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k0ka [10]
2 years ago
6

Which represents the inverse of the function f(x) = 4x?

Mathematics
1 answer:
nekit [7.7K]2 years ago
5 0

Answer:

\displaystyle h(x) = \frac{1}{4}x

Step-by-step explanation:

Sinse in this case you are dealing with direct variation, all you are doing is taking the <em>multiplicative</em><em> </em><em>inverse</em><em> </em>of 4, which is ¼:

\displaystyle f(x) = 4x \rightarrow f^{-1}(x) = \frac{1}{4}x

This option is not an answer choise. Perhaps there was a typographical errour.

I am joyous to assist you at any time.

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A heavy rope, 50 ft long, weighs 0.6 lb/ft and hangs over the edge of a building 120 ft high. Approximate the required work by a
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Exercise (a)

The work done in pulling the rope to the top of the building is 750 lb·ft

Exercise (b)

The work done in pulling half the rope to the top of the building is 562.5 lb·ft

Step-by-step explanation:

Exercise (a)

The given parameters of the rope are;

The length of the rope = 50 ft.

The weight of the rope = 0.6 lb/ft.

The height of the building = 120 ft.

We have;

The work done in pulling a piece of the upper portion, ΔW₁ is given as follows;

ΔW₁ = 0.6Δx·x

The work done for the second half, ΔW₂, is given as follows;

ΔW₂ = 0.6Δx·x + 25×0.6 × 25 =  0.6Δx·x + 375

The total work done, W = W₁ + W₂ = 0.6Δx·x + 0.6Δx·x + 375

∴ We have;

W = 2 \times \int\limits^{25}_0 {0.6 \cdot x} \, dx + 375= 2 \times \left[0.6 \cdot \dfrac{x^2}{2} \right]^{25}_0 + 375 = 750

The work done in pulling the rope to the top of the building, W = 750 lb·ft

Exercise (b)

The work done in pulling half the rope is given by W₂ as follows;

W_2 =  \int\limits^{25}_0 {0.6 \cdot x} \, dx + 375= \left[0.6 \cdot \dfrac{x^2}{2} \right]^{25}_0 + 375 = 562.5

The work done in pulling half the rope, W₂ = 562.5 lb·ft

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