Free Math Worksheets

- Write a rule for each relationship. - Check whether equal increases in the input produce equal increases in the output. - Compare what happens when the input is doubled.

1. The taxi relationship is linear. It has a fixed initial value of \(\$3.50\) and a constant rate of \(\$1.80\) per mile, so \(C(x)=1.80x+3.50\). 2. The patio relationship is not linear. Its area is \(A(s)=s^2\). For example, doubling the side length multiplies the area by \(4\), not by \(2\).

1) Linear, because the cost increases at a constant rate from a fixed starting value. 2) Not linear, because \(A=s^2\).

- Identify each year as an input and its charging-port count as the corresponding output. - Choose a vertical scale large enough to include the greatest value. - Subtract the 2019 value from the 2020 value. - Compare the change from each year to the next.

1. The points are \((2017, 10700)\), \((2018, 15600)\), \((2019, 23900)\), \((2020, 39500)\), and \((2021, 52200)\). A vertical scale from \(0\) to about \(55{,}000\) is appropriate. 2. The increase from \(2019\) to \(2020\) is \(39{,}500-23{,}900=15{,}600\). 3. The yearly first differences are \(4900\), \(8300\), \(15{,}600\), and \(12{,}700\). Because these differences are not equal, the data do not show a constant yearly increase.

a) \((2017, 10700)\), \((2018, 15600)\), \((2019, 23900)\), \((2020, 39500)\), and \((2021, 52200)\); a vertical scale from \(0\) to about \(55{,}000\) b) \(15{,}600\) charging ports c) No. The first differences are not constant.

- Substitute each speed into the model or read the corresponding graph value. - Compare the second stopping distance with twice the first. - Divide the larger distance by the smaller to find the growth factor.

1. At \(30\,\text{mph}\), \(s(30)=30+\frac{30^2}{20}=30+45=75\,\text{ft}\). 2. At \(60\,\text{mph}\), \(s(60)=60+\frac{60^2}{20}=60+180=240\,\text{ft}\). 3. Twice the first distance would be \(150\,\text{ft}\), not \(240\,\text{ft}\). The ratio is \(\frac{240}{75}=3.2\), so doubling speed multiplies the model's stopping distance by \(3.2\), not by \(2\).

a) \(75\,\text{ft}\) at \(30\,\text{mph}\); \(240\,\text{ft}\) at \(60\,\text{mph}\) b) No. The ratio is \(3.2\).

- A linear function graphs as a line; a quadratic function graphs as a parabola. - Intersections represent equal heights. - Evaluate both functions at \(t=10\) and subtract.

1. The function \(h_1(t)=15t\) is linear, so it matches the straight line. The function \(h_2(t)=t^2\) is quadratic, so it matches the parabola. 2. The graphs intersect at \(t=0\) and \(t=15\). Substitution verifies both times: \(h_1(0)=h_2(0)=0\), and \(h_1(15)=15\cdot15=225\) while \(h_2(15)=15^2=225\). After the start, the heights are equal again at \(15\) minutes, when both are \(225\,\text{cm}\). 3. \(h_1(10)=15\cdot10=150\,\text{cm}\) and \(h_2(10)=10^2=100\,\text{cm}\). The difference is \(50\,\text{cm}\).

a) Tank 1: straight line; Tank 2: parabola b) \(t=0\) and \(t=15\) minutes; after the start, \(15\) minutes at \(225\,\text{cm}\) c) \(50\,\text{cm}\)

- Compare the height increase during each equal time interval. - With constant inflow, decide whether a wider container makes the water level rise faster or more slowly.

1. In container A, the height increases by exactly \(2.0\,\text{cm}\) every \(10\) seconds. With a constant inflow, this constant rate of height increase indicates a uniform cross section, like a cylinder. 2. The graph for A is a line through the origin. The graph for B increases but becomes less steep. 3. The successive height increases for B are \(3.5\), \(2.5\), \(1.5\), \(1.0\), and \(0.5\) centimeters. Because the same volume produces a smaller rise as the water gets higher, container B becomes wider toward the top.

a) Container A, because its height increases by a constant \(2.0\,\text{cm}\) every \(10\) seconds. b) A is linear; B is increasing and becomes less steep. c) Container B becomes wider toward the top because the height increases by progressively smaller amounts.

- Multiply each quantity by the applicable unit price. - Compare the total costs for \(9\) and \(11\) shirts directly. - A proportional relation must have one constant unit rate over its entire domain.

