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4 years ago

10 PTS! How do you solve:

Mathematics

1 answer:

kotykmax [81]4 years ago

7 0

The length of the open pit is Area divided by width

A/w = L

L = (12x^2 + 8x)/4x

L = 12x^2 / 4x + 8x/4x

L = 3x + 2

The length of the back part that has to be sealed is 3x + 2 + 2 + 2 = 3x + 8

Area of the White part and the shaded part = L * W

W = 10x -1

L = 3x + 8

Area of the White Part and the Shaded Part = (10x - 1)(3x +8)

The Area of the Shaded Part + White Part = the total area.

Area of the Shaded Part + 12x^2 + 8x = (3x + 8) * (10x - 1)

Expand the right hand side.

Area of the Shaded Part + 12x^2 + 8x = 30x^2 - 3x + 80x - 8 Combine

Area of the Shaded Part + 12x^2 + 8x = 30x^2 + 77x - 8

Answer

Subtract 12x^2 + 8x From both sides.

Area of the Shaded Part = 30x^2 - 12x + 77x - 8x -8

Area of the Shaded Part = 18x^2 + 69x - 8 Answer

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============================================

Explanation:

In the sequence ABC, the letter B is the second letter and A is the first.

So the corresponding segment to BA will have its first slot be the second letter of DEF, and its second slot be the first letter of DEF. That's how I'm getting ED for the answer. It is the same as DE because it's the same segment.

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Hoochie [10]

Answer with explanation:

The given statement is which we have to prove by the principal of Mathematical Induction

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$2^{k}>k$

---------------------------------------(1)

3⇒Now, we will prove that the mathematical statement is true for, n=k+1.

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