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 \(f(x) = -\frac{3}{x}\).

- 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 b) neither c) inverse variation d) neither

- 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 Table B: direct variation

- 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) \(t = \frac{12}{n}\)

- Find the total number of robot-hours required for the job. - The total amount of work remains constant. - Divide the total robot-hours by the target time.

1. Find the total work in robot-hours: \(6 \cdot 10 = 60\) robot-hours. 2. Divide the total work by the target time: \(60 \div 4 = 15\). 3. Therefore, \(15\) robots must be used.

A total of \(15\) robots must be used.

- Identify the quantity that stays constant during the filling process. - Decide whether using fewer pipes makes the filling time longer or shorter. - Use the constant product of the number of pipes and the time. - Solve for the unknown quantity in each part.

1. Find the constant product of the number of pipes and the filling time: \(4 \cdot 18 = 72\) pipe-minutes. 2. For a), with \(3\) pipes, the time is \(72 \div 3 = 24\,\text{min}\). 3. For b), to fill the pool in \(12\,\text{min}\), the number of pipes is \(72 \div 12 = 6\).

a) With \(3\) pipes, filling takes \(24\,\text{min}\). b) A total of \(6\) pipes are needed.

- Find the constant number of person-hours required for the cleanup. - Divide that constant by each proposed group size. - Compare each calculated time with the corresponding claim.

1. The cleanup requires \(4 \cdot 6 = 24\) person-hours. 2. With \(8\) people, the time would be \(24 \div 8 = 3\,\text{h}\), so the first claim is correct. 3. With \(12\) people, the time would be \(24 \div 12 = 2\,\text{h}\), not \(1.5\,\text{h}\). Therefore, the second claim is incorrect.

The first claim is correct: \(8\) people would need \(3\,\text{h}\). The second claim is incorrect: \(12\) people would need \(2\,\text{h}\), not \(1.5\,\text{h}\).

- Identify the quantity that remains unchanged. - Find the total amount of feed available. - If the supply must last longer, the daily amount must decrease.

1. Find the total amount of feed: \(30 \cdot 12 = 360\,\text{lb}\). 2. Divide the fixed supply by the new number of days: \(360 \div 45 = 8\,\text{lb}\) per day.

The owner may use an average of \(8\,\text{lb}\) of feed per day.

- Find the fixed total amount of work or liquid in each part. - Divide the fixed total by the new number of workers or the new bottle size. - In both situations, increasing one quantity decreases the other.

1. For a), the roofing job requires \(3 \cdot 12 = 36\) roofer-hours. With \(4\) roofers, the time is \(36 \div 4 = 9\,\text{h}\). 2. For b), the tank contains \(24 \cdot 24 = 576\,\text{fl oz}\) of juice. Using \(16\,\text{fl oz}\) bottles gives \(576 \div 16 = 36\) bottles.

a) The \(4\) roofers would need \(9\,\text{h}\). b) The producer could fill \(36\) bottles.

- Express the total feed supply in cow-days. - Determine how many cows remain in part a). - If the supply must last longer, fewer cows can be fed.

1. The total supply is \(15 \cdot 20 = 300\) cow-days. 2. For a), after \(3\) cows are sold, \(15 - 3 = 12\) cows remain. The feed lasts \(300 \div 12 = 25\) days. 3. For b), if the feed must last \(30\) days, the number of cows is \(300 \div 30 = 10\).

a) The feed would last \(25\) days. b) The farmer could feed \(10\) cows.

- Identify the total amount of lemonade, which does not change. - First calculate the total number of fluid ounces. - Smaller bottles require a greater number of bottles.

1. Find the total amount of lemonade: \(15 \cdot 16 = 240\,\text{fl oz}\). 2. Divide the total amount by the new bottle size: \(240 \div 12 = 20\).

A total of \(20\) bottles would be needed.

- 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.

- Find the total number of painter-hours required for the job. - Fewer painters will need more time for the same amount of work.

1. The job requires \(6 \cdot 10 = 60\) painter-hours. 2. With \(4\) painters, the time is \(60 \div 4 = 15\,\text{h}\).

Four painters would need \(15\,\text{h}\).

- 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.

- 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.

- Add the current speed when traveling downstream and subtract it when traveling upstream. - Use \(\text{time}=\frac{\text{distance}}{\text{speed}}\). - Add the two travel times for the round trip.

1. Downstream, the boat's speed is \(v+c=15+3=18\) miles per hour. 2. The downstream time is \(\frac{s}{v+c}=\frac{36}{18}=2\) hours. 3. Upstream, the boat's speed is \(v-c=15-3=12\) miles per hour. 4. The upstream time is \(\frac{s}{v-c}=\frac{36}{12}=3\) hours. 5. The round-trip time is \(2+3=5\) hours.

a) \(\frac{s}{v+c}=2\) hours b) \(\frac{s}{v-c}=3\) hours c) \(5\) hours

- In an inverse variation, the product of the two quantities is constant. - More pumps require less time for the same fixed job. - Multiplying one quantity by \(3\) divides the other by \(3\).

