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

true? is that what you're asking or do you mean something else?

step-by-step explanation:

here's what happens:

1) the graph of f   = x^2 is shifted 2 units to the right.

2) the resulting graph is reflected in the x-axis, so that the parabola now opens down.

3) this final result is transformed up by 5 units.