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OverLord2011 [107]
3 years ago
11

Find the value of x when lines w and v are parallel.

Mathematics
1 answer:
Elodia [21]3 years ago
8 0

Answer:

the answer is A

Step-by-step explanation:

when lines w and v are parallel then

65 = 4x - 3

62 = 4x

x = 15.5

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A right triangular prism is shown. A right triangular prism is shown. The right triangles have sides with lengths 15 and 8. The
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900 cubic units

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5.5(x+6 1/2)-(x+9 1/3)-(19-x)
anastassius [24]

The simplified form of the expression [5.5(x+6 1/2)-(x+9 1/3)-(19-x)] is 11x/2 + 89/12

<h3>What is the simplified form of the expression</h3>

Given the expression;

5.5(x+6 1/2) - (x+9 1/3) - (19-x)

First, we convert 6 1/2, 9 1/3 and 5.5 to an improper fraction

6 1/2 = 13/2, 9 1/3 = 28/3 and 5.5 = 11/2

So, we have

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Next, we remove the parentheses

11x/2 + 143/4 - x - 28/3 - 19 + x

11x/2 + 143/4 - 28/3 - 19

11x/2 + 317/12 - 19

11x/2 + 89/12

Therefore, the simplified form of the expression [5.5(x+6 1/2)-(x+9 1/3)-(19-x)] is 11x/2 + 89/12.

Learn more about fractions here: brainly.com/question/28039882

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3 0
1 year ago
Simplify the expression.<br><br> [(11 - 4)^3]^2 ÷ (4 + 3)^5
MAXImum [283]

Answer:

<h2>7</h2>

Step-by-step explanation:

\left[\left(11\:-\:4\right)^3\right]^2\:\div \left(4\:+\:3\right)^5\\\\\frac{\left(\left(11-4\right)^3\right)^2}{\left(4+3\right)^5}\\\\\mathrm{Subtract\:the\:numbers:}\:11-4=7\\\\=\frac{\left(7^3\right)^2}{\left(4+3\right)^5}\\\\\mathrm{Add\:the\:numbers:}\:4+3=7\\\\=\frac{\left(7^3\right)^2}{7^5}\\\\\left(7^3\right)^2=7^6\\\\=\frac{7^6}{7^5}\\\\\mathrm{Apply\:exponent\:rule}:\quad \frac{x^a}{x^b}=x^{a-b}\\\\\frac{7^6}{7^5}=7^{6-5}\\\\\mathrm{Subtract\:the\:numbers:}\:6-5=1\\\\=7

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