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 \(3\).

- Identify the fixed total amount of work. - Express that amount in worker-days. - Fewer workers require more time for the same job. - Divide the total worker-days by the known number of workers or days.

1. The entire job requires \(6 \cdot 10 = 60\) worker-days. 2. For a), \(4\) workers would need \(60 \div 4 = 15\) days. 3. For b), finishing in \(4\) days requires \(60 \div 4 = 15\) workers.

a) Four workers would need \(15\) days. b) A total of \(15\) workers would be needed.

- 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.

- Consider the numerator and denominator separately. - Think about dividing a fixed total among a changing number of people. - Test your reasoning with simple positive values.

1. With \(n\) fixed, the positive denominator \(n-3\) is constant. Therefore, \(k\) is directly proportional to \(G\): increasing the total cost increases the cost per paying person. 2. With \(G\) fixed, increasing \(n\) increases the number of paying people, \(n-3\). Dividing the same total cost among more people decreases \(k\).

1) \(k\) increases. 2) \(k\) decreases.

- 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) \(a = -4\)

- Start at each animal count on the x-axis and move to the curve. - Read the corresponding number of days from the y-axis. - Multiply the coordinates of each point. - Interpret the constant product in the feeding context.

1. From the graph, the values are \(y = 20\), \(10\), \(8\), \(5\), and \(4\) for \(x = 2\), \(4\), \(5\), \(8\), and \(10\), respectively. 2. The products are \(2 \cdot 20 = 40\), \(4 \cdot 10 = 40\), \(5 \cdot 8 = 40\), \(8 \cdot 5 = 40\), and \(10 \cdot 4 = 40\). 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 \(f(2) = \frac{-2}{2} = -1\).

- In an inverse variation, the product of the two variables is constant. - Use the known number of workers and time to find the constant. - Locate \(t = 3\) on the graph and read the corresponding value of \(n\).

1. For inverse variation, the product \(nt\) is constant. Here, \(k = 4 \cdot 6 = 24\) worker-hours. 2. Therefore, \(t(n) = \frac{24}{n}\). 3. At \(t = 3\), the graph gives \(n = 8\). Algebraically, \(3 = \frac{24}{n}\), so \(n = 8\).

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

- Use the relationship distance equals rate times time. - With distance fixed, time varies inversely with speed. - Locate \(40\) on the horizontal axis and read the corresponding time.

1. Since distance equals speed times time, \(60 = vt\). Therefore, \(t = \frac{60}{v}\). 2. Substitute \(v = 40\): \(t = \frac{60}{40} = 1.5\) hours. 3. The trip takes \(1.5\) hours, or \(1\) hour \(30\) minutes.

\(1.5\) hours, or \(90\) minutes

- 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

- If more excavators work, does the time increase or decrease? - Check whether the product of the number of excavators and the time remains constant. - First determine how long one excavator would take.

1. 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 \(4 \cdot 15 = 60\). 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) \(t = \frac{60}{n}\)

- Use one complete pair to find the constant product. - Divide the constant product by the known value to find its partner. - As one variable decreases, the other increases proportionally.

1. Use the known pair \((4, 10)\) to find the constant product: \(k = 4 \cdot 10 = 40\). 2. When \(x = 2\), \(b = \frac{40}{2} = 20\). 3. When \(y = 5\), \(a = \frac{40}{5} = 8\). 4. When \(x = 16\), \(c = \frac{40}{16} = 2.5\). 5. When \(x = 40\), \(d = \frac{40}{40} = 1\).

\(a = 8\), \(b = 20\), \(c = 2.5\), and \(d = 1\)

- 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.

- Divide the fixed total cost by the number of students. - Choose two values of \(n\) for which one is twice the other. - Consider what happens to a quotient when its divisor doubles.

1. Divide the fixed total cost by each number of students: for \(n = 12\), \(k = 480 \div 12 = \text{€}40\); for \(n = 15\), \(k = 480 \div 15 = \text{€}32\); for \(n = 20\), \(k = 480 \div 20 = \text{€}24\); for \(n = 24\), \(k = 480 \div 24 = \text{€}20\); and for \(n = 32\), \(k = 480 \div 32 = \text{€}15\). 2. The cost per student is \(k = \frac{480}{n}\). 3. For example, with \(12\) students, each pays \(\text{€}40\). With \(24\) students, twice as many attend and each pays \(\text{€}20\), which is half as much. The claim is correct.

a) For \(n = 12\), \(\text{€}40\); for \(n = 15\), \(\text{€}32\); for \(n = 20\), \(\text{€}24\); for \(n = 24\), \(\text{€}20\); and for \(n = 32\), \(\text{€}15\). b) \(k = \frac{480}{n}\) c) The claim is correct. For example, doubling the number of students from \(12\) to \(24\) reduces the cost per student from \(\text{€}40\) to \(\text{€}20\).

- First express side \(a\) in terms of \(x\). - Use the area formula for Rectangle B. - Substitute the given value of \(x\) into the conditions for both rectangles.

