Algebra 1

Published by Prentice Hall
ISBN 10: 0133500403
ISBN 13: 978-0-13350-040-0

Chapter 4 - An Introduction to Functions - 4-5 Writing a Function Rule - Practice and Problem-Solving Exercises - Page 266: 26


1. The function for this relationship would be h(j)= 10 - $\frac{10}{64}$j 2. The height, after 47 ounces have been poured is 2.65 inches. 3. The height after 32 ounces or half of pitcher has been poured is 5 inches. 4. To decrease height by 1 in you would pour out $\frac{1}{10}$ of the juice.

Work Step by Step

1. The function for this relationship would be h(j)= 10 - $\frac{10}{64}$j because we know that the initial volume of the jug is 10 inches so we are subtracting from 10 inches. To find the slope or rate with which the height decreases per one ounce we divided the initial height 10 inches by the total ounces of juice which is 64 oz to say that the jug is decreasing by $\frac{10}{64}$ in/oz. 2. To find height after 47 ounces are poured out we will plug in 47 for j. h(47)= 10 - ($\frac{10}{64}$$\times$47) = 2.65 inches of height 3. To find height after half of the juice is poured out which is when 32 ounces are poured out, we will plug in 32 for j. h(32)= 10 - ($\frac{10}{64}$$\times$32) = 10-5 = 5 inches of height 4. To find what fraction of the juice in container must be dropped to decreases height by 1 in, we will find the amount of juice that needs to be poured out. To solve this the final height is 9 as you decreases 1 in from initial height of 10 inches. h(j)= 10 - ($\frac{10}{64}$$\times$j) 9 = 10 - ($\frac{10}{64}$$\times$j) ($\frac{10}{64}$$\times$j) = 10-9 ($\frac{10}{64}$$\times$j) = 1 j = ($\frac{64}{10}$$\times$1) I =6.4 oz Fraction = $\frac{6.4}{64}$ = $\frac{1}{10}$ So $\frac{1}{10}$ which is 6.4 ounces from the whole amount of 64 ounces is poured out to decrease the height by 1 in.
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