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.

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