1. Since side \(a\) is \(2\,\text{cm}\) longer than \(x\), \(a = x + 2\). 2. Rectangle B has area \(24\,\text{cm}^2\), so \(ab = 24\). Substitute \(a = x + 2\): \((x + 2)b = 24\). Therefore, \(b = \frac{24}{x + 2}\). 3. When \(x = 4\), \(b = \frac{24}{4 + 2} = 4\). Thus, \(b = 4\,\text{cm}\). 4. For Rectangle A, \(4y = 24\), so \(y = 6\,\text{cm}\). 5. The difference is \(6 - 4 = 2\,\text{cm}\), so \(b\) is \(2\,\text{cm}\) shorter than \(y\).

a) \(b = \frac{24}{x + 2}\) b) \(b = 4\,\text{cm}\) c) \(y = 6\,\text{cm}\). Side \(b\) is \(2\,\text{cm}\) shorter than side \(y\).

- Decide whether more horses make the hay last longer or for less time. - Write an equation using the constant number of horse-days. - Translate “at least \(18\) days” into an inequality. - Remember that the number of horses must be a whole number.

1. Let \(n\) be the number of horses and \(d\) the number of days. The constant product is \(nd = 120\), so \(d = \frac{120}{n}\). 2. Evaluate the formula: for \(n = 2\), \(d = 60\); for \(n = 4\), \(d = 30\); for \(n = 6\), \(d = 20\); for \(n = 8\), \(d = 15\); for \(n = 10\), \(d = 12\); and for \(n = 12\), \(d = 10\). 3. For the hay to last at least \(18\) days, solve \(\frac{120}{n} \ge 18\). Because \(n > 0\), this gives \(120 \ge 18n\), so \(n \le \frac{120}{18} \approx 6.67\). 4. The number of horses must be a whole number, so the greatest possible number is \(6\).

a) <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> b) The owner can feed at most \(6\) horses.

- Find the total number of employee-hours required for the job. - Use the constant product to find the time for each crew size. - Translate “less than \(3\) hours” into a strict inequality. - The number of employees must be a whole number.

1. The constant product is \(3 \cdot 8 = 24\) employee-hours. If \(m\) is the number of employees and \(t\) is the time in hours, then \(t = \frac{24}{m}\). 2. Evaluate the formula: for \(m = 2\), \(t = 12\); for \(m = 4\), \(t = 6\); for \(m = 5\), \(t = 4.8\); and for \(m = 6\), \(t = 4\). 3. To finish in less than \(3\,\text{h}\), solve \(\frac{24}{m} < 3\). Because \(m > 0\), this gives \(24 < 3m\), so \(m > 8\). 4. The number of employees must be a whole number, so at least \(9\) employees are needed. With \(9\) employees, the model gives \(\frac{24}{9} \approx 2.67\,\text{h}\).

a) <table> <thead> <tr><th>Number of employees</th><th>Time in hours</th></tr> </thead> <tbody> <tr><td>\(2\)</td><td>\(12\)</td></tr> <tr><td>\(4\)</td><td>\(6\)</td></tr> <tr><td>\(5\)</td><td>\(4.8\)</td></tr> <tr><td>\(6\)</td><td>\(4\)</td></tr> </tbody> </table> b) At least \(9\) employees are needed.

- Use the relationship among distance, speed, and time. - Translate “at most \(2.5\) hours” into an inequality. - For part c), calculate both travel times before finding their difference. - Convert the difference from hours to minutes.

1. Use \(t = \frac{s}{v}\). Since the distance is \(45\,\text{mi}\), \(t = \frac{45}{v}\). 2. For a), the times are \(45 \div 10 = 4.5\,\text{h}\), \(45 \div 12 = 3.75\,\text{h}\), \(45 \div 15 = 3\,\text{h}\), and \(45 \div 18 = 2.5\,\text{h}\). 3. For b), \(\frac{45}{v} \le 2.5\). Because \(v > 0\), this gives \(45 \le 2.5v\), so \(v \ge 18\,\text{mi/h}\). 4. For c), the original time is \(45 \div 12 = 3.75\,\text{h}\). The new speed is \(12 + 3 = 15\,\text{mi/h}\), so the new time is \(45 \div 15 = 3\,\text{h}\). The time saved is \(3.75 - 3 = 0.75\,\text{h}\), or \(0.75 \cdot 60 = 45\,\text{min}\).

a) At \(10\,\text{mi/h}\): \(4.5\,\text{h}\); at \(12\,\text{mi/h}\): \(3.75\,\text{h}\); at \(15\,\text{mi/h}\): \(3\,\text{h}\); at \(18\,\text{mi/h}\): \(2.5\,\text{h}\). b) The cyclist must average at least \(18\,\text{mi/h}\). c) The cyclist saves \(45\,\text{min}\).

- 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 volume is fixed, so increasing the amount in each cup decreases the number of cups that can be filled. This is an inverse variation. 2. The product of the number of cups and the volume in each cup equals the total volume: \(nv = 384\). Therefore, \(n = \frac{384}{v}\). 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 number of cups decreases as the amount in each cup increases. 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 \(22\,\text{m}\).

