Due in 43 minutes. Due Wed 01/31/2018 11 You\'ve already done this problem. Scor

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Due in 43 minutes. Due Wed 01/31/2018 11 You've already done this problem. Score on last attempt8 out of 10 Score in gradebook:8 out of 10 Reattempt this question Question with last attempt is displayed for your review only Jeff is watersking at Lake Pleasant. His distance from the island in the center of the lake increases at a constant speed of 43.8 feet per minute. 5.2 minutes after Jeff started waterskiing he 233.86 feet from the island. a. How much will Jeff's distance from the island change as the number of minutes since he started watersking increases from 5.2 to 12.5 minates? feet Preview b. What is Jeff's distance from the island 12.5 minutes after he started waterskiing? feet Preview 553.6 c. How much does Jeff's distance from the island change when the number of minutes since he started watersking decreases from 5.7 to 5.1 minutes? 26.28 d. What was Jeft's distance from the island before he started watersking? #feet Preview feet Preview e. Define a formula that gives Jeff's distance from the island, k, in terms of the number of minutes that have elapsed since he started waterskiing, t k-(43 6.1) Preview Post this question to forum cBo

Explanation / Answer

Speed = 43.8 ft/min
When t = 5.2, d = 233.86 ft

a)
So, d = st + C

d = 43.8t + C

Using (5.2 , 233.86), we have :
233.86 = 43.8*5.2 + C
C = 6.1

So, d= 43.8t + 6.1

When t = 5.2, d = 233.86
When t = 12.5, d = 553.6

So, change is :
553.6 - 233.86
i.e 319.74

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b)
d(12.5) = 533.6 as found above

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c)
When t = 5.7, d = 255.76
When t = 5.1, d = 229.48
So, the change in distance is :
- 26.28

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d)
We alrady found y-intercept c = 6.1
So, 6.1

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e)
Already done above
d= 43.8t + 6.1

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