For t > 30, L (t), the linear approximation to A at t = 30, is a better model for the amount of grass clippings remaining in the bin. Use L (t) to predict the time at which there will be 0.5 pound of grass clippings remaining in the bin. Show the work that leads to your answer.

Question

Mathematics

Jason Velazquez

21 March, 23:37

For t > 30, L (t), the linear approximation to A at t = 30, is a better model for the amount of grass clippings remaining in the bin. Use L (t) to predict the time at which there will be 0.5 pound of grass clippings remaining in the bin. Show the work that leads to your answer.

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Answers (1)

Know the Answer?

Fletcher Quinn · Answer 1

35 days

Step-by-step explanation:

We know that L (t) is the tangent to the line A (t) at t = 30

At t = 30:

A (30) ≈ 0.783 → (30, 0.783) = (t, A (t))

A¹ (30) ≈ - 0.56

let - 0.56 = M

For a linear function, a straight line equation applies which leads to

(y-y₁) = m (x-x₁) ... 1

So for this problem we have;

A (t) - 0.783 = - 0.56 (t - 30)

when there is 0.5 pounds of grass, A (t) - 0.5, then using the equation (1)

(0.5 - 0.783) = - 0.56 (t-30)

t = 35.054

≈ 35 days