Free math worksheets

- Find a point on the blue graph that lies exactly on grid lines. - If \((x, y)\) is on \(y = \frac{a}{x}\), how can you calculate \(a\)? - Check your result with a second point.

1. Choose an exact point on the graph, such as \(P(1, -3)\) or \(Q(3, -1)\). 2. Using \(P(1, -3)\), substitute into \(f(x) = \frac{a}{x}\): \(-3 = \frac{a}{1}\). 3. Therefore, \(a = -3\). 4. Check with \(Q(3, -1)\): \(f(3) = \frac{-3}{3} = -1\). 5. The function rule is \(f(x) = -\frac{3}{x}\).

The function rule is \(f(x) = -\frac{3}{x}\).

- Choose a nonzero point whose coordinates are easy to read from the graph. - In a direct variation, compare output to input using the same multiplicative factor. - Recall the required intercept of every graph of \(y=kx\).

1. A clear point on the graph is \((2,6)\). Thus, \(k=\frac{y}{x}=\frac{6}{2}=3\). 2. The graph is a straight line through the origin, which is the graph form of a direct variation \(y=kx\).

a) \(k=3\) b) The graph is a straight line through the origin.

- For each situation, consider what happens to the second quantity when the first quantity doubles. - For direct variation, check whether the ratio of the second quantity to the first is constant. - For inverse variation, check whether the product of the two quantities is constant. - A relationship can follow a rule without being a direct or inverse variation.

1. a) Direct variation: With a fixed price per muffin, doubling the number of muffins doubles the total cost. The ratio of cost to number of muffins is constant. 2. b) Neither: A child's height does not vary directly or inversely with age. For example, a 10-year-old is not generally twice as tall as a 5-year-old. 3. c) Inverse variation: For a fixed amount of water, doubling the number of identical, noninterfering pumps cuts the draining time in half. The product of the number of pumps and the time is constant. 4. d) Neither: The area is \(A=s^2\). Doubling \(s\) multiplies the area by \(4\), so the relationship is not direct or inverse variation.

a) Direct variation: the cost-to-muffin ratio is constant. b) Neither: height is not a constant multiple of age and does not have a constant age-height product. c) Inverse variation: for a fixed job, pump count times draining time is constant. d) Neither: \(A=s^2\), so doubling \(s\) multiplies area by \(4\), not by \(2\) or \(\frac{1}{2}\).

- In an inverse variation, what must be true about the product of each pair? - Calculate the product for every column. - Compare how one value changes when the other doubles.

1. For an inverse variation, the product \(xy\) must be constant. 2. Calculate each product: \(0.5 \cdot 40 = 20\), \(1 \cdot 20 = 20\), \(2.5 \cdot 8 = 20\), \(5 \cdot 4 = 20\), and \(10 \cdot 2 = 20\). 3. Since every product equals \(20\), the table represents an inverse variation.

Yes. The table represents an inverse variation because \(xy = 20\) for every pair.

- Ask what happens to the second quantity when the first quantity doubles. - Does the second quantity increase or decrease? - Look for a fixed total, such as a fixed amount of work or a fixed distance. - Check whether the product of the two quantities remains constant.

1. For a), the relationship is an inverse variation. If the number of painters doubles, the time is cut in half, provided the painters do not interfere with one another. The total amount of work stays constant, so the product of the number of painters and the time is constant. 2. For b), the relationship is not an inverse variation. It is a direct variation because the total cost increases in proportion to the number of tickets. The cost per ticket is constant, not the product of the two quantities. 3. For c), the relationship is an inverse variation. The travel time is \(t = \frac{30}{v}\), where \(v\) is the cyclist’s speed in miles per hour. Because the distance is fixed, doubling the speed cuts the travel time in half, and \(v \cdot t = 30\,\text{mi}\).

a) Inverse variation, because the total amount of work is fixed. b) Not an inverse variation; it is a direct variation because the cost per ticket is fixed. c) Inverse variation, because the travel distance is fixed.

- What stays constant when you multiply the coordinates of an ordered pair in an inverse variation? - How can you use the one known ordered pair to find the constant of variation? - Once you know the constant, how can you find each missing \(y\)-value?

1. Use the known ordered pair to find the constant of variation: \(k = xy = 10 \cdot 12 = 120\). 2. Substitute each \(x\)-value into \(y = \frac{120}{x}\): \(y = 24, 15, 12, 8, 6, 4, 3\), respectively. 3. The equation is \(y = \frac{120}{x}\). 4. The graph of an inverse variation of this form is a hyperbola.

a) The completed table is: <table> <tbody> <tr><td>\(x\)</td><td>\(5\)</td><td>\(8\)</td><td>\(10\)</td><td>\(15\)</td><td>\(20\)</td><td>\(30\)</td><td>\(40\)</td></tr> <tr><td>\(y\)</td><td>\(24\)</td><td>\(15\)</td><td>\(12\)</td><td>\(8\)</td><td>\(6\)</td><td>\(4\)</td><td>\(3\)</td></tr> </tbody> </table> b) \(y = \frac{120}{x}\) c) The graph is a hyperbola.

- Use the relationship among distance, speed, and time. - The product of speed and time is the fixed distance. - Recall how one variable changes when the other is multiplied by a factor in an inverse variation.

1. Because distance equals speed times time, \(v \cdot t = 120\). Solving for time gives \(t = \frac{120}{v}\). 2. Evaluate the formula for each speed: \(120 \div 20 = 6\), \(120 \div 30 = 4\), \(120 \div 40 = 3\), \(120 \div 60 = 2\), \(120 \div 80 = 1.5\), \(120 \div 100 = 1.2\), and \(120 \div 120 = 1\). 3. This is an inverse variation. When the speed is cut in half, the travel time doubles.

a) \(t = \frac{120}{v}\) b) <table> <tr><td>\(v\) (in \(\text{mi/h}\))</td><td>\(20\)</td><td>\(30\)</td><td>\(40\)</td><td>\(60\)</td><td>\(80\)</td><td>\(100\)</td><td>\(120\)</td></tr> <tr><td>\(t\) (in \(\text{h}\))</td><td>\(6\)</td><td>\(4\)</td><td>\(3\)</td><td>\(2\)</td><td>\(1.5\)</td><td>\(1.2\)</td><td>\(1\)</td></tr> </table> c) The travel time doubles.

- What must be constant for all ordered pairs in an inverse variation? - Calculate the product of the two coordinates in each ordered pair. - If the products are equal, that common value is the constant of variation.

1. Test for inverse variation by calculating \(xy\) for each ordered pair. 2. The products are \(4 \cdot 15 = 60\), \(2 \cdot 30 = 60\), \(10 \cdot 6 = 60\), \(12 \cdot 5 = 60\), and \(2.5 \cdot 24 = 60\). 3. Because every product equals \(60\), the ordered pairs represent an inverse variation with \(a = 60\). 4. The equation is \(y = \frac{60}{x}\).

Yes. Every ordered pair has the constant product \(xy = 60\), so the equation is \(y = \frac{60}{x}\).

- Test both criteria for each table: Is \(\frac{y}{x}\) constant? Is \(xy\) constant? - One pair that does not match the others is enough to rule out a type of variation. - In Table A, notice that \(y\) increases by \(2\) whenever \(x\) increases by \(1\). Does that alone make the relationship a direct variation?

1. For Table A, the ratios are not constant because \(\frac{5}{1}=5\) but \(\frac{7}{2}=3.5\). The products are also not constant because \(1\cdot5=5\) but \(2\cdot7=14\). Therefore, Table A represents neither type of variation. 2. For Table B, \(\frac{2.5}{1}=\frac{5}{2}=\frac{10}{4}=\frac{20}{8}=2.5\). Because \(\frac{y}{x}\) is constant, Table B represents a direct variation.

Table A: neither, because neither \(\frac{y}{x}\) nor \(xy\) is constant. Table B: direct variation, because \(\frac{y}{x}=2.5\) for every listed pair.

- Decide whether the project takes more or less time when more students help. - Find the time by dividing the total student-hours by the number of students. - Check whether the product of the two quantities is the same for every pair.

1. Divide the total of \(12\) student-hours by each number of students: \(12 \div 1 = 12\), \(12 \div 2 = 6\), \(12 \div 3 = 4\), \(12 \div 4 = 3\), and \(12 \div 6 = 2\). 2. The relationship is an inverse variation because the product of the number of students and the time is always \(12\): \(nt = 12\). 3. Solving \(nt = 12\) for \(t\) gives \(t = \frac{12}{n}\).

a) <table> <tr><td>Number of students \(n\)</td><td>\(1\)</td><td>\(2\)</td><td>\(3\)</td><td>\(4\)</td><td>\(6\)</td></tr> <tr><td>Time \(t\) in \(\text{h}\)</td><td>\(12\)</td><td>\(6\)</td><td>\(4\)</td><td>\(3\)</td><td>\(2\)</td></tr> </table> b) It is an inverse variation because \(1 \cdot 12 = 2 \cdot 6 = 3 \cdot 4 = 4 \cdot 3 = 6 \cdot 2 = 12\). c) \(t = \frac{12}{n}\)

- Identify the fixed amount of work in robot-hours. - Write a constant-product equation before substituting the rush-order time. - The requested number is the total robot count.

1. The fixed work is \(6\cdot10=60\) robot-hours, so \(tn=60\) and \(t(n)=\frac{60}{n}\). 2. Set \(t=4\): \(4n=60\), so \(n=15\).

a) \(tn=60\), so \(t(n)=\frac{60}{n}\). b) \(15\) robots.

- Identify the constant product of pipe count and filling time. - Use the same inverse model for both a forward and a reverse question. - Check that more pipes correspond to less time.

1. The fixed work gives \(tn=18\cdot4=72\), so \(t(n)=\frac{72}{n}\). 2. With \(n=3\), \(t=\frac{72}{3}=24\) minutes. 3. With \(t=12\), \(12n=72\), so \(n=6\).

a) \(tn=72\), so \(t(n)=\frac{72}{n}\). b) \(24\,\text{min}\). c) \(6\) pipes.

- Use the original group size and time to determine the constant product. - Check each claim by evaluating the same inverse model. - For the factor argument, compare the new group size with the original group size multiplicatively.

1. The fixed work is \(4\cdot6=24\) person-hours, so \(t(n)=\frac{24}{n}\). 2. \(t(8)=3\,\text{h}\), so the first claim is correct. \(t(12)=2\,\text{h}\), so the second claim is false. 3. Tripling the number of people divides the time by \(3\). Starting from \(6\,\text{h}\), the predicted time is \(2\,\text{h}\), consistent with the model calculation.

a) \(t(n)=\frac{24}{n}\) b) The \(8\)-person claim is correct: \(3\,\text{h}\). The \(12\)-person claim is false: the model gives \(2\,\text{h}\). c) Tripling the group size divides the time by \(3\).

- Identify the fixed total amount of feed. - Daily use and duration form a constant product. - Use the required duration as the input to the inverse model.

1. The fixed supply is \(30\cdot12=360\) pounds, so \(ad=360\). Thus \(a(d)=\frac{360}{d}\). 2. For \(d=45\), \(a=\frac{360}{45}=8\) pounds per day.

a) \(ad=360\), so \(a(d)=\frac{360}{d}\). b) \(8\,\text{lb}\) per day.

- In each situation, identify the product that stays fixed when one quantity changes. - Express the unknown quantity as the fixed product divided by the input quantity. - Substitute the new input only after you have written the inverse-variation function.

1. For part a, the fixed amount of work is \(3\cdot12=36\) roofer-hours, so \(rt=36\) and \(t(r)=\frac{36}{r}\). 2. For \(r=4\), \(t(4)=\frac{36}{4}=9\,\text{h}\). 3. For part b, the tank contains \(24\cdot24=576\,\text{fl oz}\), so \(vN=576\) and \(N(v)=\frac{576}{v}\). 4. For \(v=16\), \(N(16)=\frac{576}{16}=36\).

a) \(t(r)=\frac{36}{r}\); \(9\,\text{h}\) b) \(N(v)=\frac{576}{v}\); \(36\) bottles

- Express the fixed feed supply in cow-days. - Use the same inverse model for a forward and a reverse question. - Account for the cows sold before evaluating part b).

1. The fixed supply is \(15\cdot20=300\) cow-days, so \(D(n)=\frac{300}{n}\). 2. After selling \(3\) cows, \(n=12\), so \(D=\frac{300}{12}=25\) days. 3. For \(D=30\), \(30n=300\), so \(n=10\) cows.

a) \(D(n)=\frac{300}{n}\). b) \(25\) days. c) \(10\) cows.

- Identify the total volume that stays fixed. - Bottle count and bottle size have a constant product. - Evaluate the model at the new bottle size.

1. The batch contains \(15\cdot16=240\) fluid ounces, so \(Ns=240\) and \(N(s)=\frac{240}{s}\). 2. For \(s=12\), \(N=\frac{240}{12}=20\).

a) \(N(s)=\frac{240}{s}\). b) \(20\) bottles.

- Identify the total volume of the tank. - Use the product of flow rate and time. - Explain what remains constant when the flow rate changes.

