Question: User: 10. The table shows the number of packages of peanut butter crackers $y$ that can be made using various amounts of peanut butter $x$
Peanut butter (lb)
$\frac{1}{2}$ $1$ $2$ $3$
Cracker packages
$2$ $4$ $8$ $12$
A. Show that the relationship is proportional.
B. Write an equation to represent the relationship, and find the amount of peanut butter used to make $25$ cracker packages.
11. The relationship between adult dog weight $x$ in pounds and the daily recommended amount of dog food $y$ in cups is proportional.
Dog weight (lb)
$10$ $40$ $50$ $100$
Dog food (cups)
$\frac{1}{6}$ $\frac{2}{3}$ $\frac{5}{6}$ $1 \frac{2}{3}$
Write an equation for the relationship. How much dog food is recommended for a $25$-pound adult dog?
For Exercises $12$-$15$, find each unit rate.
12. $\frac{5}{8}$ mile in $\frac{1}{4}$ hour
13. $68$ for $8 \frac{1}{2}$ hours
14. $1 \frac{1}{4}$ cup of flour per $\frac{1}{3}$ cup of butter
15. $2 \frac{1}{2}$ miles in $\frac{3}{4}$ hour
1. **Problem 10A:** Show that the relationship between peanut butter $x$ and cracker packages $y$ is proportional.
2. To check proportionality, verify if the ratio $\frac{y}{x}$ is constant for all given pairs.
3. Calculate ratios:
$$\frac{2}{\frac{1}{2}} = 4, \quad \frac{4}{1} = 4, \quad \frac{8}{2} = 4, \quad \frac{12}{3} = 4$$
4. Since all ratios equal $4$, the relationship is proportional.
5. **Problem 10B:** Write the equation and find peanut butter for $25$ packages.
6. The proportional relationship is $y = kx$ where $k = 4$.
7. So, $$y = 4x$$
8. To find $x$ when $y=25$:
$$25 = 4x$$
9. Divide both sides by $4$:
$$x = \frac{25}{4}$$
10. Simplify:
$$x = 6.25$$
11. **Problem 11:** Relationship between dog weight $x$ and dog food $y$ is proportional.
12. Calculate ratio $\frac{y}{x}$ for given data:
$$\frac{\frac{1}{6}}{10} = \frac{1}{60}, \quad \frac{\frac{2}{3}}{40} = \frac{2}{3} \times \frac{1}{40} = \frac{1}{60}, \quad \frac{\frac{5}{6}}{50} = \frac{5}{6} \times \frac{1}{50} = \frac{1}{60}, \quad \frac{\frac{5}{3}}{100} = \frac{5}{3} \times \frac{1}{100} = \frac{1}{60}$$
13. Ratio is constant $k = \frac{1}{60}$.
14. Equation:
$$y = \frac{1}{60} x$$
15. For $x=25$ pounds:
$$y = \frac{1}{60} \times 25 = \frac{25}{60} = \frac{5}{12}$$
16. Recommended dog food is $\frac{5}{12}$ cups.
17. **Problem 12:** Find unit rate for $\frac{5}{8}$ mile in $\frac{1}{4}$ hour.
18. Unit rate = distance/time:
$$\frac{\frac{5}{8}}{\frac{1}{4}} = \frac{5}{8} \times \frac{4}{1} = \frac{20}{8} = 2.5$$ miles per hour.
19. **Problem 13:** Find unit rate for $68$ in $8 \frac{1}{2}$ hours.
20. Convert $8 \frac{1}{2} = \frac{17}{2}$ hours.
21. Unit rate:
$$\frac{68}{\frac{17}{2}} = 68 \times \frac{2}{17} = 8$$ per hour.
22. **Problem 14:** Find unit rate for $1 \frac{1}{4}$ cups flour per $\frac{1}{3}$ cup butter.
23. Convert $1 \frac{1}{4} = \frac{5}{4}$ cups.
24. Unit rate:
$$\frac{\frac{5}{4}}{\frac{1}{3}} = \frac{5}{4} \times 3 = \frac{15}{4} = 3.75$$ cups flour per cup butter.
25. **Problem 15:** Find unit rate for $2 \frac{1}{2}$ miles in $\frac{3}{4}$ hour.
26. Convert $2 \frac{1}{2} = \frac{5}{2}$ miles.
27. Unit rate:
$$\frac{\frac{5}{2}}{\frac{3}{4}} = \frac{5}{2} \times \frac{4}{3} = \frac{20}{6} = \frac{10}{3} \approx 3.33$$ miles per hour.