- First determine the total cost of the ping-pong table. - The total cost does not change when the number of contributors changes. - For part b), determine how many people remain. - Part b) asks for the increase, not just the new contribution.

1. Find the total cost: \(12 \cdot \$15.00 = \$180.00\). 2. For a), divide the total cost among \(18\) people: \(\$180.00 \div 18 = \$10.00\) per person. 3. For b), the new number of participants is \(12 - 3 = 9\). 4. The new contribution is \(\$180.00 \div 9 = \$20.00\) per person. 5. The increase is \(\$20.00 - \$15.00 = \$5.00\).

a) Each person would pay \(\$10.00\). b) Each remaining person’s contribution increases by \(\$5.00\).

- Find the total number of landscaper-hours required for the project. - Determine how much work is completed during the first \(3\) hours. - Divide the remaining work by the size of the new crew.

1. The entire project requires \(4 \cdot 9 = 36\) landscaper-hours. 2. During the first \(3\,\text{h}\), the original crew completes \(4 \cdot 3 = 12\) landscaper-hours of work. 3. The remaining work is \(36 - 12 = 24\) landscaper-hours. 4. After the additional landscapers arrive, the crew has \(4 + 2 = 6\) people. 5. The remaining work takes \(24 \div 6 = 4\,\text{h}\). 6. The total project time is \(3 + 4 = 7\,\text{h}\).

The entire project will take \(7\,\text{h}\).

- Express the entire job in truck-days. - Determine how much work is completed before the breakdown. - Find how long the remaining trucks need for the unfinished work. - Compare the actual total time with the original plan.

1. The entire job requires \(3 \cdot 20 = 60\) truck-days. 2. During the first \(4\) days, the trucks complete \(3 \cdot 4 = 12\) truck-days of work. 3. The remaining work is \(60 - 12 = 48\) truck-days. 4. After the breakdown, \(3 - 1 = 2\) trucks remain. 5. The remaining work takes \(48 \div 2 = 24\) days. 6. The actual total time is \(4 + 24 = 28\) days. 7. The delay is \(28 - 20 = 8\) days.

Completion is delayed by \(8\) days.

- Find the total number of worker-hours required for the harvest. - Determine the total number of workers needed for a \(5\)-hour harvest. - The question asks for the number of additional workers, not the total number.

1. The harvest requires \(5 \cdot 8 = 40\) worker-hours. 2. To complete the harvest in \(5\,\text{h}\), the total number of workers needed is \(40 \div 5 = 8\). 3. The owner already has \(5\) workers, so the number of additional workers is \(8 - 5 = 3\).

The owner must hire \(3\) additional workers.

- Find the total number of landscaper-hours required for the job. - Use inverse variation to calculate the time for each group size. - For part b), consider workspace and coordination limits in the park.

1. The job requires \(3 \cdot 12 = 36\) landscaper-hours. 2. For a), \(4\) landscapers would need \(36 \div 4 = 9\,\text{h}\). 3. For b), the inverse-variation model predicts \(36 \div 40 = 0.9\,\text{h}\), or \(0.9 \cdot 60 = 54\,\text{min}\). Thus, the claim is mathematically correct under the model. 4. In reality, \(40\) people working in the same area might interfere with one another. Travel time, limited workspace, and coordination could reduce the group’s efficiency.

a) Four landscapers would need \(9\,\text{h}\). b) The model predicts \(54\,\text{min}\), which is less than one hour. In reality, limited space and coordination could prevent \(40\) landscapers from working independently at full efficiency.

- 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: direct variation Situation 2: inverse variation b) \(40\) minutes

- 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 Situation 2: inverse variation; \(15\) days

- 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.

- First find the total length of twine on the roll. - Determine the new length of each piece. - The total amount of twine remains fixed. - Divide the total length by the length of one new piece.

1. Find the total length of the roll: \(25 \cdot 8 = 200\,\text{ft}\). 2. The new length of each piece is \(8 - 3 = 5\,\text{ft}\). 3. Divide the total length by the new piece length: \(200 \div 5 = 40\).

The gardener could cut \(40\) 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.

- Find the fixed total amount of heating oil. - Increasing daily use makes the supply last for fewer days. - Divide the total supply by the number of days or the daily use, depending on the unknown.

1. The total supply is \(12 \cdot 80 = 960\,\text{gal}\). 2. For a), at \(16\,\text{gal}\) per day, the supply lasts \(960 \div 16 = 60\) days. 3. For b), to last \(120\) days, the daily use can be at most \(960 \div 120 = 8\,\text{gal}\).

a) The supply will last \(60\) days. b) The greatest daily use is \(8\,\text{gal}\).

- Express the original food supply in person-days. - Subtract the amount consumed during the first \(3\) days. - Divide the remaining person-days by the new group size.

