) and \(g(0)=4\).