Free math worksheets

- Compare consecutive output changes for equal input changes. - A linear function has a constant rate of change.

1. Each time \(x\) increases by \(1\), the output increases by \(3\): \(7-4=3\), \(10-7=3\), and \(13-10=3\). 2. A constant rate of change means the points lie on a straight line, so the function is linear.

Linear. The output changes by a constant \(3\) for every increase of \(1\) in \(x\).

- 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 \(A=s^2\).

- What must be true about the slopes between pairs of points if all three points lie on one line? - Find the slope between two different pairs of points. - Use change in \(y\) divided by change in \(x\).

1. The slope from \(P\) to \(Q\) is \(\frac{3 - 5}{2 - (-2)} = -\frac{1}{2}\). 2. The slope from \(Q\) to \(R\) is \(\frac{1 - 3}{5 - 2} = -\frac{2}{3}\). 3. Since \(-\frac{1}{2} \neq -\frac{2}{3}\), the slope is not constant, so the three points are not collinear.

No. The slopes are \(-\frac{1}{2}\) and \(-\frac{2}{3}\), so the three points do not lie on the same line.

- What must be true about the graph of a proportional relationship? - Consider the role of \(b\) in \(y=mx+b\). - What output does a proportional function have when \(x=0\)?

1. A linear function has the form \(y=mx+b\). A proportional function has the form \(y=kx\), which is the special case \(b=0\). 2. Therefore, the statement is false. Only linear functions whose graphs pass through the origin are proportional. 3. For example, \(y=2x+5\) is linear but not proportional because its y-intercept is \(5\).

The statement is false. A linear function \(y=mx+b\) is proportional only when \(b=0\). For example, \(y=2x+5\) is linear but not proportional.

- 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. - For the increase, compare the two years named in the question. - To test for a constant yearly increase, compare each value with the one immediately before it.

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. - For part b, compare the second stopping distance with twice the first and form the requested ratio. - For part c, use the shape of the graph or compare the formula with the general form of a linear function; do not rely only on the doubling test.

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 stopping distance by \(3.2\), not by \(2\). 4. The function is nonlinear. Its formula contains a \(v^2\) term rather than having the form \(mv+b\), and its graph is curved rather than a straight line. The doubling result in part b by itself would not prove nonlinearity, because a linear function with a nonzero initial value also need not double its output when the input doubles.

a) \(75\,\text{ft}\) at \(30\,\text{mph}\); \(240\,\text{ft}\) at \(60\,\text{mph}\) b) No. The ratio is \(3.2\). c) Nonlinear; the formula contains a \(v^2\) term and the graph is curved rather than a straight line.

- 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 how much the area changes when the side length increases by the same amount each time. - A linear relationship has a constant rate of change and appears as a straight-line pattern.

1. As the side length increases by \(1\,\text{cm}\) each time, the increases in area are \(3\), \(5\), \(7\), \(9\), and \(11\,\text{cm}^2\), so the rate of change is not constant. 2. The plotted points curve upward rather than lying on one straight line. 3. Therefore, the relationship is nonlinear.

The relationship is nonlinear because equal increases in side length do not produce equal increases in area, and the plotted points do not lie on one straight line.

- Compare the changes in \(y\) over equal changes in \(x\). - Ask what must stay constant for a relationship to be linear.

1. A linear relationship has a constant rate of change, so equal changes in \(x\) produce equal changes in \(y\). 2. Here the changes in \(y\) are \(2,4,6,8\), which are not constant. 3. Therefore, the relationship is nonlinear.

The relationship is nonlinear because equal changes in \(x\) do not produce a constant change in \(y\).

- 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 unit price that applies to that order size. - Compare the total costs for \(9\), \(10\), and \(11\) shirts carefully around the pricing break. - A proportional relation has one constant unit rate, while a linear relation has one constant rate of change across its full domain. - Think about which input values are actually possible when the input counts shirts.

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. It is not linear over its full domain because the change in total price is not constant; for example, the total drops by \(\$18.00\) from \(10\) to \(11\) shirts instead of continuing at one constant rate. 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 neither proportional nor linear over its full domain because it does not have one constant rate of change. Its graph is discrete because only whole-number quantities of shirts are allowed.

- Compare the slopes between two pairs of points. - Use change in \(y\) divided by change in \(x\), even when the coordinates are decimals. - What does it mean if the two slopes are different?