1. The original supply contains \(25 \cdot 12 = 300\) person-days of food. 2. During the first \(3\) days, the group consumes \(25 \cdot 3 = 75\) person-days. 3. The remaining supply is \(300 - 75 = 225\) person-days. 4. The new group has \(25 + 5 = 30\) people. 5. The remaining food lasts \(225 \div 30 = 7.5\) more days.

The remaining food will last \(7.5\) more days.

- 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.

- Express the entire project in worker-days. - Determine how much work the full crew completes in the first \(6\) days. - Divide the remaining work by the number of workers who remain. - Add the elapsed time to the time needed for the remaining work.

1. The entire job requires \(4 \cdot 15 = 60\) worker-days. 2. During the first \(6\) days, the full crew completes \(4 \cdot 6 = 24\) worker-days of work. 3. The remaining work is \(60 - 24 = 36\) worker-days. 4. After one worker leaves, \(4 - 1 = 3\) workers remain. 5. The remaining work takes \(36 \div 3 = 12\) days. 6. The total construction time is \(6 + 12 = 18\) days.

The total construction time is \(18\) days.

- Express the entire food supply in person-days. - Determine how much food remains after the first \(4\) days. - Find how long the remaining group can use that supply. - Compare the new remaining duration with the originally planned remaining duration.

1. The original supply contains \(24 \cdot 10 = 240\) person-days of food. 2. During the first \(4\) days, the group consumes \(24 \cdot 4 = 96\) person-days. 3. The remaining supply is \(240 - 96 = 144\) person-days. 4. After \(6\) people leave, \(24 - 6 = 18\) people remain. 5. The remaining supply lasts \(144 \div 18 = 8\) days. 6. Under the original plan, \(10 - 4 = 6\) days would have remained. Therefore, the food lasts \(8 - 6 = 2\) days longer than planned.

The food will last \(2\) days longer than originally planned.

- Express the entire filling job in pump-hours. - Subtract the work completed during the first \(2\) hours. - Divide the remaining pump-hours among \(4\) pumps. - The question asks only for the time after the fourth pump is turned on.

1. Filling the tank requires \(3 \cdot 8 = 24\) pump-hours. 2. During the first \(2\,\text{h}\), the original pumps complete \(3 \cdot 2 = 6\) pump-hours of work. 3. The remaining work is \(24 - 6 = 18\) pump-hours. 4. With \(4\) pumps operating, the remaining time is \(18 \div 4 = 4.5\,\text{h}\).

After the fourth pump is turned on, filling will take \(4.5\,\text{h}\), or \(4\,\text{h}\) \(30\,\text{min}\), more.

- Find the fixed total amount of flour. - Use the constant product of daily use and number of days.

1. The total supply is \(12 \cdot 20 = 240\,\text{lb}\). 2. For a), at \(15\,\text{lb}\) per day, the supply lasts \(240 \div 15 = 16\) days. 3. For b), to last at least \(16\) days, daily use can be at most \(240 \div 16 = 15\,\text{lb}\).

a) The supply will last \(16\) days. b) At most \(15\,\text{lb}\) may be used each day.

- Express the whole project in worker-days. - Determine how much work is completed during the first \(3\) days. - Divide the remaining work by the number of workers who remain. - Focus on the additional time after the workers leave.

1. The entire project requires \(8 \cdot 15 = 120\) worker-days. 2. During the first \(3\) days, the team completes \(8 \cdot 3 = 24\) worker-days of work. 3. The remaining work is \(120 - 24 = 96\) worker-days. 4. After \(2\) workers leave, \(8 - 2 = 6\) workers remain. 5. The remaining workers need \(96 \div 6 = 16\) more days.

The remaining workers will need \(16\) more days.

- Evaluate the function carefully, especially for negative inputs. - What happens to \(f(x)\) as \(x\) approaches \(0\) from either side? - In a graphing tool, enter both functions on the same coordinate plane. - The solutions occur where the two graphs intersect.

1. Evaluate \(f(x) = \frac{2}{x}\) at the requested inputs: \(f(-4) = -0.5\), \(f(-2) = -1\), \(f(-1) = -2\), \(f(-0.5) = -4\), \(f(0.5) = 4\), \(f(1) = 2\), \(f(2) = 1\), and \(f(4) = 0.5\). 2. Enter \(f(x) = \frac{2}{x}\) and \(g(x) = x - 1\) in a graphing tool and view both graphs on the same coordinate plane. 3. The graphs intersect at \((-1, -2)\) and \((2, 1)\). 4. The solutions are the \(x\)-coordinates of the intersection points: \(x = -1\) and \(x = 2\).

a) The value table is: <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) The graphs intersect at \((-1, -2)\) and \((2, 1)\). c) \(x = -1\) and \(x = 2\)

- Find the fixed volume of the tank. - A greater flow rate produces a shorter filling time. - In part b), calculate the new time and then find the difference from the original time.