1. The tank holds \(10 \cdot 60 = 600\,\text{gal}\). 2. At \(15\,\text{gal/min}\), the filling time is \(600 \div 15 = 40\,\text{min}\). 3. The relationship is an inverse variation because the tank volume is fixed, so the product of flow rate and filling time remains constant.

a) Filling will take \(40\,\text{min}\). b) It is an inverse variation because the product of flow rate and time equals the fixed tank volume.

- Find the total number of horse-days in the supply. - Divide that fixed amount by \(60\) days. - Describe what happens to the duration when the number of horses increases.

1. The total supply is \(18 \cdot 40 = 720\) horse-days. 2. For a), the number of horses that can be fed for \(60\) days is \(720 \div 60 = 12\). 3. For b), the relationship is an inverse variation because the product of the number of horses and the number of days remains constant. For example, doubling the number of horses would cut the duration in half.

a) The supply could feed at most \(12\) horses for \(60\) days. b) It is an inverse variation because the product of the number of horses and the number of days is constant.

- Express the fixed job in painter-hours. - Write time as an inverse function of painter count. - Check that fewer painters give a longer time.

1. The fixed work is \(6\cdot10=60\) painter-hours, so \(tn=60\) and \(t(n)=\frac{60}{n}\). 2. For \(n=4\), \(t=\frac{60}{4}=15\) hours.

a) \(t(n)=\frac{60}{n}\). b) \(15\,\text{h}\).

- Find the total number of pump-hours required to fill the pool. - Consider whether more pumps should make the job take more or less time. - Identify the type of variation by determining what product remains constant.

1. Filling the pool requires \(2 \cdot 15 = 30\) pump-hours. 2. For a), \(3\) pumps would need \(30 \div 3 = 10\,\text{h}\). 3. For b), using more pumps increases the total flow rate, so the fixed pool volume is filled in less time. The number of pumps and the filling time form an inverse variation.

a) Three pumps would need \(10\,\text{h}\). b) The claim is incorrect. More pumps reduce the time needed, so the relationship is an inverse variation.

- For the table, divide \(2\) by each listed input and keep track of the sign. - On the supplied graph, look for points that lie on both curves. - The x-coordinate of an intersection is an input for which the two function values are equal.

1. Evaluating \(f(x)=\frac{2}{x}\) gives \(f(-4)=-0.5\), \(f(-2)=-1\), \(f(-1)=-2\), \(f(-0.5)=-4\), \(f(0.5)=4\), \(f(1)=2\), \(f(2)=1\), and \(f(4)=0.5\). 2. From the graph, the curves intersect at \((-1,-2)\) and \((2,1)\). 3. An equation \(f(x)=g(x)\) is satisfied at the x-coordinates of the intersection points, so the solutions are \(x=-1\) and \(x=2\).

a) <table><tbody><tr><td>\(x\)</td><td>\(-4\)</td><td>\(-2\)</td><td>\(-1\)</td><td>\(-0.5\)</td><td>\(0.5\)</td><td>\(1\)</td><td>\(2\)</td><td>\(4\)</td></tr><tr><td>\(f(x)\)</td><td>\(-0.5\)</td><td>\(-1\)</td><td>\(-2\)</td><td>\(-4\)</td><td>\(4\)</td><td>\(2\)</td><td>\(1\)</td><td>\(0.5\)</td></tr></tbody></table> b) \((-1,-2)\) and \((2,1)\) c) \(x=-1\) and \(x=2\). At an intersection, the two graphs have the same y-value for the same x-value, so \(f(x)=g(x)\), which is exactly \(\frac{2}{x}=x-1\).

- Find the total number of student-hours required. - Consider what happens to the time when more students help. - Identify the quantity that remains constant.

1. The decorating requires \(6 \cdot 4 = 24\) student-hours. 2. For a), \(8\) students would need \(24 \div 8 = 3\,\text{h}\). 3. For b), the amount of work is fixed, so the product of the number of students and the time remains constant. Increasing the number of students decreases the time by the corresponding factor.

a) Eight students would need \(3\,\text{h}\). b) It is an inverse variation because the product of the number of students and the time is constant at \(24\) student-hours.

- In an inverse variation, what is true about the product of each pair? - Use the general form \(y = \frac{k}{x}\). - Substitute the given value of \(x\) into the equation.

1. Multiply the coordinates of any pair: \(2.5 \cdot 16 = 40\), \(4 \cdot 10 = 40\), and \(5 \cdot 8 = 40\). Therefore, \(k = 40\). 2. An inverse variation has the form \(y = \frac{k}{x}\), so the equation is \(y = \frac{40}{x}\). 3. For \(x = 20\), \(y = \frac{40}{20} = 2\).

1. \(k = 40\) 2. \(y = \frac{40}{x}\) 3. \(y = 2\)

- What quantity is constant for all ordered pairs in an inverse variation? - How are \(x\), \(f(x)\), and \(a\) related? - How can you rearrange the equation to isolate the unknown quantity?

1. Use the point to find the constant of variation: \(a = x f(x) = 1.25 \cdot 8 = 10\). Thus, \(f(x) = \frac{10}{x}\). 2. Evaluate the function: \(f(5) = \frac{10}{5} = 2\) and \(f(0.1) = \frac{10}{0.1} = 100\). 3. Solve \(20 = \frac{10}{x}\): \(20x = 10\), so \(x = \frac{10}{20} = 0.5\).

a) \(a = 10\); \(f(x) = \frac{10}{x}\) b) \(f(5) = 2\); \(f(0.1) = 100\) c) \(x = 0.5\)

- Ask whether doubling one quantity halves the other. - Check whether the product of the two quantities would remain constant. - Consider what happens when one quantity becomes very large or very small.

1. For a), the total pizza area is fixed. Doubling the number of equal slices halves the area of each slice, so the relationship is an inverse variation. 2. For b), at a fixed price per pound, doubling the number of pounds doubles the total price. This is a direct variation, not an inverse variation. 3. For c), the distance is fixed, so \(v \cdot t = 20\,\text{mi}\). Doubling the speed halves the travel time, so the relationship is an inverse variation. 4. For d), trees do not grow at a constant rate throughout their lives, and neither a constant product nor a constant ratio describes age and height. This is not an inverse variation.

a) Inverse variation, because the total pizza area is fixed. b) Not an inverse variation; it is a direct variation. c) Inverse variation, because the travel distance is fixed. d) Not an inverse variation.

- Separate the boat's still-water speed from its actual speed relative to the shore. - For a fixed distance, consider the product of travel time and actual speed. - Determine the actual speed in each direction before evaluating the inverse model.

1. For fixed distance \(s\), \(Tq=s\), so \(T(q)=\frac{s}{q}\). Thus travel time varies inversely with actual speed. 2. Downstream, \(q=v+c=18\), so \(T=\frac{36}{18}=2\) hours. 3. Upstream, \(q=v-c=12\), so \(T=\frac{36}{12}=3\) hours. 4. The round-trip time is \(2+3=5\) hours.

a) \(Tq=s\), so \(T(q)=\frac{s}{q}\). b) Downstream: \(q=18\) and \(T=2\,\text{h}\). Upstream: \(q=12\) and \(T=3\,\text{h}\). c) \(5\,\text{h}\).

- In an inverse variation, identify the constant product of pump count and draining time. - Use that constant to evaluate several pump counts with one model. - For the factor-change explanation, compare the constant product before and after the pump count is scaled.

1. The constant product is \(k = 3 \cdot 16 = 48\) pump-hours, so \(T(n) = \frac{48}{n}\). 2. The times are \(T(2) = 24\,\text{h}\), \(T(4) = 12\,\text{h}\), \(T(6) = 8\,\text{h}\), and \(T(8) = 6\,\text{h}\). 3. Tripling the number of pumps divides the time by \(3\), so the time becomes one-third of its original value.

1) \(k = 48\) pump-hours and \(T(n) = \frac{48}{n}\) 2) For \(2\) pumps: \(24\,\text{h}\); for \(4\) pumps: \(12\,\text{h}\); for \(6\) pumps: \(8\,\text{h}\); for \(8\) pumps: \(6\,\text{h}\). 3) The time is divided by \(3\).

- Express the wall project as a fixed number of worker-days. - Use the same inverse model for both a forward and reverse evaluation. - Keep worker count and day count in their correct roles.

1. The fixed work is \(6\cdot10=60\) worker-days, so \(D(n)=\frac{60}{n}\). 2. For \(n=4\), \(D=15\) days. 3. For \(D=4\), \(4n=60\), so \(n=15\) workers.

a) \(D(n)=\frac{60}{n}\). b) \(15\) days. c) \(15\) workers.

- What equation relates \(x\), \(y\), and \(k\) for this function type? - How can you rearrange an equation to isolate the unknown? - What must be true when a point lies on a graph?

1. Substitute \(P(4, -3)\) into the rule: \(-3 = \frac{k}{4}\). Multiplying by \(4\) gives \(k = -12\), so \(g(x) = -\frac{12}{x}\). 2. Evaluate the function: \(g(-1.5) = \frac{-12}{-1.5} = 8\). 3. Solve \(12 = -\frac{12}{x}\): \(12x = -12\), so \(x = -1\).

a) \(k = -12\); \(g(x) = -\frac{12}{x}\) b) \(g(-1.5) = 8\) c) \(x = -1\)

- Identify the total cost that remains fixed. - Divide the total cost by the number of students. - Use the constant product to write the function. - Compare the price before and after doubling the group size.

1. With \(24\) students, each pays \(\$360.00 \div 24 = \$15.00\). 2. The function is \(y = \frac{360}{x}\). 3. The table values are \(\$360.00 \div 12 = \$30.00\), \(\$360.00 \div 15 = \$24.00\), \(\$360.00 \div 20 = \$18.00\), \(\$360.00 \div 30 = \$12.00\), and \(\$360.00 \div 40 = \$9.00\). 4. Doubling the number of students halves the price per student because \(xy = 360\) remains constant.

1. Each student pays \(\$15.00\). 2. \(y = \frac{360}{x}\) 3. <table> <tr><td>Number of students \(x\)</td><td>\(12\)</td><td>\(15\)</td><td>\(20\)</td><td>\(30\)</td><td>\(40\)</td></tr> <tr><td>Price per student \(y\)</td><td>\(\$30.00\)</td><td>\(\$24.00\)</td><td>\(\$18.00\)</td><td>\(\$12.00\)</td><td>\(\$9.00\)</td></tr> </table> 4. Doubling the number of students halves the price per student.

- Separate the total number of people from the number who actually share the cost. - For direct variation, look for a constant ratio between the dependent and independent quantities. - For inverse variation, look for a constant product; check which pair of variables actually has that property.

1. With \(p\) fixed, \(k=\frac{1}{p}G\), so \(k\) varies directly with \(G\). The constant of variation is \(\frac{1}{p}\). 2. With \(G\) fixed, \(k=\frac{G}{p}\), so \(k\) varies inversely with \(p\). The constant product is \(pk=G\). 3. It is not an inverse variation in \(n\), because the fixed-product equation is \((n-3)k=G\), not \(nk=G\). The inverse input is the number of paying people, \(p=n-3\).

1) Direct variation: \(k=\frac{1}{p}G\) 2) Inverse variation: \(k=\frac{G}{p}\), with \(pk=G\) 3) No. With \(G\) fixed, \((n-3)k=G\); \(k\) is inversely proportional to the number of paying people \(p=n-3\), not to \(n\).

- Choose a point on each graph whose coordinates can be read exactly from the grid. - How can you use a point on \(y = \frac{a}{x}\) to find \(a\)? - Rearrange the equation to express \(a\) in terms of \(x\) and \(y\).

1. Read an exact point on the blue graph, such as \((1, 2)\). 2. Substitute into \(y = \frac{a}{x}\): \(2 = \frac{a}{1}\), so \(a = 2\). 3. Read an exact point on the red graph, such as \((2, -2)\). 4. Substitute into \(y = \frac{a}{x}\): \(-2 = \frac{a}{2}\), so \(a = -4\).

a) \(a = 2\) b) \(a = -4\)

- Use only plotted whole-number animal counts; the number of animals is discrete. - Read the y-coordinate paired with each requested x-coordinate. - Compare the products of the coordinate pairs and interpret the common product in context.

1. From the plotted points, the values are \(y=20\), \(10\), \(8\), \(5\), and \(4\) for \(x=2\), \(4\), \(5\), \(8\), and \(10\), respectively. 2. The products are \(2\cdot20=40\), \(4\cdot10=40\), \(5\cdot8=40\), \(8\cdot5=40\), and \(10\cdot4=40\). 3. Every product is \(40\), confirming an inverse variation. 4. The constant product represents \(40\) one-animal daily portions of feed.

a) <table><tr><td>Number of animals \(x\)</td><td>\(2\)</td><td>\(4\)</td><td>\(5\)</td><td>\(8\)</td><td>\(10\)</td></tr><tr><td>Number of days \(y\)</td><td>\(20\)</td><td>\(10\)</td><td>\(8\)</td><td>\(5\)</td><td>\(4\)</td></tr></table> b) Every product \(xy\) equals \(40\). c) The supply contains \(40\) one-animal daily portions.