1. The slope from \(A\) to \(B\) is \(\frac{9 - 4}{4 - 1.5} = \frac{5}{2.5} = 2\). 2. The slope from \(B\) to \(C\) is \(\frac{13.5 - 9}{6 - 4} = \frac{4.5}{2} = 2.25\). 3. Since \(2 \neq 2.25\), the points are not collinear. Three noncollinear points form a triangle.

The points form a triangle because the slopes \(2\) and \(2.25\) are different.

- 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 \(x\), and separately multiply the entire original function by \(2\). - If the two expressions are different for some input, that is enough to rule out proportionality. - To confirm proportionality, look for a rule of the form \(f(x)=kx\), not merely one property that happens to hold.

1. a) \(f(2x)=0.8(2x)=1.6x\), and \(2f(x)=2(0.8x)=1.6x\). The doubling property holds. Also, \(f(x)=0.8x\) has the form \(kx\), so it is proportional. It is linear. 2. b) \(f(2x)=1.6x+2\), while \(2f(x)=1.6x+4\). The doubling property fails, so the function is not proportional. It is linear because it has the form \(mx+b\). 3. c) \(f(2x)=(2x)^2=4x^2\), while \(2f(x)=2x^2\). For \(x\ne0\), the doubling property fails, 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 \(x\ne0\). The function is not proportional and is nonlinear.

- Compare average rates over more than one time interval. - Look at whether the temperature drops by equal amounts in equal times. - For the final question, separate what the measured trend supports from a numerical value the table never gives.

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 \(\frac{74-84}{15-10}=-2\,\text{°F}\) per minute. 3. The measurements suggest that the cocoa may be approaching a temperature below the latest measured value of \(74\,\text{°F}\). The exact limiting or room temperature cannot be determined from these measurements alone without additional information or a cooling model.

a) No. The average rates of change are not constant. b) The cocoa cools quickly at first and more slowly later. c) The data suggest a limiting temperature below \(74\,\text{°F}\), but they do not determine its exact value.

- 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 \((2, -1)\): \(-1=-2\cdot2+b\), 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.

- What geometric object is the solution set of one linear equation in two variables? - Compare the slopes between two pairs of points. - What must be true about those slopes if all three points lie on one line?

1. The solution set of a nondegenerate linear equation in two variables is a line, so the three points would have to be collinear. 2. The slope from \((-2, 7)\) to \((1, 1)\) is \(\frac{1 - 7}{1 - (-2)} = -2\). 3. The slope from \((1, 1)\) to \((4, -4)\) is \(\frac{-4 - 1}{4 - 1} = -\frac{5}{3}\). 4. Since the slopes are different, the three points are not collinear and cannot all satisfy the same nondegenerate linear equation.

No. The slopes are \(-2\) and \(-\frac{5}{3}\), so the three points are not collinear.

- 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 \((2, 4)\)

- 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.

- What special point must a proportional graph contain? - Look at where the graph crosses the y-axis. - Choose two points and compare the horizontal and vertical changes. - Check whether the same rate of change holds elsewhere on the line.

1. A proportional relationship must graph as a line through \((0, 0)\). This graph crosses the y-axis at \(2\), so Maya is incorrect. 2. Using \((0, 2)\) and \((2, 3)\), the change in \(y\) is \(1\) when the change in \(x\) is \(2\). The slope is \(\frac{1}{2}=0.5\), and it is constant along the line. Therefore, Jordan is correct.

Maya is incorrect because the line does not pass through \((0, 0)\). Jordan is correct because the constant slope is \(0.5\), so every increase of \(2\) in \(x\) produces an increase of \(1\) in \(y\).

- Write each area formula and substitute the fixed height. - To test proportionality, look for a constant multiple of \(r\) with no added constant. - To classify linearity, compare each formula with the form \(mr+b\); remember that a relationship can be linear without being proportional.

1. The lateral surface area is \(L(r)=2\pi rh\). With \(h=10\), \(L(r)=20\pi r\). This has the form \(mr\) with a constant rate, so it is proportional and linear. 2. The base area is \(B(r)=\pi r^2\). The variable is squared, so the relationship does not have the form \(mr+b\). It is not proportional and is nonlinear. For example, doubling \(r\) multiplies \(B\) by \(4\), but the nonlinear classification follows from the squared variable rather than from nonproportionality alone.

a) Proportional and linear, because \(L(r)=20\pi r\). b) Not proportional and nonlinear, 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.