Find the surface area of a rectangular prism with a height of 6 inches, a width of 7 inches, and a length of 13 inches.

Mathematics

1 answer:

madam [21]3 years ago

5 0

The answer is B)422 ,you're welcome :)

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The point (5, -2) is rotated 90° about the origin. Its image is:

Morgarella [4.7K]

Image below, rotation for point (5,-2) gives you (2,5).

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Which of the tables represents a function?

Damm [24]

Input is domain and output is co-domain.
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Prove that (Root of Sec A - 1 / Root of Sec A + 1) + (Root of Sec A + 1 / Root of Sec A - 1) = 2 cosec A

iVinArrow [24]

Answer:

The answer is below

Step-by-step explanation:

We need to prove that:

(Root of Sec A - 1 / Root of Sec A + 1) + (Root of Sec A + 1 / Root of Sec A - 1) = 2 cosec A.

Firstly, 1 / cos A = sec A, 1 / sin A = cosec A and tanA = sinA / cosA.

Also, 1 + tan²A = sec²A; sec²A - 1 = tan²A

$\frac{\sqrt{secA-1} }{\sqrt{secA+1} } +\frac{\sqrt{secA+1} }{\sqrt{secA-1} } =\frac{(\sqrt{secA-1)}(\sqrt{secA-1})+(\sqrt{secA+1)}(\sqrt{secA+1}) }{(\sqrt{secA+1})(\sqrt{secA-1}) } \\\\=\frac{secA-1+(secA+1)}{\sqrt{sec^2A-secA+secA-1} } \\\\=\frac{2secA}{\sqrt{sec^2A-1} } \\\\=\frac{2secA}{\sqrt{tan^2A} } \\\\=\frac{2secA}{tanA} \\\\=\frac{2*\frac{1}{cosA} }{\frac{sinA}{cosA} }\\\\= 2*\frac{1}{cosA}*\frac{cosA}{sinA}\\\\=2*\frac{1}{sinAA}\\\\=2cosecA$

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2 years ago

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pychu [463]

-3(b - 7)

Distributive property

-3*b = -3b

-3*7 = -21

-3b + 21

Answer: -3b + 21

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2 years ago

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EastWind [94]

Answer: (4x+12) - (2x-5)

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