- Choose a point on the graph whose coordinates lie exactly on grid intersections. - What product stays constant for a function of this form? - For the point check, substitute the coordinates into the relationship \(xy = a\). - Use the function rule you found to calculate each missing table value.

1. Read an exact point from the graph, such as \((2, 6)\) or \((4, 3)\). Since \(a = xy\), \(a = 2 \cdot 6 = 12\). 2. Check point \(P\): \(2.5 \cdot 4.8 = 12\). The product equals \(a\), so the point lies on the graph. 3. Use \(f(x) = \frac{12}{x}\): \(f(0.5) = 24\), \(f(1.5) = 8\), \(f(5) = 2.4\), and \(f(10) = 1.2\).

a) \(a = 12\) b) Yes, because \(2.5 \cdot 4.8 = 12\). c) The completed table is: <table> <tr><td>\(x\)</td><td>\(0.5\)</td><td>\(1.5\)</td><td>\(5\)</td><td>\(10\)</td></tr> <tr><td>\(f(x)\)</td><td>\(24\)</td><td>\(8\)</td><td>\(2.4\)</td><td>\(1.2\)</td></tr> </table>

- How are \(x\), \(y\), and \(a\) related for a function of this form? - Choose an easy-to-read point on each graph. - Multiply the coordinates of each selected point to find \(a\). - For part b), substitute the value of \(a\) for \(f\) and the given input into the function rule.

1. For each graph, read an exact point and use \(a = xy\). 2. For \(f\), the point \((1, 15)\) gives \(a_f = 1 \cdot 15 = 15\). 3. For \(g\), the point \((2, 5)\) gives \(a_g = 2 \cdot 5 = 10\). 4. For \(h\), the point \((5, 1)\) gives \(a_h = 5 \cdot 1 = 5\). 5. Since \(f(x) = \frac{15}{x}\), \(f(6) = \frac{15}{6} = 2.5\).

a) \(a_f = 15\), \(a_g = 10\), and \(a_h = 5\) b) \(f(6) = 2.5\)

- Look for points marked exactly at grid intersections. - Multiply the coordinates of a point on the graph. What does that product represent? - Use the quadrants of the graph to check the sign of \(a\).

1. One clearly marked point is \(P(1, -2)\). 2. Substitute into \(y = \frac{a}{x}\): \(-2 = \frac{a}{1}\), so \(a = -2\). 3. Check with \(Q(2, -1)\): \(f(2) = \frac{-2}{2} = -1\), which agrees with the graph.

a) Using \(P(1, -2)\), \(a = -2\). b) The point \(Q(2, -1)\) confirms the result because \(f(2) = \frac{-2}{2} = -1\).

- Use the known worker count and time to identify the constant product. - Write the time as the constant divided by the number of workers. - Because workers are counted in whole numbers, use the discrete plotted points when reading the graph.

1. For inverse variation, the product \(nt\) is constant. Here, \(k=4\cdot6=24\) worker-hours. 2. Therefore, \(t(n)=\frac{24}{n}\). 3. At \(t=3\), the plotted point has \(n=8\). Algebraically, \(3=\frac{24}{n}\), so \(n=8\).

\(t(n)=\frac{24}{n}\); \(8\) workers are needed.

- With distance fixed, relate speed and time using a constant product. - Compare the two requested speeds by a multiplication factor, not only by subtraction. - After writing the model, evaluate it at the final speed and convert units if useful.

1. Fixed distance gives \(vt=60\), so \(t(v)=\frac{60}{v}\). The constant of variation is \(60\). 2. The graph gives \(t(20)=3\,\text{h}\) and \(t(40)=1.5\,\text{h}\). Doubling the speed from \(20\) to \(40\) halves the time from \(3\) to \(1.5\), which is the multiplicative behavior of inverse variation. 3. \(t(80)=\frac{60}{80}=0.75\,\text{h}=45\,\text{min}\).

a) \(t(v)=\frac{60}{v}\), with \(k=60\) b) \(t(20)=3\,\text{h}\) and \(t(40)=1.5\,\text{h}\); doubling speed halves time. c) \(0.75\,\text{h}\), or \(45\,\text{min}\)

- In which quadrants should the graph lie when the numerator is positive? - What happens to the outputs as the absolute value of \(x\) becomes larger? - Substitute simple values such as \(x = 1\) and \(x = 2\), then compare the resulting points with the graphs. - Recall the typical shape of a reciprocal-function graph.

1. The constant of variation is positive, so the graph must lie in Quadrants I and III. 2. Test easy inputs: \(f(1) = 4\) and \(f(2) = 2\). 3. Graph A lies in Quadrants I and III and passes through \((1, 4)\) and \((2, 2)\). Graph B has a negative constant, Graph C is linear, and Graph D has only positive outputs. 4. Therefore, the correct graph is A.

A

- Use a complete input-output pair to identify the constant multiplier. - Write the direct-variation equation before solving the reverse pair. - Check that every completed pair has the same output-to-input ratio.

1. For a direct variation, \(y=kx\). Using \((2,7)\), \(k=\frac{7}{2}=3.5\). 2. The equation is \(y=3.5x\). 3. For \(y=35\), solve \(35=3.5x\), giving \(x=10\).

a) \(k=3.5\) and \(y=3.5x\) b) \(x=10\)

- Direct variation has no nonzero starting amount; the output scales from zero with the input. - The constant of variation is the amount of output per one unit of input. - Once the equation is written, substitute the requested length.

1. The cost is always the same constant rate per foot and is \(0\) when \(x=0\), so the relationship has the form \(C=kx\). 2. The constant of variation is the unit price, \(k=2.40\), so \(C=2.40x\). 3. For \(7.5\) feet, \(C=2.40(7.5)=18\), so the cost is \(\$18.00\).

a) The cost has a constant ratio \(\frac{C}{x}=2.40\) and passes through \((0,0)\). b) \(k=2.40\), so \(C=2.40x\). c) \(\$18.00\)

- Think about what happens to the completion time when the number of equal-rate excavators increases. - Look for a quantity involving excavators and hours that stays fixed for the same total job. - Use the given four-excavator case to determine the constant for the model.

1. Under the stated equal-rate assumptions, increasing the number of excavators decreases the required time so that the product of the two quantities stays constant. Therefore, the relationship is an inverse variation. 2. The constant product is \(4\cdot15=60\). Thus, \(nt=60\). 3. Divide \(60\) by each number of excavators: for \(n=1\), \(t=60\); for \(n=2\), \(t=30\); for \(n=6\), \(t=10\); and for \(n=10\), \(t=6\). 4. Solving \(nt=60\) for \(t\) gives \(t=\frac{60}{n}\).

a) The relationship is an inverse variation, with constant product \(60\). b) <table><tr><td>Number of excavators \(n\)</td><td>\(1\)</td><td>\(2\)</td><td>\(6\)</td><td>\(10\)</td></tr><tr><td>Time \(t\) in hours</td><td>\(60\)</td><td>\(30\)</td><td>\(10\)</td><td>\(6\)</td></tr></table> c) \(t=\frac{60}{n}\)

- Check whether every pair has the same product. - Identify the product that appears most often. - Use the constant product to correct the inconsistent value.

1. Calculate the products: \(3 \cdot 20 = 60\), \(5 \cdot 12 = 60\), \(6 \cdot 10 = 60\), \(12 \cdot 6 = 72\), and \(15 \cdot 4 = 60\). 2. The pair \((12, 6)\) is inconsistent because its product is not \(60\). 3. For \(x = 12\), the corrected value is \(y = \frac{60}{12} = 5\).

a) The pair \((12, 6)\) is incorrect. b) The correct \(y\)-value is \(5\).

- Multiply the two values in each column. - An inverse variation has the same product for every pair. - Use the constant product to find the time for \(8\) landscapers.

1. Calculate the products: \(3 \cdot 16 = 48\), \(6 \cdot 8 = 48\), and \(12 \cdot 4 = 48\). Because every product is the same, the relationship is an inverse variation. 2. The constant product is \(48\). In context, this means the job requires \(48\) landscaper-hours, so one landscaper working alone would need \(48\,\text{h}\). 3. Since \(xy = 48\), substitute \(x = 8\): \(y = \frac{48}{8} = 6\). A crew of \(8\) landscapers would need \(6\,\text{h}\).

1. \(3 \cdot 16 = 6 \cdot 8 = 12 \cdot 4 = 48\), so the relationship is an inverse variation. 2. The constant product is \(48\). It represents \(48\) landscaper-hours of work. 3. A crew of \(8\) landscapers would need \(6\,\text{h}\).

- What must be true about the product \(xy\) for every ordered pair in an inverse variation? - How are \(x\), \(y\), and the constant of variation related? - Once you know the constant product, how can you find a missing coordinate?

1. Check whether each product \(xy\) has the same value: \(2 \cdot 24 = 48\), \(4 \cdot 12 = 48\), \(6 \cdot 8 = 48\), and \(16 \cdot 3 = 48\). 2. Because all four products equal \(48\), the points represent the same inverse variation with \(k = 48\). 3. The equation is \(y = \frac{48}{x}\). 4. For \(E(x, 1.5)\), solve \(1.5x = 48\): \(x = 48 \div 1.5 = 32\).

a) Yes. Each point has the constant product \(xy = 48\). b) \(y = \frac{48}{x}\) c) \(x = 32\), so the point is \(E(32, 1.5)\).

- How can you find \(k\) from one known ordered pair? - What quantity stays constant in an inverse variation? - Compare the outputs for \(x = 2\) and \(x = 6\). What factor relates the inputs, and what factor relates the outputs?

1. Use the known ordered pair to find the constant: \(k = xy = 4 \cdot 18 = 72\). 2. Substitute the requested inputs into \(y = \frac{72}{x}\): \(y = 36, 24, 12, 8,\) and \(6\), respectively. 3. If an input \(x\) is replaced by \(3x\), then the new output is \(\frac{k}{3x} = \frac{1}{3}\left(\frac{k}{x}\right)\). Therefore, tripling \(x\) divides \(y\) by \(3\).

a) \(k = 72\) b) The completed table is: <table> <tbody> <tr><td>\(x\)</td><td>\(2\)</td><td>\(3\)</td><td>\(6\)</td><td>\(9\)</td><td>\(12\)</td></tr> <tr><td>\(y\)</td><td>\(36\)</td><td>\(24\)</td><td>\(12\)</td><td>\(8\)</td><td>\(6\)</td></tr> </tbody> </table> c) The \(y\)-value is divided by \(3\), because the product \(xy\) must remain constant.

- Translate the fixed supply into a constant product of horses and days. - Write the inverse equation before filling the table. - For the threshold, translate “at least \(18\) days” into an inequality and remember that the horse count is a whole number.

1. The fixed supply is \(120\) horse-days, so \(nd=120\) and \(d(n)=\frac{120}{n}\). 2. Evaluating gives \(d(2)=60\), \(d(4)=30\), \(d(6)=20\), \(d(8)=15\), \(d(10)=12\), and \(d(12)=10\). 3. The condition is \(\frac{120}{n}\ge18\). For positive \(n\), this gives \(n\le\frac{120}{18}\approx6.67\). Therefore, the greatest whole-number group size is \(6\).

a) \(d(n)=\frac{120}{n}\) b) <table><thead><tr><th>Number of horses</th><th>Number of days</th></tr></thead><tbody><tr><td>\(2\)</td><td>\(60\)</td></tr><tr><td>\(4\)</td><td>\(30\)</td></tr><tr><td>\(6\)</td><td>\(20\)</td></tr><tr><td>\(8\)</td><td>\(15\)</td></tr><tr><td>\(10\)</td><td>\(12\)</td></tr><tr><td>\(12\)</td><td>\(10\)</td></tr></tbody></table> c) \(6\) horses

- Use the original crew and time to find the constant product. - Write time as a function of crew size before evaluating the requested cases. - Translate “less than \(3\) hours” as a strict inequality and enforce a whole-number crew size.

1. The fixed work is \(3\cdot8=24\) employee-hours, so \(nt=24\) and \(t(n)=\frac{24}{n}\). 2. The times are \(t(2)=12\), \(t(4)=6\), \(t(5)=4.8\), and \(t(6)=4\) hours. 3. The strict deadline requires \(\frac{24}{n}<3\). For positive \(n\), this gives \(n>8\), so the minimum whole number is \(9\).

a) \(t(n)=\frac{24}{n}\) b) \(t(2)=12\,\text{h}\), \(t(4)=6\,\text{h}\), \(t(5)=4.8\,\text{h}\), \(t(6)=4\,\text{h}\) c) \(9\) employees

- Fixed distance means the product of speed and time is constant. - Write the inverse model before substituting any speeds. - For the deadline, translate “at most” into an inequality. - For the time saved, compare the two model outputs and then convert hours to minutes.