1. The tank volume is \(12 \cdot 45 = 540\,\text{gal}\). 2. For a), at \(18\,\text{gal/min}\), the filling time is \(540 \div 18 = 30\,\text{min}\). 3. For b), the new flow rate is \(12 + 3 = 15\,\text{gal/min}\). 4. The new filling time is \(540 \div 15 = 36\,\text{min}\). 5. The reduction is \(45 - 36 = 9\,\text{min}\).

a) Filling will take \(30\,\text{min}\). b) The filling time is reduced by \(9\,\text{min}\).

- 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 \(xy\) 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.

- Calculate the product of each pair in both tables. - Identify the table with a constant product. - Divide the constant product by \(8\) to find the missing value.

1. For Table A, the known products are \(1 \cdot 12 = 12\), \(2 \cdot 6 = 12\), \(3 \cdot 4 = 12\), and \(6 \cdot 2 = 12\). The product is constant, so Table A represents an inverse variation. 2. For Table B, the products begin \(1 \cdot 2 = 2\) and \(2 \cdot 4 = 8\), 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 \(1.5\).

- Multiplying one variable by a factor divides the other by the same factor in an inverse variation. - Use the constant product to find an unknown partner value. - Check that each pair has product \(40\).

1. The constant of variation is \(k = 2.5 \cdot 16 = 40\). 2. In an inverse variation, multiplying \(x\) by \(4\) divides \(y\) by \(4\). Therefore, the new value is \(16 \div 4 = 4\). 3. When \(y = 10\), use \(xy = 40\): \(10x = 40\), so \(x = 4\).

a) \(k = 40\) b) The \(y\)-value is divided by \(4\), so the new value is \(4\). c) \(x = 4\)

- Add the riders' speeds because they are moving toward each other. - Use the relationship \(\text{time}=\frac{\text{distance}}{\text{rate}}\). - Express Maya's speed in terms of \(v\) before finding the new closing rate.

1. Lucas rides at \(v\) miles per hour, and Maya rides at \(v+4\) miles per hour. 2. Because they ride toward each other, their closing rate is \(v+(v+4)=2v+4\) miles per hour. 3. The meeting time is \(t=\frac{d}{2v+4}\). 4. For \(d=48\) and \(v=10\), \(t=\frac{48}{2\cdot 10+4}=\frac{48}{24}=2\) hours. 5. If Maya rides at half of Lucas's speed, the closing rate is \(v+\frac{v}{2}=\frac{3v}{2}\), so \(t=\frac{2d}{3v}\).

a) \(t=\frac{d}{2v+4}\) hours b) \(2\) hours c) \(t=\frac{2d}{3v}\) hours

- Find the total amount of water from the original daily use and number of days. - Subtract \(w\) from the original daily use. - Find the new duration before comparing it with the original duration. - “Additional time” means new duration minus original duration.

1. The total water supply is \(tv\) liters. 2. The reduced daily use is \(v-w\) liters. 3. The new duration is \(\frac{tv}{v-w}\) days. 4. The additional time is \(\frac{tv}{v-w}-t\). 5. Substituting the values gives \(\frac{20\cdot 60}{60-12}-20=\frac{1200}{48}-20=25-20=5\) days.

The expression is \(\frac{tv}{v-w}-t\). For the given values, the water lasts \(5\) additional days.

- Find the planned number of lines per day. - Express the actual number of lines and actual number of days. - Divide each total by its corresponding time. - Subtract the planned rate from the actual rate.

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 find the tank's volume. - Add \(s\) to the original pumping rate. - Find the new filling time before subtracting it from the original time.

1. The tank's volume is \(tr\) gallons. 2. The upgraded rate is \(r+s\) gallons per hour. 3. The upgraded pump's filling time is \(\frac{tr}{r+s}\) hours. 4. The time saved is \(t-\frac{tr}{r+s}\). 5. For the given values, \(10-\frac{10\cdot 1200}{1200+300}=10-\frac{12{,}000}{1500}=10-8=2\) hours.

a) \(t-\frac{tr}{r+s}\) hours b) \(2\) hours c) It is the number of hours the upgraded pump needs to fill the tank.

- Find the total number of meal packs. - Determine how many campers remain. - Divide the total number of packs by the number used each day. - Replace the numerical quantities with the given variables for part b).

1. The group has \(24\cdot 10=240\) meal packs. 2. After \(4\) campers cancel, \(24-4=20\) campers remain. 3. The food lasts \(240\div 20=12\) days. 4. This is \(12-10=2\) additional days. 5. In general, there are \(nt\) meal packs and \(n-k\) campers, so \(T=\frac{nt}{n-k}\).

a) \(2\) additional days b) \(T=\frac{nt}{n-k}\) days

- Find the actual water volume when the aquarium is one-third full. - Adding pure water changes the volume but not the amount of salt. - Multiply the target concentration by the full volume. - Subtract the amount of salt already present.