1. Ten shirts cost \(10(\$15.00)=\$150.00\). Eleven shirts cost \(11(\$12.00)=\$132.00\). Therefore, buying \(11\) shirts costs \(\$18.00\) less than buying \(10\). 2. Nine shirts cost \(9(\$15.00)=\$135.00\). Eleven shirts cost \(\$132.00\), so the customer would receive two additional shirts and spend \(\$3.00\) less. 3. For whole-number quantities, \(C(n)=15n\) for \(1\le n\le10\), and \(C(n)=12n\) for \(n\ge11\). The table is: <table><tr><td>Shirts</td><td>\(1\)</td><td>\(2\)</td><td>\(3\)</td><td>\(4\)</td><td>\(5\)</td><td>\(6\)</td><td>\(7\)</td><td>\(8\)</td><td>\(9\)</td><td>\(10\)</td><td>\(11\)</td><td>\(12\)</td><td>\(13\)</td><td>\(14\)</td><td>\(15\)</td></tr><tr><td>Total price</td><td>\(\$15.00\)</td><td>\(\$30.00\)</td><td>\(\$45.00\)</td><td>\(\$60.00\)</td><td>\(\$75.00\)</td><td>\(\$90.00\)</td><td>\(\$105.00\)</td><td>\(\$120.00\)</td><td>\(\$135.00\)</td><td>\(\$150.00\)</td><td>\(\$132.00\)</td><td>\(\$144.00\)</td><td>\(\$156.00\)</td><td>\(\$168.00\)</td><td>\(\$180.00\)</td></tr></table> The unit price is not constant across the full domain, so the relation is not proportional. The relation is discrete because fractional shirts are not ordered.

a) \(10\) shirts cost \(\$150.00\); \(11\) shirts cost \(\$132.00\). The larger order costs \(\$18.00\) less. b) \(9\) shirts cost \(\$135.00\), so ordering \(11\) shirts for \(\$132.00\) saves \(\$3.00\). c) <table><tr><td>Shirts</td><td>\(1\)</td><td>\(2\)</td><td>\(3\)</td><td>\(4\)</td><td>\(5\)</td><td>\(6\)</td><td>\(7\)</td><td>\(8\)</td><td>\(9\)</td><td>\(10\)</td><td>\(11\)</td><td>\(12\)</td><td>\(13\)</td><td>\(14\)</td><td>\(15\)</td></tr><tr><td>Total price</td><td>\(\$15.00\)</td><td>\(\$30.00\)</td><td>\(\$45.00\)</td><td>\(\$60.00\)</td><td>\(\$75.00\)</td><td>\(\$90.00\)</td><td>\(\$105.00\)</td><td>\(\$120.00\)</td><td>\(\$135.00\)</td><td>\(\$150.00\)</td><td>\(\$132.00\)</td><td>\(\$144.00\)</td><td>\(\$156.00\)</td><td>\(\$168.00\)</td><td>\(\$180.00\)</td></tr></table> The relation is not proportional because the price per shirt changes from \(\$15.00\) to \(\$12.00\), and it is discrete because only whole-number quantities are allowed.

- Compare the change in output with the change in input between pairs of known points. - What must be constant if all the points lie on one line? - Once you know the slope, how can you find the function value at \(x = 0\)?

1. Compare rates of change. From \((1, 5)\) to \((2, 8)\), the slope is \(\frac{8 - 5}{2 - 1} = 3\). From \((2, 8)\) to \((4, 14)\), the slope is \(\frac{14 - 8}{4 - 2} = 3\). The equal slopes show that the relationship is linear. 2. Use \((1, 5)\) with slope \(3\): \(5 = 3 \cdot 1 + b\), so \(b = 2\). Thus \(f(x) = 3x + 2\). 3. Then \(f(0) = 2\). 4. For the last column, solve \(26 = 3x + 2\). This gives \(x = 8\).

The values represent the linear function \(f(x) = 3x + 2\). The missing value is \(f(0) = 2\), and the missing x-value paired with \(26\) is \(x = 8\).

- Substitute \(2x\) everywhere the original function has \(x\). - Separately multiply the entire original function by \(2\). - Compare the resulting expressions for all values of \(x\).

1. a) \(f(2x)=0.8(2x)=1.6x\), and \(2f(x)=2(0.8x)=1.6x\). The property holds, so the function is proportional. It is linear because it has the form \(mx+b\). 2. b) \(f(2x)=1.6x+2\), while \(2f(x)=1.6x+4\). The property does not hold, so the function is not proportional. It is still linear because it has the form \(mx+b\). 3. c) \(f(2x)=(2x)^2=4x^2\), while \(2f(x)=2x^2\). For \(x\ne 0\), the property does not hold, so the function is not proportional. It is nonlinear because the variable is squared.

a) The doubling property holds. The function is proportional and linear. b) The doubling property does not hold. The function is not proportional, but it is linear. c) The doubling property does not hold for \(x\ne 0\). The function is not proportional and is nonlinear.

- Compare average rates over more than one interval. - Look at whether the temperature drops by equal amounts in equal times. - Consider the temperature the drink appears to be approaching.