1. Fixed distance gives \(vt=45\), so \(t(v)=\frac{45}{v}\), an inverse variation. 2. \(t(10)=4.5\,\text{h}\), \(t(12)=3.75\,\text{h}\), \(t(15)=3\,\text{h}\), and \(t(18)=2.5\,\text{h}\). 3. The condition \(\frac{45}{v}\le2.5\) gives \(v\ge18\), so the minimum speed is \(18\,\text{mi/h}\). 4. Increasing from \(12\) to \(15\,\text{mi/h}\) changes the time from \(3.75\) to \(3\) hours. The savings are \(0.75\,\text{h}=45\,\text{min}\).

a) \(t(v)=\frac{45}{v}\) b) \(4.5\,\text{h}\), \(3.75\,\text{h}\), \(3\,\text{h}\), and \(2.5\,\text{h}\), respectively c) \(18\,\text{mi/h}\) d) \(45\,\text{min}\)

- Identify the quantity that remains fixed. - Consider what happens to the number of cups when each cup receives more punch. - Relate the total volume, the number of cups, and the volume in each cup.

1. The total punch volume is fixed at \(384\,\text{fl oz}\), so the number of cups \(n\) and volume per cup \(v\) satisfy the constant product \(nv=384\). A constant product is the defining structure of an inverse variation. 2. Solving for the number of cups gives \(n=\frac{384}{v}\). 3. For \(v=6\), \(n=\frac{384}{6}=64\). 4. For \(v=8\), \(n=\frac{384}{8}=48\).

a) This is an inverse variation because the fixed total gives the constant product \(nv=384\). b) \(n=\frac{384}{v}\) c) At \(6\,\text{fl oz}\) per cup, \(64\) cups can be filled. At \(8\,\text{fl oz}\) per cup, \(48\) cups can be filled.

- Start with the area formula for a rectangle. - Think about how one side must change when the other side increases but the area stays fixed. - Test the effect of tripling one side with a numerical example.

1. The area formula is \(A = ab\). Since \(A = 48\), the side lengths satisfy \(ab = 48\), so the relationship is an inverse variation. 2. Possible ordered pairs include \((4, 12)\), \((6, 8)\), and \((2, 24)\), because the product of the coordinates in each pair is \(48\). 3. If the original dimensions satisfy \(ab = 48\), tripling the length gives \((3a)b_{\text{new}} = 48\). Since \(ab = 48\), it follows that \(3ab_{\text{new}} = ab\), so \(b_{\text{new}} = \frac{b}{3}\). The width is divided by \(3\).

a) The relationship is an inverse variation because \(ab = 48\) is constant. b) Possible ordered pairs are \((4, 12)\), \((6, 8)\), and \((2, 24)\). c) The width is divided by \(3\), because tripling one factor requires dividing the other factor by \(3\) to keep the product constant.

- In part a, think about what happens when the copier runs for twice as long. - In part c, the total number of pages is fixed. How does the time change when the print rate increases? - Use a constant ratio to identify direct variation and a constant product to identify inverse variation.

1. For part a, multiply the rate by the time: \(45\cdot4=180\), \(45\cdot10=450\), and \(45\cdot15=675\). 2. The number of pages is \(P=45t\), so the ratio \(\frac{P}{t}=45\) is constant. This is a direct variation. 3. For part c, divide the fixed number of pages by each print rate: \(540\div30=18\), \(540\div45=12\), and \(540\div60=9\). 4. For part d, rate \(r\) and time \(t\) satisfy \(rt=540\). Their product is constant, so the relationship is an inverse variation.

a) \(180\) pages, \(450\) pages, and \(675\) pages b) direct variation c) \(18\) minutes, \(12\) minutes, and \(9\) minutes d) <table><tbody><tr><td>Print rate (pages per minute)</td><td>\(30\)</td><td>\(45\)</td><td>\(60\)</td></tr><tr><td>Time (minutes)</td><td>\(18\)</td><td>\(12\)</td><td>\(9\)</td></tr></tbody></table> The relationship is an inverse variation.

- Worker-hours are the product of the number of workers and the number of hours each worker works. - For the same job, decide whether adding workers makes the duration longer or shorter. - In part c, each room adds the same amount of work.

1. Divide the fixed \(120\) worker-hours by the number of workers: \(120\div2=60\), \(120\div3=40\), \(120\div5=24\), and \(120\div8=15\). 2. The duration is \(d=\frac{120}{p}\). Because \(pd=120\), duration varies inversely with the number of workers. 3. Multiply \(15\) worker-hours per room by the number of rooms: \(15\cdot4=60\), \(15\cdot6=90\), and \(15\cdot10=150\). 4. The total worker-hours \(H\) satisfy \(H=15r\), where \(r\) is the number of rooms. This is a direct variation.

a) \(60\) hours, \(40\) hours, \(24\) hours, and \(15\) hours b) \(d=\frac{120}{p}\) c) \(60\) worker-hours, \(90\) worker-hours, and \(150\) worker-hours d) direct variation

- Divide the fixed total cost by the number of students. - Use an inequality to represent “no more than \(\text{€}18\).” - When the result for the number of students is not a whole number, decide which direction to round so the cost limit is met.

1. Divide the fixed cost by each group size: \(480 \div 20 = \text{€}24\), \(480 \div 24 = \text{€}20\), and \(480 \div 30 = \text{€}16\). 2. The cost per student is \(p = \frac{480}{n}\). 3. The condition is \(\frac{480}{n} \le 18\). Since \(n > 0\), multiply by \(n\) to get \(480 \le 18n\), so \(n \ge \frac{480}{18} = \frac{80}{3} \approx 26.67\). 4. The number of students must be a whole number. With \(26\) students, each would pay about \(\text{€}18.46\), which is too much; with \(27\) students, each would pay about \(\text{€}17.78\). Therefore, at least \(27\) students must attend.

a) With \(20\) students: \(\text{€}24\); with \(24\) students: \(\text{€}20\); with \(30\) students: \(\text{€}16\). b) \(p = \frac{480}{n}\) c) At least \(27\) students must attend.

- Begin with the area formula for a rectangle. - In an inverse variation, multiplying one quantity by a factor divides the other by the same factor. - Calculate the perimeter before and after changing the dimensions in a specific example.

1. The dimensions must satisfy \(xy = 24\). Four possible ordered pairs are \((1, 24)\), \((2, 12)\), \((3, 8)\), and \((4, 6)\). 2. Solving for the width gives \(y = \frac{24}{x}\), so the relationship is an inverse variation. If the length is tripled, the width is divided by \(3\). 3. Test the perimeter claim using \((x, y) = (4, 6)\). The original perimeter is \(2 \cdot (4 + 6) = 20\,\text{m}\). Doubling the length and halving the width gives dimensions \((8, 3)\), whose perimeter is \(2 \cdot (8 + 3) = 22\,\text{m}\). Since \(20 \ne 22\), the claim is false.

a) One possible set is \((1, 24)\), \((2, 12)\), \((3, 8)\), and \((4, 6)\). b) \(y = \frac{24}{x}\), so this is an inverse variation. Tripling the length divides the width by \(3\). c) No. For example, a \(4\,\text{m}\) by \(6\,\text{m}\) rectangle has perimeter \(20\,\text{m}\), while an \(8\,\text{m}\) by \(3\,\text{m}\) rectangle has perimeter \(22\,\text{m}\).

- Recover the fixed total cost first. - Write contribution per person as an inverse function of group size. - For the final comparison, distinguish the new contribution from the amount of increase.

1. The fixed cost is \(12\cdot15=180\) dollars, so \(cn=180\) and \(c(n)=\frac{180}{n}\). 2. For \(n=18\), \(c=\frac{180}{18}=10\), so each person pays \(\$10.00\). 3. If three leave, \(n=9\), so \(c=\frac{180}{9}=20\). The contribution increases by \(20-15=5\) dollars.

a) \(cn=180\), so \(c(n)=\frac{180}{n}\). b) \(\$10.00\). c) The contribution increases by \(\$5.00\).

- Express food supply in person-days. - After some food has been used, form a new constant product for the remaining supply. - Keep the full-supply model separate from the remaining-supply model.

1. The full supply is \(6\cdot20=120\) person-days, so \(D(n)=\frac{120}{n}\). 2. In the first \(4\) days, \(6\cdot4=24\) person-days are used, leaving \(96\) person-days. For the remaining food, \(R(n)=\frac{96}{n}\). 3. With \(8\) people, \(R(8)=12\), so the remaining food lasts \(12\) more days. 4. For the full original supply, \(D(8)=\frac{120}{8}=15\) days.

a) \(D(n)=\frac{120}{n}\). b) \(R(n)=\frac{96}{n}\); the remaining food lasts \(12\) more days. c) \(15\) days.

- Express the whole job in landscaper-hours. - Subtract the work completed before the crew size changes. - Apply inverse variation to the remaining fixed amount of work.

1. The full job requires \(4\cdot9=36\) landscaper-hours, so \(T(n)=\frac{36}{n}\). 2. The first \(3\) hours use \(4\cdot3=12\) landscaper-hours, leaving \(24\) landscaper-hours. Thus the remaining-time model is \(R(n)=\frac{24}{n}\). 3. With \(6\) landscapers, \(R(6)=4\) hours. The total project time is \(3+4=7\) hours.

a) \(T(n)=\frac{36}{n}\). b) \(24\) landscaper-hours remain, so \(R(n)=\frac{24}{n}\). c) \(7\,\text{h}\).

- Use truck-days to represent each fixed amount of work. - The breakdown changes the constant for the remaining-work model, not the inverse structure. - Compare the new total duration with the original planned duration.

1. The full job is \(3\cdot20=60\) truck-days, so \(T(n)=\frac{60}{n}\). 2. The first \(4\) days complete \(3\cdot4=12\) truck-days, leaving \(48\) truck-days. Thus \(R(n)=\frac{48}{n}\). 3. With \(2\) trucks, the remaining time is \(24\) days, so the total time is \(4+24=28\) days. 4. The delay is \(28-20=8\) days.

a) \(T(n)=\frac{60}{n}\). b) \(R(n)=\frac{48}{n}\). c) The project is delayed by \(8\) days.

- Find the constant amount of work in worker-hours. - Write the inverse model before solving the reverse question. - Distinguish the total crew size from the number newly hired.

1. The harvest requires \(5\cdot8=40\) worker-hours, so \(t(n)=\frac{40}{n}\). 2. Set \(t=5\): \(5n=40\), so \(n=8\) workers are needed in total. 3. The owner already has \(5\) workers, so \(8-5=3\) additional workers are needed.

a) \(t(n)=\frac{40}{n}\). b) \(8\) workers in total. c) \(3\) additional workers.

- Identify the constant product of crew size and completion time. - Use the same inverse model for both crew sizes. - Separate what the mathematical model predicts from whether its assumptions remain realistic.

1. The job requires \(3\cdot12=36\) landscaper-hours, so \(t(n)=\frac{36}{n}\). 2. With \(4\) landscapers, \(t=\frac{36}{4}=9\) hours. 3. With \(40\) landscapers, the model gives \(t=\frac{36}{40}=0.9\) hour, or \(54\) minutes, so the mathematical claim is correct under the model. 4. In reality, limited workspace, travel, and coordination can prevent all workers from contributing independently at the assumed rate.

a) \(t(n)=\frac{36}{n}\). b) \(9\,\text{h}\). c) The model predicts \(0.9\,\text{h}=54\,\text{min}\), so the mathematical claim is true, but real crowding and coordination can invalidate the constant-rate assumption.

- When one quantity doubles, decide whether the other quantity doubles or is cut in half. - What product remains constant in the vegetable-preparation situation?

1. Situation 1 is a direct variation. Because the amount of water per serving is constant, doubling the number of servings doubles the amount of water. 2. Situation 2 is an inverse variation. Under the stated assumptions, doubling the number of helpers cuts the preparation time in half. 3. The fixed amount of work is \(4\cdot60=240\) helper-minutes. 4. Six helpers need \(\frac{240}{6}=40\) minutes.

a) Situation 1 is direct variation because water per serving is constant. Situation 2 is inverse variation because helper count times preparation time is constant for the fixed job. b) \(40\,\text{min}\), since \(4\cdot60=240\) helper-minutes and \(240\div6=40\).

- Find the total number of landscaper-hours required for the job. - Use the constant product to calculate each unknown value. - Explain the relationship by describing what remains constant.

1. The job requires \(4 \cdot 12 = 48\) landscaper-hours. 2. For a), \(6\) landscapers would need \(48 \div 6 = 8\,\text{h}\). 3. For b), completing the job in \(4\,\text{h}\) requires \(48 \div 4 = 12\) landscapers. 4. For c), the product of the number of landscapers and the time is constant at \(48\). Therefore, multiplying the number of workers by a factor divides the time by the same factor.

a) Six landscapers would need \(8\,\text{h}\). b) A total of \(12\) landscapers would be needed. c) It is an inverse variation because the product of the number of landscapers and the time remains constant.

- Analyze the two situations separately. - In Situation 1, identify the constant rate in miles per gallon. - In Situation 2, first find the total number of horse-days of feed.