1. The current water volume is \(\frac{V}{3}\) liters, so the concentration is \(\frac{k}{V/3}=\frac{3k}{V}\) grams per liter. 2. After adding pure water, the salt amount remains \(k\) grams and the volume is \(V\) liters, so the new concentration is \(\frac{k}{V}\). 3. A concentration of \(m\) grams per liter in \(V\) liters requires \(mV\) grams of salt. 4. The additional salt needed is \(mV-k\) grams.

a) \(\frac{3k}{V}\) grams per liter b) \(\frac{k}{V}\) grams per liter c) \(mV-k\) grams

- Think about how quickly the distance between the robots changes. - For a lap, the faster robot must gain one full circumference. - Consider the motion from Beta's point of view.

1. In opposite directions, their relative speed is \(v_{\alpha}+v_{\beta}\). 2. They first meet after covering one circumference together, so \(t_O=\frac{C}{v_{\alpha}+v_{\beta}}\). 3. In the same direction, Alpha must gain one full circumference on Beta. Their relative speed is \(v_{\alpha}-v_{\beta}\). 4. Therefore, \(t_S=\frac{C}{v_{\alpha}-v_{\beta}}\). 5. The speeds add when the robots move toward each other, but only their speed difference closes the gap when they move in the same direction.

a) \(t_O=\frac{C}{v_{\alpha}+v_{\beta}}\) b) \(t_S=\frac{C}{v_{\alpha}-v_{\beta}}\) c) Opposite-direction speeds add, while same-direction catching depends on the difference in speeds.

- 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. - Set the original horse-days equal to the new horse-days. - In an inverse variation, halving one quantity doubles the other.

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) \(R(12, 3)\)

- Find the total number of worker-days required. - Subtract \(k\) from the original number of days. - Divide the fixed amount of work by the new number of days. - Distinguish between the total crew and the additional workers.

1. The project requires \(nd\) worker-days. 2. The shorter schedule is \(d-k\) days. 3. The number of workers needed is \(\frac{nd}{d-k}\). 4. For the given values, \(\frac{12\cdot 20}{20-5}=\frac{240}{15}=16\) workers. 5. The number of additional workers is \(16-12=4\).

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

- 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 \(8\,\text{km/h}\).

- Read the coordinates from the graph carefully. - In an inverse variation, the product of the two coordinates is constant. - Interpret the constant product as the total cost. - Multiplying the number of participants by \(3\) divides the amount per person by \(3\).

1. From the graph, when \(x = 3\), \(y = \$8\); when \(x = 6\), \(y = \$4\). 2. Since \(xy = a\), use \((3, 8)\): \(a = 3 \cdot 8 = 24\). 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 \(3 \cdot 8 = 24\) and \(6 \cdot 4 = 24\). Tripling the number of participants 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, \(3 \cdot 8 = 6 \cdot 4 = 24\). 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 \(r\).

- Express the original supply in person-days. - Subtract the amount used during the first \(4\) days. - Divide the remaining supply by the new group size. - Distinguish between the additional duration and the total duration.

1. The original supply contains \(10 \cdot 12 = 120\) person-days. 2. During the first \(4\) days, the group uses \(10 \cdot 4 = 40\) person-days. 3. The remaining supply is \(120 - 40 = 80\) person-days. 4. After \(2\) people leave, \(10 - 2 = 8\) people remain. 5. The remaining supplies last \(80 \div 8 = 10\) more days. 6. The total duration is \(4 + 10 = 14\) days.

a) The remaining supplies will last \(10\) more days. b) The group can be supplied for \(14\) days in total.

- Express the entire filling job in pipe-hours. - Determine how much work the two pipes complete during the first \(6\) hours. - Divide the remaining work by the one pipe that remains.

1. Filling the tank requires \(2 \cdot 18 = 36\) pipe-hours. 2. During the first \(6\,\text{h}\), the two pipes complete \(2 \cdot 6 = 12\) pipe-hours of work. 3. The remaining work is \(36 - 12 = 24\) pipe-hours. 4. With only one pipe operating, the remaining time is \(24 \div 1 = 24\,\text{h}\).

After one pipe is shut off, filling will take \(24\) more hours.

- Interpret a point on the graph in context. - Multiply the number of people by the cost per person to find the fixed total. - Write an inequality for a cost of no more than \(\$8\). - The number of participants must be a whole number.

1. From the graph, when \(x = 4\), \(y = \$25\); when \(x = 5\), \(y = \$20\). 2. The total cost is the constant product: \(4 \cdot 25 = 100\), which agrees with \(5 \cdot 20 = 100\). Therefore, the rental costs \(\$100\) in total. 3. For the cost to be no more than \(\$8\), solve \(\frac{100}{x} \le 8\). Since \(x > 0\), \(100 \le 8x\), so \(x \ge 12.5\). 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.

- Start with the cylinder volume formula and isolate \(h\). - Replace \(r\) with \(\frac{r}{2}\) and simplify the squared factor. - Substitute \(r=3.2\) into the given lateral-area function.

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) \(221.875\,\text{cm}^2\)

- First determine the total number of line-minutes required. - Use that constant product to find the time for any number of lines. - For part c), consider physical limits such as space and the rate at which liquid can flow through pipes.