1. From \(0\) to \(2\) minutes, the average rate is \(\frac{140-170}{2-0}=-15\,\text{°F}\) per minute. From \(2\) to \(4\) minutes, it is \(\frac{118-140}{4-2}=-11\,\text{°F}\) per minute. Because the rates differ, the relation is not linear. 2. The cocoa cools rapidly at first and then more slowly. For example, from minute \(10\) to minute \(15\), the average rate is only \(\frac{74-84}{15-10}=-2\,\text{°F}\) per minute. 3. The measurements suggest that the cocoa is approaching a temperature near or below \(74\,\text{°F}\). A room temperature around \(68\)–\(72\,\text{°F}\) is plausible, but finitely many measurements do not determine the limiting temperature without an additional cooling model.

a) No. The average rates of change are not constant. b) The cocoa cools quickly at first and more slowly later. c) About \(68\)–\(72\,\text{°F}\) is plausible, but the exact room temperature cannot be determined from these measurements alone.

- What must be true about the ratios \(\frac{y}{x}\) in a proportional relationship? - What point must every proportional relationship contain? - How can you use changes in \(y\) and changes in \(x\) to test for a constant rate of change?

1. For Table 1, check the ratios \(\frac{y}{x}\): \(\frac{2}{1.5}=\frac{4}{3}\), \(\frac{4}{3}=\frac{4}{3}\), \(\frac{6}{4.5}=\frac{4}{3}\), and \(\frac{8}{6}=\frac{4}{3}\). The ratio is constant, so the relationship is proportional. Every proportional relationship is also linear. 2. Table 2 contains the point \((0, 3)\), so it is not proportional because a proportional relationship must contain \((0, 0)\). 3. For Table 2, calculate the rate of change between consecutive points: \(\frac{3-(-1)}{0-(-2)}=2\), \(\frac{7-3}{2-0}=2\), and \(\frac{11-7}{4-2}=2\). The rate of change is constant, so the relationship is linear.

Table 1 is proportional and linear because every ratio \(\frac{y}{x}\) equals \(\frac{4}{3}\). Table 2 is linear because its rate of change is always \(2\), but it is not proportional because it contains \((0, 3)\) rather than \((0, 0)\).

- How can you compare the slopes between pairs of points? - Once you know the slope, how can you find the y-intercept? - What must the y-intercept be for a proportional relationship?

1. Compare slopes between consecutive points: \(\frac{5 - 11}{-1 - (-4)} = -2\), \(\frac{-1 - 5}{2 - (-1)} = -2\), and \(\frac{-7 - (-1)}{5 - 2} = -2\). Since the rate of change is constant, the relationship is linear. 2. Use slope \(-2\) and point \((2, -1)\): \(-1 = -2 \cdot 2 + b\), so \(b = 3\). Thus \(y = -2x + 3\). 3. A proportional relationship has the form \(y = kx\) and therefore has y-intercept \(0\). Here the y-intercept is \(3\), so the relationship is not proportional.

a) The slope between every pair of consecutive points is \(-2\), so the function is linear. b) \(y = -2x + 3\) c) No. The y-intercept is \(3\), not \(0\), so the relationship is not proportional.

- Connect cross-sectional area with how quickly water height changes. - A straight line represents a constant rate of change. - In a container that widens, each additional unit of height requires more water.

1. A cylinder has the same cross-sectional area at every height. With a constant inflow, its water height rises at a constant rate, producing a linear graph. Therefore, Graph \(f\) represents Container A. 2. Container B has a larger cross-sectional area higher up. Each additional unit of height requires more water than the previous one, so the water height rises more slowly over time. The graph therefore flattens, as Graph \(g\) does.

a) Graph \(f\): Container A; Graph \(g\): Container B b) The cylinder's height increases at a constant rate, while the widening container's height increases at a decreasing rate.

- Write each area formula and substitute the fixed height. - Check whether \(r\) appears only as a first-power factor. - Compare the results when the radius changes from \(1\) to \(2\).

1. The lateral surface area is \(L(r)=2\pi rh\). With \(h=10\), \(L(r)=20\pi r\), so it is proportional to \(r\). 2. The base area is \(B(r)=\pi r^2\). Since the radius is squared, doubling \(r\) multiplies the area by \(4\), not \(2\). Therefore, base area is not proportional to radius.

a) Proportional, because \(L(r)=20\pi r\). b) Not proportional, because \(B(r)=\pi r^2\).

- A straight line represents a constant rate of change in height. - A wider horizontal layer contains more water and therefore takes longer to drain through. - Compare how each graph's steepness changes over time.

1. Graph \(h_2\) decreases slowly at first and then becomes increasingly steep, so the water height decreases faster over time. Graph \(h_3\) decreases steeply at first and then flattens, so the water height decreases more slowly over time. 2. Graph \(h_1\) matches Container 1 because a cylinder has constant cross-sectional area, producing a constant rate of decrease in height. Graph \(h_2\) matches Container 2 because the wide upper portion makes the height drop slowly at first, while the narrower lower portion makes it drop faster later. Graph \(h_3\) matches Container 3 because the narrow upper portion makes the height drop quickly at first, while the wider lower portion makes it drop more slowly later.

a) \(h_2\) decreases slowly and then faster; \(h_3\) decreases quickly and then more slowly. b) \(h_1\) matches 1; \(h_2\) matches 2; \(h_3\) matches 3.