1. Situation 1 is a direct variation because fuel used is proportional to distance when fuel efficiency is constant. 2. The car uses \(\frac{375}{25}=15\) gallons. 3. Situation 2 is an inverse variation because the fixed supply represents \(10\cdot12=120\) horse-days of feed. 4. After \(2\) horses leave, \(8\) horses remain, so the feed lasts \(\frac{120}{8}=15\) days.

Situation 1: direct variation; \(15\) gallons. Fuel used is a constant multiple of distance: \(\frac{g}{d}=\frac{1}{25}\). Situation 2: inverse variation; \(15\) days. The fixed feed gives \(10\cdot12=120\) horse-days, so \(120\div8=15\).

- Decide what stays constant in each situation. - For part a, find the cost per pound. - For part b, find the total number of painter-hours required.

1. a) The price varies directly with the weight because the price per pound is constant. 2. The unit price is \(\$5.40\div3=\$1.80\) per pound. Therefore, \(7\) pounds cost \(\$1.80\cdot7=\$12.60\). 3. b) The time varies inversely with the number of painters because the amount of work is fixed. 4. The job requires \(4\cdot6=24\) painter-hours, so three painters need \(24\div3=8\) hours.

a) direct variation; \(\$12.60\) b) inverse variation; \(8\) hours

- Identify what remains fixed in each situation. - In part a, use the fixed ratio of water to concentrate. - In part b, divide the fixed total volume by the volume of one cup.

1. a) The amount of water varies directly with the amount of concentrate because the mixing ratio is fixed. 2. Each cup of concentrate requires \(10\div2=5\) cups of water, so \(5\) cups of concentrate require \(5\cdot5=25\) cups of water. 3. b) The number of cups varies inversely with the capacity of each cup because the total volume is fixed. 4. The number of \(3\)-fluid-ounce cups is \(120\div3=40\).

a) \(25\) cups of water b) \(40\) sample cups c) Part a: direct variation, because the water-to-concentrate ratio is constant. Part b: inverse variation, because cup capacity times number of cups is constant.

- Determine how many days the remaining feed would have lasted the original group. - Express the remaining supply in horse-days. - Divide by the new number of horses. - Decide whether fewer horses make a fixed supply last longer or for less time.

1. After \(5\) days, the feed would still last the original \(12\) horses for \(20 - 5 = 15\) days. 2. The remaining supply is \(12 \cdot 15 = 180\) horse-days. 3. After \(3\) horses are sold, \(12 - 3 = 9\) horses remain. 4. The remaining feed lasts \(180 \div 9 = 20\) more days. 5. This uses an inverse variation because, for a fixed amount of feed, decreasing the number of horses increases the number of days, while the product of horses and days remains constant.

The remaining feed will last the \(9\) horses for \(20\) more days. The relationship is an inverse variation because the product of the number of horses and the number of days is constant.

- Divide the fixed bus cost by each number of students. - For part b), calculate the cost per student both before and after the cancellations. - The question asks for the increase, so subtract the original cost from the new cost.

1. For a), divide the total cost by each group size: \(\$540.00 \div 20 = \$27.00\), \(\$540.00 \div 25 = \$21.60\), and \(\$540.00 \div 30 = \$18.00\). 2. For b), the original cost per student is \(\$540.00 \div 27 = \$20.00\). 3. After \(3\) cancellations, \(27 - 3 = 24\) students remain. 4. The new cost per student is \(\$540.00 \div 24 = \$22.50\). 5. The increase is \(\$22.50 - \$20.00 = \$2.50\).

a) <table> <thead> <tr><th>Number of students</th><th>Cost per student</th></tr> </thead> <tbody> <tr><td>\(20\)</td><td>\(\$27.00\)</td></tr> <tr><td>\(25\)</td><td>\(\$21.60\)</td></tr> <tr><td>\(30\)</td><td>\(\$18.00\)</td></tr> </tbody> </table> b) The cost increases by \(\$2.50\) per remaining student.

- Find the fixed total length of twine. - Use piece length as the inverse-variation input. - Determine the new piece length before evaluating the model.

1. The roll is \(25\cdot8=200\) feet long, so \(N\ell=200\) and \(N(\ell)=\frac{200}{\ell}\). 2. The new piece length is \(8-3=5\) feet, so \(N(5)=\frac{200}{5}=40\).

a) \(N(\ell)=\frac{200}{\ell}\). b) \(40\) pieces.

- Multiply the number of horses by the number of days in each case. - Use the result from part a) to check the claim in part b). - For part c), consider whether every horse consumes exactly the same amount and whether all feed is used.

1. The products are \(12 \cdot 20 = 240\) and \(15 \cdot 16 = 240\). The value \(240\) represents \(240\) horse-days of feed, meaning the supply would feed one horse for \(240\) days under the model. 2. For \(80\) horses, the model gives \(240 \div 80 = 3\) days, so the farmer’s calculation is correct. 3. In reality, horses may eat different amounts, some feed may be wasted or spoil, and managing a much larger herd may change how efficiently the feed is distributed.

a) The constant product is \(240\), representing \(240\) horse-days of feed. b) The calculation is correct because \(80 \cdot 3 = 240\). c) Different feeding needs, waste, spoilage, or distribution problems could make the actual duration different.

- Identify the fixed total amount of oil. - Write the constant-product relation between daily use and duration. - Use the same model in opposite directions for parts b) and c).

1. The fixed supply is \(12\cdot80=960\) gallons, so \(Du=960\) and \(D(u)=\frac{960}{u}\). 2. For \(u=16\), \(D=\frac{960}{16}=60\) days. 3. For \(D=120\), \(120u=960\), so \(u=8\) gallons per day.

a) \(D(u)=\frac{960}{u}\). b) \(60\) days. c) \(8\,\text{gal}\) per day.

- Identify the total cost that remains fixed. - Divide the total cost by each group size. - Compare the products \(np\) and observe what happens when \(n\) doubles.

1. Use \(p = \frac{480}{n}\). The costs are \(\$40.00\), \(\$30.00\), \(\$24.00\), \(\$20.00\), \(\$15.00\), and \(\$12.00\), respectively. 2. The relationship is an inverse variation because \(np = 480\), so the product is constant. 3. When the number of students doubles, the cost per student is cut in half.

1. <table> <tr><td>Number of students \(n\)</td><td>\(12\)</td><td>\(16\)</td><td>\(20\)</td><td>\(24\)</td><td>\(32\)</td><td>\(40\)</td></tr> <tr><td>Cost per student \(p\)</td><td>\(\$40.00\)</td><td>\(\$30.00\)</td><td>\(\$24.00\)</td><td>\(\$20.00\)</td><td>\(\$15.00\)</td><td>\(\$12.00\)</td></tr> </table> 2. It is an inverse variation with a constant product. 3. Doubling the number of students halves the cost per student.

- Separate the food already consumed from the food that remains. - Use the remaining person-days as the constant in the inverse model. - Compare the new remaining duration with the remaining duration in the original plan.

1. The original supply is \(24\cdot10=240\) person-days. The first \(4\) days use \(24\cdot4=96\), leaving \(144\) person-days. 2. Therefore \(R(n)=\frac{144}{n}\). 3. After \(6\) people leave, \(n=18\), so \(R(18)=8\) days. The original plan had \(6\) days remaining, so the food lasts \(2\) days longer.

a) \(144\) person-days. b) \(R(n)=\frac{144}{n}\). c) \(2\) days longer.

- Identify the total amount of flour that remains fixed as the daily use changes. - Write the number of days as the fixed total divided by the daily use before substituting values. - For part c, “at least \(20\) days” places an upper limit on the daily amount used.

1. The fixed supply is \(20\cdot12=240\,\text{lb}\), so \(ud=240\) and \(d(u)=\frac{240}{u}\). 2. For \(u=15\), \(d(15)=\frac{240}{15}=16\) days. 3. To last at least \(20\) days, require \(\frac{240}{u}\ge20\) with \(u>0\). 4. This gives \(240\ge20u\), so \(u\le12\). The greatest daily use is \(12\,\text{lb}\).

a) \(ud=240\); \(d(u)=\frac{240}{u}\) b) \(16\) days c) \(12\,\text{lb}\) per day

- Find the fixed amount of work that remains after the staffing change. - Use worker count as the inverse-variation input for the remaining job. - The question asks for time after the workers leave, not total elapsed project time.

1. The whole project is \(8\cdot15=120\) worker-days. The first \(3\) days complete \(8\cdot3=24\), leaving \(96\) worker-days. 2. Therefore \(R(n)=\frac{96}{n}\). 3. Six workers remain, so \(R(6)=\frac{96}{6}=16\) days.

a) \(96\) worker-days. b) \(R(n)=\frac{96}{n}\). c) \(16\) more days.

- Use a complete pair to find the constant product. - Divide the constant product by the known value to find its partner. - Check that every completed pair has the same product.

1. Use the known pair \((3, 16)\) to find the constant product: \(k = 3 \cdot 16 = 48\). 2. Apply \(xy = 48\) to each missing entry. 3. When \(x = 4\), \(y = \frac{48}{4} = 12\). 4. When \(y = 8\), \(x = \frac{48}{8} = 6\). 5. When \(x = 12\), \(y = \frac{48}{12} = 4\). 6. The final pair is consistent because \(24 \cdot 2 = 48\).

<table> <tr><td>\(x\)</td><td>\(3\)</td><td>\(4\)</td><td>\(6\)</td><td>\(12\)</td><td>\(24\)</td></tr> <tr><td>\(y\)</td><td>\(16\)</td><td>\(12\)</td><td>\(8\)</td><td>\(4\)</td><td>\(2\)</td></tr> </table> Each pair satisfies \(xy = 48\).

- Calculate the product \(xy\) for every pair in each table. - Compare what happens when \(x\) doubles in different parts of Table B. - Think of other decreasing relationships that do not have a constant product.

1. For Table A, calculate the products: \(1 \cdot 120 = 120\), \(2 \cdot 60 = 120\), \(3 \cdot 40 = 120\), \(5 \cdot 24 = 120\), and \(8 \cdot 15 = 120\). Table A represents an inverse variation. 2. For Table B, the products are \(1 \cdot 120 = 120\), \(2 \cdot 60 = 120\), \(3 \cdot 30 = 90\), and \(4 \cdot 15 = 60\). Because the products are not constant, Table B does not represent an inverse variation. 3. In both tables, \(y\) decreases as \(x\) increases. That pattern alone does not establish inverse variation; the products \(xy\) must be constant for all pairs.

Only Table A represents an inverse variation because every pair has \(xy = 120\). Table B has products \(120\), \(120\), \(90\), and \(60\). A decreasing pattern alone is insufficient; inverse variation requires a constant product.

- Calculate \(xy\) for every pair in each table. - Compare inverse variation with direct variation. - Use the constant product to write the inverse-variation equation.

1. For Data Set A, the products are \(3 \cdot 20 = 60\), \(5 \cdot 12 = 60\), and \(10 \cdot 6 = 60\). The ratios are not constant because \(\frac{20}{3} \ne \frac{12}{5}\). 2. For Data Set B, the products are not constant, but the ratios are: \(\frac{7.5}{3} = \frac{12.5}{5} = \frac{25}{10} = 2.5\). Thus, Data Set B represents a direct variation. 3. Data Set A represents an inverse variation because \(xy = 60\). Its equation is \(y = \frac{60}{x}\).

1. Data Set A has a constant product of \(60\) and does not have a constant ratio. Data Set B does not have a constant product and has a constant ratio of \(2.5\). 2. Data Set A represents an inverse variation. 3. \(y = \frac{60}{x}\)

- In an inverse variation, what happens to one variable when the other is multiplied by a factor? - What product stays constant? - How are the scale factors for the input and output related?

1. The input changes from \(4\) to \(12\), so it is multiplied by \(3\). In an inverse variation, the output is divided by the same factor. Therefore, \(f(12) = 9 \div 3 = 3\). 2. Find the constant: \(a = x f(x) = 4 \cdot 9 = 36\). Checking gives \(f(12) = \frac{36}{12} = 3\). 3. Multiplying the input by \(4\) divides the output by \(4\). Dividing the input by \(10\) multiplies the output by \(10\).

a) \(f(12) = 3\), because multiplying the input by \(3\) divides the output by \(3\). b) \(a = 36\), and \(f(12) = \frac{36}{12} = 3\). c) Multiplying the input by \(4\) divides the output by \(4\). Dividing the input by \(10\) multiplies the output by \(10\).

- Identify the total amount of money that remains fixed. - Divide the budget by each price per liter. - Use the constant product to write the equation.

1. Divide the fixed budget by each price per liter: \(120 \div 4 = 30\), \(120 \div 6 = 20\), \(120 \div 8 = 15\), \(120 \div 10 = 12\), and \(120 \div 12 = 10\). 2. Since the product of the price \(x\) and the amount of paint \(y\) is \(120\), the equation is \(y = \frac{120}{x}\). 3. The constant \(k = 120\) represents the fixed budget of \(\text{€}120\).

a) <table> <tr><td>Price per liter</td><td>\(\text{€}4\)</td><td>\(\text{€}6\)</td><td>\(\text{€}8\)</td><td>\(\text{€}10\)</td><td>\(\text{€}12\)</td></tr> <tr><td>Paint in liters</td><td>\(30\)</td><td>\(20\)</td><td>\(15\)</td><td>\(12\)</td><td>\(10\)</td></tr> </table> b) \(y = \frac{120}{x}\). The constant \(k = 120\) represents the fixed budget in euros.

