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3 years ago

What is the solution to the following equation?

2 answers:

4 0

2(3x - 7) + 18 = 10

Subtract 18 from both sides:

2(3x - 7) = -8

divide both sides by 2

3x - 7 = -4

add 7 to both sides

3x = 3

divide both sides by 3:

x = 1

So your answer is A) 1

olga2289 [7]3 years ago

3 0

2(3x - 7) + 18 = 10
Subtract 18 from both sides.
2(3x - 7) = -8
Distribute the side with parenthesis.
2(3x) + 2(-7) = -8
6x -14 = -8
Add 14 to both sides.
6x = 6
Divide both sides by 6.
x = 1

I hope this helps!

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Answer:

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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