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14 February, 12:54

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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  1. 14 February, 13:00
    0
    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
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