- Compare a consistent multiplicative quantity across several known pairs in each table. - Once you identify the inverse-variation table, use its constant relationship with the pair containing the missing value. - Check that the completed pair follows the same pattern as the other pairs in that table.

1. For Table A, the known products are \(1\cdot12=12\), \(2\cdot6=12\), \(3\cdot4=12\), and \(6\cdot2=12\). The product is constant, so Table A represents an inverse variation. 2. For Table B, the products begin \(1\cdot2=2\) and \(2\cdot4=8\), so they are not constant. Table B is a direct variation because \(\frac{y}{x}=2\). 3. In Table A, \(8y=12\), so \(y=\frac{12}{8}=1.5\).

a) Table A represents an inverse variation because \(xy=12\) for every known pair. b) The missing value is \(1.5\).

- First identify the rate at which the distance between the riders closes. - For a fixed distance, compare travel time with the actual closing rate rather than with one rider's speed alone. - An inverse-variation model has a constant product between the two varying quantities.

1. The closing rate is \(q=v+(v+4)=2v+4\) miles per hour. 2. With distance \(d\) fixed, \(t=\frac{d}{q}\), so \(t\) varies inversely with the closing rate \(q\). 3. Substituting \(q=2v+4\) gives \(t=\frac{d}{2v+4}\). 4. For \(d=48\) and \(v=10\), \(t=\frac{48}{24}=2\) hours. 5. If Maya rides at \(\frac{v}{2}\), the closing rate is \(q=v+\frac{v}{2}=\frac{3v}{2}\), so \(t=\frac{d}{q}=\frac{2d}{3v}\).

a) \(q=2v+4\) and \(t(q)=\frac{d}{q}\); for fixed \(d\), \(qt=d\) is constant. b) \(t=\frac{d}{2v+4}\); \(t=2\,\text{h}\). c) \(q=\frac{3v}{2}\) and \(t=\frac{2d}{3v}\).

- Identify the fixed total amount of water first. - Use daily use as one of the two varying quantities. - Write the constant-product relationship before substituting the reduced daily-use expression.

1. The fixed water supply is \(tv\) liters. 2. Therefore \(Du=tv\), so \(D(u)=\frac{tv}{u}\). This is inverse variation in \(D\) and \(u\). 3. After the reduction, \(u=v-w\), so the new duration is \(\frac{tv}{v-w}\). 4. The additional time is \(\frac{tv}{v-w}-t\). 5. For the given values, \(\frac{20\cdot60}{48}-20=25-20=5\) days.

a) \(Du=tv\), so \(D(u)=\frac{tv}{u}\). b) \(\frac{tv}{v-w}-t\) days. c) \(5\) additional days.

- Express the planned daily rate as a quotient of planned work and planned time. - Determine how both the total work and total time change in the actual situation. - Form the difference between the two rational rate expressions before substituting values.

1. The planned daily rate is \(\frac{a}{b}\). 2. The actual total is \(a+c\) lines, and the actual time is \(b-d\) days. 3. The actual daily rate is \(\frac{a+c}{b-d}\). 4. The difference in rates is \(\frac{a+c}{b-d}-\frac{a}{b}\). 5. Substituting the values gives \(\frac{1500+300}{20-5}-\frac{1500}{20}=\frac{1800}{15}-75=120-75=45\) lines per day.

The expression is \(\frac{a+c}{b-d}-\frac{a}{b}\). For the given values, the programmer writes \(45\) more lines per day than planned.

- Use the original time and rate to identify the fixed tank volume. - Treat the actual pumping rate as the independent varying quantity. - Write the inverse model before replacing the rate by \(r+s\).

1. The fixed tank volume is \(tr\) gallons, so \(Tq=tr\). 2. Thus \(T(q)=\frac{tr}{q}\), an inverse variation between filling time and pumping rate. 3. With the upgraded rate \(q=r+s\), the new time is \(\frac{tr}{r+s}\), so the time saved is \(t-\frac{tr}{r+s}\). 4. For the given values, \(10-\frac{12000}{1500}=10-8=2\) hours. 5. The expression \(\frac{tr}{r+s}\) is the upgraded pump's filling time.

a) \(Tq=tr\), so \(T(q)=\frac{tr}{q}\). b) \(t-\frac{tr}{r+s}\) hours. c) \(2\) hours; \(\frac{tr}{r+s}\) is the upgraded filling time.

- Identify the quantity that stays fixed when the group size changes. - Use the number actually attending as the inverse-variation input. - In the general case, express that attending number in terms of \(n\) and \(k\).

1. The fixed supply is \(24\cdot10=240\) meal packs, so \(Tp=240\) and \(T(p)=\frac{240}{p}\). 2. If \(4\) campers cancel, \(p=20\), so \(T=\frac{240}{20}=12\) days, which is \(2\) additional days. 3. In general the fixed supply is \(nt\) packs and the attending group is \(p=n-k\). 4. Therefore \(T=\frac{nt}{n-k}\).

a) \(Tp=240\), so \(T(p)=\frac{240}{p}\). b) \(2\) additional days. c) \(T=\frac{nt}{n-k}\).

- Keep the amount of dissolved salt fixed while the water volume changes. - Ask what product remains constant in the concentration-volume relationship. - For the target concentration, first determine the total salt mass required in the full volume.

1. Concentration is salt amount divided by water volume, so \(Cw=k\) and \(C(w)=\frac{k}{w}\). Thus \(C\) varies inversely with \(w\) while the salt amount is fixed. 2. At one-third capacity, \(w=\frac{V}{3}\), so \(C=\frac{k}{V/3}=\frac{3k}{V}\). At full capacity, \(w=V\), so \(C=\frac{k}{V}\). 3. A concentration of \(m\) grams per liter in \(V\) liters requires \(mV\) grams of salt, so the amount to add is \(mV-k\) grams.

a) \(Cw=k\), so \(C(w)=\frac{k}{w}\). b) One-third full: \(\frac{3k}{V}\) grams per liter; full: \(\frac{k}{V}\) grams per liter. c) \(mV-k\) grams.

- Identify the fixed distance that must be closed before either meeting occurs. - Use relative speed as the varying rate in the inverse model. - Decide whether the gap closes through the sum or the difference of the two speeds.

1. One full circumference \(C\) must be closed, so \(tq=C\) and \(t(q)=\frac{C}{q}\). Thus time varies inversely with relative speed. 2. In opposite directions, the separation closes at \(q=v_{\alpha}+v_{\beta}\), so \(t_O=\frac{C}{v_{\alpha}+v_{\beta}}\). 3. In the same direction, Alpha gains on Beta at \(q=v_{\alpha}-v_{\beta}\), so \(t_S=\frac{C}{v_{\alpha}-v_{\beta}}\). 4. Opposite-direction speeds combine, while same-direction catching depends on the difference in speeds.

a) \(tq=C\), so \(t(q)=\frac{C}{q}\). b) \(q=v_{\alpha}+v_{\beta}\), so \(t_O=\frac{C}{v_{\alpha}+v_{\beta}}\). c) \(q=v_{\alpha}-v_{\beta}\), so \(t_S=\frac{C}{v_{\alpha}-v_{\beta}}\). d) Opposite-direction closing uses the sum of speeds; same-direction catching uses their difference.

- Consider what happens to a product when one factor is doubled and the other is halved. - Pay attention to the square on the variable in the denominator. - Write the new expression as a multiple of the original expression.

1. Replacing \(x\) with \(2x\) and \(y\) with \(\frac{1}{2}y\) gives \(T_{\text{new}}=\frac{(2x)(\frac{1}{2}y)}{z^2}=\frac{xy}{z^2}=T\). Thus, \(T\) is unchanged. 2. Because \(z^2\) is in the denominator, increasing \(z\) increases the denominator and decreases \(T\). 3. Replacing \(z\) with \(2z\) gives \(T_{\text{new}}=\frac{xy}{(2z)^2}=\frac{xy}{4z^2}=\frac{1}{4}T\).

a) \(T\) stays the same. b) \(T\) decreases. c) \(T\) is multiplied by \(\frac{1}{4}\).

- Identify the product that represents the fixed amount of hay. - Relate the original horse-days to the new horse-days before solving for the new duration. - After deriving the equation, substitute the changed horse count and compare the result with the original duration.

1. The amount of hay is fixed, so the product of the number of horses and the number of days is constant. 2. The original and new situations satisfy \(pd = nx\). 3. Solving for \(x\) gives \(x = \frac{pd}{n}\). 4. If \(n = \frac{p}{2}\), then \(x = \frac{pd}{p/2} = 2d\). Halving the number of horses doubles the number of days because the relationship is an inverse variation.

a) \(pd = nx\), or \(x = \frac{pd}{n}\) b) The supply lasts \(2d\) days, so the number of days doubles.

- How are \(x\), \(y\), and \(k\) related in an inverse variation? - How can substitution show whether a point satisfies a function rule? - What happens to a fraction when a variable in its denominator is multiplied by \(4\)? - Try expressing the new output as a multiple of the original output.

1. Substitute \(P(4, 1.5)\) into the inverse-variation rule: \(1.5 = \frac{k}{4}\), so \(k = 1.5 \cdot 4 = 6\). 2. For \(x = -0.5\), the rule gives \(y = \frac{6}{-0.5} = -12\). Therefore, \(Q\) lies on the graph. 3. Let \(y_R = \frac{k}{x_R}\). Since \(x_S = 4x_R\), \(y_S = \frac{k}{4x_R} = \frac{1}{4}\left(\frac{k}{x_R}\right) = \frac{1}{4}y_R\). 4. Thus, the \(y\)-coordinate of \(S\) is one-fourth of the \(y\)-coordinate of \(R\).

a) \(k = 6\) b) Yes. Since \(\frac{6}{-0.5} = -12\), point \(Q\) lies on the graph. c) The \(y\)-coordinate of \(S\) is \(\frac{1}{4}\) of the \(y\)-coordinate of \(R\).

- Express the total prize using \(n\) and \(x\). - Divide the same total prize among \(m\) people. - Substitute \(m = 4n\) and simplify.

1. In the original situation, the total prize is \(G = nx\). 2. When \(m\) people share the prize, each receives \(y = \frac{G}{m}\). Substituting \(G = nx\) gives \(y = \frac{nx}{m}\). 3. If \(m = 4n\), then \(y = \frac{nx}{4n} = \frac{x}{4}\). Each person receives one-fourth of the original amount.

a) \(y = \frac{nx}{m}\) b) When \(m = 4n\), \(y = \frac{x}{4}\), so the amount per person is divided by \(4\).

- Express each changed variable as a factor times its original value. - To keep \(m\) constant, keep the ratio \(\frac{n}{p}\) constant. - In part c, look for a common factor that cancels.

1. Replacing \(n\) with \(2n\) and \(p\) with \(3p\) gives \(m_{\text{new}}=\frac{k(2n)}{3p}=\frac{2}{3}m\). 2. To keep \(\frac{n}{p}\) unchanged when \(n\) becomes \(\frac{1}{4}n\), \(p\) must also become \(\frac{1}{4}p\). 3. Replacing \(n\) with \(cn\) and \(p\) with \(cp\) gives \(m_{\text{new}}=\frac{k(cn)}{cp}=\frac{kn}{p}=m\), because the positive factor \(c\) cancels.

a) \(m\) is multiplied by \(\frac{2}{3}\). b) \(p\) must be reduced to one-fourth of its original value. c) \(\frac{k(cn)}{cp}=\frac{kn}{p}=m\)

- Begin with the area formula for a rectangle. - Determine which values are possible for a side length. - Recall the graph shape of \(y = \frac{k}{x}\). - Interpret the product of the length and width.

1. Since \(xy = 18\), when \(x = 4.5\), \(y = \frac{18}{4.5} = 4\,\text{m}\). 2. Solving the area equation for \(y\) gives \(y(x) = \frac{18}{x}\). Since a length must be positive, the contextual domain is \(x > 0\). 3. For \(x > 0\), the graph is the first-quadrant branch of a hyperbola. 4. The products are \(2 \cdot 9 = 18\) and \(6 \cdot 3 = 18\). The constant product represents the fixed area of \(18\,\text{m}^2\).

1. \(y = 4\,\text{m}\) 2. \(y(x) = \frac{18}{x}\), with \(x > 0\) 3. The graph is the first-quadrant branch of a hyperbola. 4. Both products equal \(18\,\text{m}^2\), the area of the flower bed.

- Which positive factor pairs have a product of \(36\)? - What relationship between the coordinates holds on the line \(y = x\)? - Write the condition for point \(R\) as an equation, then substitute it into the function rule.