1. Find the constant product: \(4 \cdot 90 = 360\) line-minutes. 2. For a), with \(3\) lines, the time is \(360 \div 3 = 120\,\text{min}\). 3. For b), with \(100\) lines, the model gives \(360 \div 100 = 3.6\,\text{min}\). Since \(3.6 < 4\), the trainee is correct according to the inverse-variation model. 4. For c), the model assumes every line can operate at full capacity without affecting the others. In reality, the tank’s pipes might not supply liquid quickly enough, or there might not be enough space for all \(100\) lines.

a) The process would take \(120\,\text{min}\). b) The trainee’s calculation is correct according to the model because \(360 \div 100 = 3.6\) and \(3.6 < 4\). c) Possible reasons include limited pipe capacity, insufficient space, or interference among the bottling lines.

- Express the entire food supply in person-days. - Subtract the amount eaten during the first \(4\) days. - Divide the remaining person-days by the new group size. - For part b), use the entire original supply rather than the amount remaining.

1. The original supply contains \(6 \cdot 20 = 120\) person-days of food. 2. During the first \(4\) days, the original group consumes \(6 \cdot 4 = 24\) person-days. 3. The remaining supply is \(120 - 24 = 96\) person-days. 4. The new group has \(6 + 2 = 8\) people, so the remaining food lasts \(96 \div 8 = 12\) more days. 5. For b), if \(8\) people had shared the entire original supply, it would have lasted \(120 \div 8 = 15\) days.

a) The remaining food will last \(12\) more days. b) The original supply would have lasted \(15\) days.

- Determine how much of the project remains after \(2\) hours. - Determine how much time remains in the original schedule. - Find the current team’s hourly rate and the hourly rate required to finish on time. - Compare the two rates to determine how the team size must change.

1. The current team paints \(\frac{1}{4}\) of the walls in \(2\,\text{h}\), so its rate is \(\frac{1}{4} \div 2 = \frac{1}{8}\) of the project per hour. 2. The remaining work is \(1 - \frac{1}{4} = \frac{3}{4}\) of the walls. 3. The remaining time is \(5 - 2 = 3\,\text{h}\). 4. To finish \(\frac{3}{4}\) of the walls in \(3\,\text{h}\), the team must paint at a rate of \(\frac{3}{4} \div 3 = \frac{1}{4}\) of the project per hour. 5. The required rate, \(\frac{1}{4}\) per hour, is twice the current rate, \(\frac{1}{8}\) per hour. 6. Because the work rate is proportional to the number of equally productive painters, the number of people must double: \(4 \cdot 2 = 8\).

A total of \(8\) people must work on the project from that point forward.

- Use either given reading schedule to find the total number of pages. - Recall how many hours are in half a day. - Consider whether a mathematical model can be extended without limit to human performance.

1. The book has \(20 \cdot 15 = 300\) pages; this agrees with \(30 \cdot 10 = 300\). 2. To finish in \(4\) days, Lucas must read \(300 \div 4 = 75\) pages per day. 3. At \(600\) pages per day, the model gives \(300 \div 600 = 0.5\) day. However, \(0.5\) day is \(12\,\text{h}\), not \(2\,\text{h}\). 4. The model also assumes an arbitrarily high reading rate without accounting for comprehension or fatigue. Reading and understanding \(300\) pages in \(2\,\text{h}\) is not realistic for most readers.

a) The book has \(300\) pages. b) Lucas must read \(75\) pages per day. c) Mathematically, half a day is \(12\,\text{h}\), not \(2\,\text{h}\). In context, the model also ignores limits on reading speed, concentration, and comprehension.

- 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 \(16.67\%\).

- 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.

- 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. - Use a common denominator to add the rates. - The time for one whole job is the reciprocal of the combined rate. - In part d), replace \(y\) with \(x\) and simplify.

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 \(y=3\), \(\frac{6\cdot 3}{6+3}=2\) 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 \(t\). - Find the fraction of the fence each painter completes in one hour. - Add the two work rates, then take the reciprocal to find the total time. - For part d), substitute \(t=6\) and 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.

- Divide the tank volume by a pump's rate to find its filling time. - Write the two filling-time expressions before subtracting them. - Replace \(V\) with \(2V\) and factor the resulting expression. - You may test your conclusion with convenient positive values of \(V\) and \(x\).

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 \(2\).

- First find the round-trip time with the same speed in both directions. - For the river, add the current downstream and subtract it upstream. - Compare the two total times. - Consider why a decrease in speed affects travel time more than an equal increase in speed.

1. With no current, the round-trip time is \(\frac{2s}{v}=\frac{2\cdot 120}{25}=9.6\) hours. 2. Downstream, the time is \(\frac{120}{25+5}=4\) hours. 3. Upstream, the time is \(\frac{120}{25-5}=6\) hours. 4. The river round trip takes \(4+6=10\) hours. 5. Since \(10>9.6\), the current increases the total time. The upstream loss is greater because the boat travels at the reduced speed for a longer time.

1) \(9.6\) hours 2) \(10\) hours 3) No. The river trip takes \(0.4\) hour longer, so the downstream gain does not offset the upstream loss.

- 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.