1. Positive integer points satisfy \(xy = 36\). Pair each positive divisor of \(36\) with its corresponding quotient: \((1, 36)\), \((2, 18)\), \((3, 12)\), \((4, 9)\), \((6, 6)\), \((9, 4)\), \((12, 3)\), \((18, 2)\), and \((36, 1)\). 2. On the line \(y = x\), solve \(x = \frac{36}{x}\). Then \(x^2 = 36\), so \(x = 6\) or \(x = -6\). The intersection points are \((6, 6)\) and \((-6, -6)\). 3. Use \(x = 4y\) in \(y = \frac{36}{x}\): \(y = \frac{36}{4y}\), so \(4y^2 = 36\) and \(y^2 = 9\). Since \(x > 0\), use \(y = 3\), giving \(x = 12\). Thus, \(R = (12, 3)\).

1) \((1, 36)\), \((2, 18)\), \((3, 12)\), \((4, 9)\), \((6, 6)\), \((9, 4)\), \((12, 3)\), \((18, 2)\), and \((36, 1)\) 2) \((6, 6)\) and \((-6, -6)\) 3) \(R(12, 3)\)

- Treat the total worker-days as the constant of variation. - Use completion time itself as the inverse-variation input before substituting \(d-k\). - Distinguish the total required crew from the number added.

1. The fixed work is \(nd\) worker-days, so \(N(T)T=nd\) and \(N(T)=\frac{nd}{T}\). 2. The shorter time is \(T=d-k\), so \(N=\frac{nd}{d-k}\). 3. For \(n=12\), \(d=20\), and \(k=5\), \(N=\frac{240}{15}=16\) workers. 4. The additional number is \(16-12=4\).

a) \(N(T)=\frac{nd}{T}\). b) \(\frac{nd}{d-k}\) workers. c) \(16\) workers. d) \(4\) additional workers.

- What is the general shape of a graph whose variable is in the denominator? - Use a graphing tool to display both equations on the same coordinate plane. - Make sure the reciprocal graph is viewed for both positive and negative inputs. - What do common points of the two graphs represent?

1. Enter \(y = \frac{4}{x}\) and \(y = x + 3\) in a graphing tool and display both graphs on the same coordinate plane. 2. The reciprocal graph includes points such as \((1, 4)\), \((2, 2)\), \((4, 1)\), \((-1, -4)\), \((-2, -2)\), and \((-4, -1)\). The line has slope \(1\) and \(y\)-intercept \(3\). 3. The graphs intersect at \((1, 4)\) and \((-4, -1)\). 4. These intersection points are the solutions of the system.

The graphs intersect at \((1, 4)\) and \((-4, -1)\), so these are the solutions.

- Split the equation so that each side becomes a function rule. - Remember that \(x = 0\) is not in the domain of the reciprocal function. - Use a graphing tool to find the intersection points, then identify the requested \(x\)-coordinates. - How can substitution confirm that a proposed value satisfies the original equation?

1. Use \(f(x) = \frac{6}{x}\) and \(g(x) = x + 1\). 2. Enter both functions in a graphing tool and display them on the same coordinate plane. 3. The graphs intersect at \((2, 3)\) and \((-3, -2)\), so the solutions are \(x = 2\) and \(x = -3\). 4. Check \(x = 2\): \(\frac{6}{2} = 3\) and \(2 + 1 = 3\). 5. Check \(x = -3\): \(\frac{6}{-3} = -2\) and \(-3 + 1 = -2\). Both values satisfy the original equation.

a) \(f(x) = \frac{6}{x}\) and \(g(x) = x + 1\) b) The graphs intersect at \((2, 3)\) and \((-3, -2)\), so \(x = 2\) and \(x = -3\). c) \(\frac{6}{2} = 2 + 1 = 3\), and \(\frac{6}{-3} = -3 + 1 = -2\).

- Check which quantity is shown on each axis. - Multiply each speed by its corresponding time. - Relate speed, time, and distance. - Use the total distance to solve for the speed when the time is \(1.5\,\text{h}\).

1. From the graph, the pairs are \((2, 6)\), \((3, 4)\), \((4, 3)\), and \((6, 2)\). 2. The products are \(2 \cdot 6 = 12\), \(3 \cdot 4 = 12\), \(4 \cdot 3 = 12\), and \(6 \cdot 2 = 12\). Therefore, the relationship is an inverse variation, and the constant represents a trail distance of \(12\,\text{km}\). 3. For \(t = 1.5\,\text{h}\), solve \(1.5v = 12\): \(v = 8\,\text{km/h}\). The graph contains the point \((8, 1.5)\).

a) <table> <thead> <tr><th>Speed \(v\) in \(\text{km/h}\)</th><td>\(2\)</td><td>\(3\)</td><td>\(4\)</td><td>\(6\)</td></tr> </thead> <tbody> <tr><th>Time \(t\) in \(\text{h}\)</th><td>\(6\)</td><td>\(4\)</td><td>\(3\)</td><td>\(2\)</td></tr> </tbody> </table> b) The relationship is an inverse variation because \(vt = 12\) for every pair. The trail is \(12\,\text{km}\) long. c) The required average speed is \(8\,\text{km/h}\).

- Because the number of participants is a count, read only the plotted whole-number inputs. - In an inverse variation, compare products of the coordinates rather than differences or ratios. - Interpret the constant product using the total amount being shared.

1. From the graph, when \(x=3\), \(y=\$8\); when \(x=6\), \(y=\$4\). 2. Since \(xy=a\), use \((3,8)\): \(a=3\cdot8=24\). Thus, \(f(x)=\frac{24}{x}\), and \(a=24\) represents the total gift cost of \(\$24\). 3. The pairs \((3,8)\) and \((6,4)\) satisfy \(3\cdot8=24\) and \(6\cdot4=24\). 4. In an inverse variation, multiplying the participant count by \(3\) divides the amount per person by \(3\).

a) With \(3\) people, each pays \(\$8\). With \(6\) people, each pays \(\$4\). b) \(a=24\), representing the total cost of \(\$24\). c) For example, \(3\cdot8=6\cdot4=24\). Tripling the number of participants divides the amount per person by \(3\).

- Separate each student’s statement into individual claims. - A linear relationship has a straight-line graph and a constant rate of change. - Compare the time differences for two different speed doublings. - Check whether the products \(vt\) are constant at the marked points. - Use the defining factor relationship for inverse variation.

1. Jordan’s claim that the graph is linear is false. A linear decrease has a straight-line graph with constant slope, while this graph is curved. 2. Jordan’s claim about always saving one hour is also false. Doubling the speed from \(2\,\text{km/h}\) to \(4\,\text{km/h}\) changes the time from \(6\,\text{h}\) to \(3\,\text{h}\), a savings of \(3\,\text{h}\). Doubling from \(3\,\text{km/h}\) to \(6\,\text{km/h}\) changes the time from \(4\,\text{h}\) to \(2\,\text{h}\), a savings of \(2\,\text{h}\). 3. Maya’s claim is correct. The marked points satisfy \(2 \cdot 6 = 3 \cdot 4 = 4 \cdot 3 = 6 \cdot 2 = 12\), so the relationship is an inverse variation for a fixed \(12\,\text{km}\) trail. 4. In an inverse variation, doubling the speed divides the travel time by \(2\), so the time is cut in half.

Jordan’s claims are false. The graph is curved rather than linear, and doubling the speed does not produce a constant time savings: the savings can be \(3\,\text{h}\) or \(2\,\text{h}\) in the examples shown. Maya’s claims are correct. The products satisfy \(vt = 12\), and doubling the speed halves the travel time.

- Use \(A = xy\) for each rectangle. - Choose points that lie on grid intersections. - A larger constant area places the inverse-variation curve farther from the origin.

1. For graph \(p\), the point \((2, 2)\) is on the curve, so the area is \(2 \cdot 2 = 4\,\text{cm}^2\). 2. For graph \(q\), the point \((2, 4)\) is on the curve, so the area is \(2 \cdot 4 = 8\,\text{cm}^2\). 3. For graph \(r\), the point \((4, 4)\) is on the curve, so the area is \(4 \cdot 4 = 16\,\text{cm}^2\). 4. Since \(12\) lies between \(8\) and \(16\), the graph for area \(12\,\text{cm}^2\) would lie between graphs \(q\) and \(r\).

a) Graph \(p\) represents \(4\,\text{cm}^2\), graph \(q\) represents \(8\,\text{cm}^2\), and graph \(r\) represents \(16\,\text{cm}^2\). b) The graph for \(12\,\text{cm}^2\) would lie between \(q\) and \(r\).

- Read only the plotted whole-number participant counts. - Use a point from the graph to identify the fixed total cost as a constant product. - For part c, translate “no more than \(\$8\)” into an inequality, then respect the whole-number domain.

1. From the graph, when \(x=4\), \(y=\$25\); when \(x=5\), \(y=\$20\). 2. The total cost is the constant product: \(4\cdot25=100\), which agrees with \(5\cdot20=100\). Therefore, the rental costs \(\$100\) in total. 3. For the cost to be no more than \(\$8\), solve \(\frac{100}{x}\le8\). Since \(x>0\), \(100\le8x\), so \(x\ge12.5\). 4. The number of participants must be a whole number, so at least \(13\) people are needed.

a) With \(4\) people, the cost is \(\$25\) per person. With \(5\) people, it is \(\$20\) per person. b) The total rental cost is \(\$100\). c) At least \(13\) people must attend.

- For an inverse variation, the product \(xy\) must be constant. - Choose a point on \(f\) that lies on a grid intersection. - Calculate the product for each marked point on \(g\). - Compare the products before drawing a conclusion.

1. A clearly readable point on \(f\) is \((2, 4)\). Therefore, \(k = 2 \cdot 4 = 8\), so \(f(x) = \frac{8}{x}\). 2. For the marked points on \(g\), the products are \(1 \cdot 8 = 8\), \(2 \cdot 6 = 12\), \(4 \cdot 4 = 16\), and \(10 \cdot 2 = 20\). 3. Since these products are not constant, \(g\) is not an inverse variation.

a) \(f(x) = \frac{8}{x}\) b) No. The products for \(g\) are \(8\), \(12\), \(16\), and \(20\), so they are not constant.

- Relate the fixed cylinder volume to its radius and height before isolating the requested variable. - In part b), focus on how the squared radius factor changes when the radius is scaled. - For part c), substitute the given radius into the stated area model and keep the units consistent.

1. The cylinder volume formula is \(V=\pi r^2h\). Solving for \(h\) gives \(h(r)=\frac{355}{\pi r^2}\). A radius must be positive, so the domain is \(r>0\). 2. Replacing \(r\) with \(\frac{r}{2}\) gives \(h\left(\frac{r}{2}\right)=\frac{355}{\pi(\frac{r}{2})^2}=\frac{355}{\frac{1}{4}\pi r^2}=4\cdot\frac{355}{\pi r^2}=4h(r)\). Therefore, cutting the radius in half multiplies the height by \(4\). 3. \(M(3.2)=\frac{710}{3.2}=221.875\). The lateral surface area is \(221.875\,\text{cm}^2\).

a) \(h(r)=\frac{355}{\pi r^2}\), with domain \(r>0\) b) The height is multiplied by \(4\). c) \(221.875\,\text{cm}^2\)

- Recall the required intercept for a direct-variation line. - Look for the characteristic two-branch shape of a reciprocal relationship. - A function can be linear and still fail to be a direct variation.

1. Function \(f\) is a straight line through the origin, so it is a direct variation. 2. Function \(g\) has the two-branch reciprocal shape and approaches both axes without crossing them, so it is an inverse variation. 3. Function \(h\) is linear but does not pass through the origin, so it is not a direct variation. It also does not have the reciprocal shape of an inverse variation. Therefore, it is neither.

\(f\): direct variation because its graph is a straight line through the origin. \(g\): inverse variation because it has reciprocal branches approaching the axes. \(h\): neither because it is linear but has a nonzero y-intercept, so it is not \(y=kx\), and it is not reciprocal-shaped.

- For direct variation, compare the ratio of output to input for the nonzero table entries. - Also check what happens when the input is zero. - Two lines can have the same slope but differ in whether they pass through the origin.

1. For each nonzero input, \(\frac{y}{x}=-2\), and the table also contains \((0,0)\). Therefore, \(y\) varies directly with \(x\) with \(k=-2\). 2. The equation is \(y=-2x\). 3. The graph of \(f\) passes through the origin. The graph of \(g(x)=-2x+3\) has y-intercept \(3\), so it does not have the form \(y=kx\) and is not a direct variation.

a) \(y\) varies directly with \(x\) because \(\frac{y}{x}=-2\) for every nonzero listed input and the relation contains \((0,0)\). Thus \(k=-2\). b) \(y=-2x\) c) \(g\) is not a direct variation because it does not pass through the origin; its y-intercept is \(3\).

- Use the original number of lines and time to determine the constant product. - Express time as a function of the number of operating lines before evaluating new cases. - For the extreme case, distinguish what the mathematical model predicts from whether its assumptions remain realistic.

1. The fixed work is \(4\cdot90=360\) line-minutes, so \(nT=360\) and \(T(n)=\frac{360}{n}\). The constant product shows inverse variation. 2. \(T(3)=\frac{360}{3}=120\,\text{min}\). 3. \(T(100)=\frac{360}{100}=3.6\,\text{min}\), which is less than \(4\,\text{min}\). The trainee's calculation is correct according to the model. 4. In a real plant, limits such as pipe capacity, available space, or interference among lines can prevent perfect inverse scaling.

a) \(nT=360\), so \(T(n)=\frac{360}{n}\); this is inverse variation because the product \(nT\) is constant. b) \(120\,\text{min}\) c) The model gives \(3.6\,\text{min}\), so the claim is correct under the model. d) Possible reasons include limited pipe capacity, insufficient space, or interference among lines.

- Use the observed fraction completed to recalibrate the total amount of work. - Express only the remaining work in student-hours before writing the inverse model. - The final crew size must make the remaining time equal the time left on the schedule.

1. Four students working for \(2\) hours use \(8\) student-hours to complete \(\frac14\) of the job, so the whole job requires \(4\cdot8=32\) student-hours at the observed rate. 2. Three-fourths of the job remains, corresponding to \(24\) student-hours. Therefore \(R(n)=\frac{24}{n}\). 3. There are \(3\) planned hours left. Set \(\frac{24}{n}=3\), giving \(n=8\).

a) \(32\) student-hours. b) \(R(n)=\frac{24}{n}\). c) \(8\) people in total.

- Compare the products “pages per day \(\times\) days” for the two given schedules. - Once you know what quantity stays fixed, express the number of days as that fixed quantity divided by the daily reading rate. - For the claim in part c, check the time-unit conversion separately from the question of whether the model remains realistic at an extreme input.

1. The products are \(20\cdot15=300\) and \(30\cdot10=300\), so the constant product is \(300\) pages. 2. Therefore, \(pd=300\), so the inverse-variation model is \(d(p)=\frac{300}{p}\). The constant \(300\) is the total number of pages in the book. 3. To finish in \(4\) days, \(4=\frac{300}{p}\), so \(p=75\) pages per day. 4. At \(600\) pages per day, the model gives \(d=\frac{300}{600}=0.5\) day. Half a day is \(12\,\text{h}\), not \(2\,\text{h}\). 5. The inverse model also assumes the reading rate can increase without affecting comprehension, fatigue, or other practical limits, so extrapolating it to an extreme rate is not realistic.

a) \(20\cdot15=30\cdot10=300\), so \(d(p)=\frac{300}{p}\). The constant \(300\) is the total number of pages. b) \(75\) pages per day c) Half a day is \(12\,\text{h}\), not \(2\,\text{h}\), and the model is not realistic at an arbitrarily high reading rate because it ignores human limits such as comprehension and fatigue.

- Start with the area formula for a rectangle. - Increasing \(a\) by \(20\%\) means multiplying it by \(1.2\). - Compare the new value of \(b\) with the original value using a multiplicative factor.

1. From \(ab = 72\), solve for \(b\): \(b = \frac{72}{a}\). 2. For \(a = 12\,\text{cm}\), \(b = \frac{72}{12} = 6\,\text{cm}\). 3. Increasing \(a\) by \(20\%\) multiplies it by \(1.2\), so \(a_{\text{new}} = 1.2a\). 4. Then \(b_{\text{new}} = \frac{72}{1.2a} = \frac{1}{1.2}b = \frac{5}{6}b\). 5. The decrease is \(1 - \frac{5}{6} = \frac{1}{6}\), which is \(16\frac{2}{3}\%\), or approximately \(16.67\%\).

a) \(b = \frac{72}{a}\) b) \(b = 6\,\text{cm}\) c) The side length \(b\) decreases by \(16\frac{2}{3}\%\), or approximately \(16.67\%\).

- Express both \(y\)-values in terms of \(k\). - Translate “decreases by \(8\)” into an equation involving the two \(y\)-values. - Solve the resulting equation for \(k\). - Substitute \(x = 0.5\) into the function.

1. Use \(y = \frac{k}{x}\). The two values are \(y_1 = \frac{k}{2}\) and \(y_2 = \frac{k}{10}\). 2. Because the \(y\)-value decreases by \(8\), \(\frac{k}{2} - \frac{k}{10} = 8\). 3. Combine the fractions: \(\frac{5k}{10} - \frac{k}{10} = 8\), so \(\frac{4k}{10} = 8\). Therefore, \(0.4k = 8\) and \(k = 20\). 4. The function equation is \(y = \frac{20}{x}\). 5. For \(x = 0.5\), \(y = \frac{20}{0.5} = 40\).

1. \(k = 20\) 2. \(y = \frac{20}{x}\) 3. \(y = 40\)

- Find the fraction of the pond each hose fills in one hour. - Think about how two rational rates can be combined into one rate. - The time for one whole job is related reciprocally to the combined rate. - In part d), compare the expression after the two individual times are made equal.

1. Hose A fills \(\frac{1}{x}\) of the pond per hour, and Hose B fills \(\frac{1}{y}\) per hour. 2. Together, their rate is \(\frac{1}{x}+\frac{1}{y}=\frac{x+y}{xy}\) pond per hour. 3. The time for one whole pond is the reciprocal: \(\frac{xy}{x+y}\) hours. 4. For \(x=6\) and \(y=3\), \(\frac{6\cdot3}{6+3}=2\) hours. 5. If \(x=y\), then \(\frac{x^2}{2x}=\frac{x}{2}\). Two equally fast hoses cut the filling time in half.

a) \(\frac{1}{x}+\frac{1}{y}=\frac{x+y}{xy}\) b) \(\frac{xy}{x+y}\) hours c) \(2\) hours d) \(\frac{x}{2}\); the filling time is half the time for one hose.

- Express “half as much time” in terms of \(t\). - Convert each worker's completion time into a fractional work rate. - Combine the rational rates before converting back to a completion time. - For part d), compare the combined time with Lucas's time alone.

1. Simon needs half of Lucas's time, so his time is \(\frac{t}{2}\) hours. 2. Lucas paints \(\frac{1}{t}\) of the fence per hour. Simon paints \(\frac{1}{t/2}=\frac{2}{t}\) per hour. 3. Their combined rate is \(\frac{1}{t}+\frac{2}{t}=\frac{3}{t}\) fence per hour. 4. The time for one whole fence is the reciprocal of the combined rate: \(\frac{t}{3}\) hours. 5. When \(t=6\), their combined time is \(\frac{6}{3}=2\) hours. The time saved is \(6-2=4\) hours, so Lucas is correct.

a) \(\frac{t}{2}\) hours b) \(\frac{3}{t}\) of the fence per hour c) \(\frac{t}{3}\) hours d) Lucas is correct. Together they need \(2\) hours, which saves \(4\) hours.

- Express each filling time as volume divided by rate. - Keep the two rational time expressions separate until the requested difference is clear. - For part b), examine how a common factor in the volume affects both rational terms. - Check whether your conclusion depends on a particular numerical value of \(V\).

1. Pump A needs \(t_A=\frac{V}{x}\) minutes. 2. Pump B's rate is \(x+10\) gallons per minute, so it needs \(t_B=\frac{V}{x+10}\) minutes. 3. The time difference is \(\Delta t=\frac{V}{x}-\frac{V}{x+10}\). 4. Replacing \(V\) with \(2V\) gives \(\Delta t_{\text{new}}=\frac{2V}{x}-\frac{2V}{x+10}=2\left(\frac{V}{x}-\frac{V}{x+10}\right)\). 5. Therefore, doubling the tank volume doubles the time difference.

a) \(\Delta t=\frac{V}{x}-\frac{V}{x+10}\) minutes b) The time difference doubles because replacing \(V\) with \(2V\) multiplies the entire expression by \(2\).

- Write the one-leg model using actual speed, not still-water speed. - The current changes the actual speed by the same amount in opposite directions. - Compare the two reciprocal values rather than assuming equal speed changes create equal time changes.

1. For one leg, \(Tq=120\), so \(T(q)=\frac{120}{q}\). 2. On the lake, each leg has \(q=25\), so the round trip takes \(2\cdot\frac{120}{25}=9.6\) hours. 3. On the river, the downstream speed is \(30\), giving \(4\) hours, and the upstream speed is \(20\), giving \(6\) hours. The total is \(10\) hours. 4. The gain and loss do not cancel. Because time is reciprocal in speed, equal additive changes in speed do not produce equal opposite changes in time. The river trip is \(0.4\) hour longer.

1) \(T(q)=\frac{120}{q}\). 2) \(9.6\,\text{h}\). 3) \(10\,\text{h}\). 4) No. The river trip is \(0.4\,\text{h}\) longer; reciprocal dependence makes the upstream time increase larger than the downstream time decrease.

- Use the relationship among distance, speed, and time. - Increasing a quantity by \(25\%\) means multiplying it by \(1.25\). - Express the new travel time as a multiple of the old travel time. - Convert that multiplicative factor into a percent decrease. - Check the result using one of the speeds from part 2.

1. Travel time equals distance divided by speed, so \(t(v) = \frac{180}{v}\). 2. The times are \(t(45) = \frac{180}{45} = 4\,\text{h}\), \(t(60) = \frac{180}{60} = 3\,\text{h}\), and \(t(90) = \frac{180}{90} = 2\,\text{h}\). 3. Increasing the speed by \(25\%\) multiplies it by \(1.25\). The new time is \(t_{\text{new}} = \frac{180}{1.25v} = \frac{1}{1.25}t_{\text{old}} = 0.8t_{\text{old}}\). The new time is \(80\%\) of the old time, so it decreases by \(20\%\).

1. \(t(v) = \frac{180}{v}\) 2. The times are \(4\,\text{h}\), \(3\,\text{h}\), and \(2\,\text{h}\), respectively. 3. The travel time decreases by \(20\%\).

- Represent each percent increase or decrease with a decimal factor. - Substitute the changed variables into the formula. - In part b, let the unknown change in \(z\) be a factor \(k\), then solve an equation for \(k\).

1. In part a, the new values are \(1.2x\) and \(0.8y\). 2. Therefore, \(W_{\text{new}}=\frac{1.2x}{(0.8y)z}=\frac{1.2}{0.8}W=1.5W\). 3. In part b, let \(z_{\text{new}}=kz\). Then \(2W=\frac{x}{(\frac{1}{3}y)(kz)}=\frac{3}{k}W\). 4. Solving \(2=\frac{3}{k}\) gives \(k=\frac{3}{2}=1.5\). Thus, \(z\) must be multiplied by \(1.5\).

a) \(W_{\text{new}}=1.5W\) b) Multiply \(z\) by \(1.5\), which is a \(50\%\) increase.

- How does the sign of a quotient depend on the signs of its numerator and denominator? - What happens to a fraction with a fixed numerator when its denominator becomes very large? - Rearrange the function equation to solve for \(x\). - A point lies on a graph when its coordinates satisfy the function rule.

1. Since the numerator is positive, the function is positive on \((0, \infty)\) and negative on \((-\infty, 0)\). 2. The values are \(f(0.1) = \frac{4}{0.1} = 40\), \(f(100) = \frac{4}{100} = 0.04\), and \(f(-1000) = \frac{4}{-1000} = -0.004\). As \(x \to \infty\), \(f(x) \to 0\), so the graph approaches the x-axis. 3. Solve \(0.5 = \frac{4}{x}\): \(0.5x = 4\), so \(x = 8\). 4. Since \(f(2) = \frac{4}{2} = 2\), point \(P\) lies on the graph. Because its coordinates satisfy \(x = y\), it is one of the graph's intersection points with the line \(y = x\), which is a line of symmetry for this hyperbola.

1) Positive on \((0, \infty)\); negative on \((-\infty, 0)\). 2) \(f(0.1) = 40\), \(f(100) = 0.04\), and \(f(-1000) = -0.004\). As \(x \to \infty\), the graph approaches the x-axis. 3) \(x = 8\) 4) Yes. Point \(P\) is an intersection point of the graph and the line \(y = x\), which is a line of symmetry.

- Use the sign rules for division. - To find intersections, set the two function expressions equal. - Can the square of a real number be negative? - What happens to both coordinates under a rotation of \(180^\circ\) about the origin?

1. Since the constant \(-9\) is negative, \(x\) and \(y\) have opposite signs. Thus, the graph lies in Quadrants II and IV. 2. Set the equations equal: \(-\frac{9}{x} = -x\). Then \(x^2 = 9\), so \(x = 3\) or \(x = -3\). The intersection points are \((3, -3)\) and \((-3, 3)\). 3. Set \(-\frac{9}{x} = x\). This gives \(x^2 = -9\), which has no real solution. Therefore, the graphs do not intersect in the real coordinate plane. 4. If \(y = -\frac{9}{x}\), then \(g(-x) = -\frac{9}{-x} = \frac{9}{x} = -y\). Therefore, \((-x, -y)\) lies on the graph whenever \((x, y)\) does, confirming symmetry about the origin.

1) Quadrants II and IV 2) \((3, -3)\) and \((-3, 3)\) 3) The equation leads to \(x^2 = -9\), so there is no real intersection. 4) Since \(g(-x) = -g(x)\), the graph is symmetric about the origin.