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2 years ago

11 The two trapezoids shown below are

Mathematics

1 answer:

Damm [24]2 years ago

3 0

Answer:

9.6 (D)

Step-by-step explanation:

You know that when they are similar, they have a ratio. Assume that WT is x. 6:10 is the ratio, or 3:5.

So, what number is x when x:16=3:5?

(two ways)

1. You can cross multiply

x/16 = 3:5, 3x16=5x

48=5x

x=9.6

2. You can do it by looking at the answer choices and simplifying them.

10:16=5:8

3.8:16=1.9:8

12:16=3:4

9.6:16=4.8:8=

2.4:4

0.6:1

6:10

YES

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In the base 5 number system, which number comes after 444,444?

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I think it’s 555,555

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2 years ago

The difference of 9 times a number and 2 is 67. Write an equation that can be used to find the unknown number?

pochemuha

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3 years ago

2 tan 30°<br>II<br>1 + tan- 300

shusha [124]

Question:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$

Answer:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= sin(60^{\circ})$

Step-by-step explanation:

Given

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$

Required

Simplify

In trigonometry:

$tan(30^{\circ}) = \frac{1}{\sqrt{3}}$

So, the expression becomes:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{2 * \frac{1}{\sqrt{3}}}{1 + (\frac{1}{\sqrt{3}})^2}$

Simplify the denominator

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{2 * \frac{1}{\sqrt{3}}}{1 + \frac{1}{3}}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{\frac{2}{\sqrt{3}}}{1 + \frac{1}{3}}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{\frac{2}{\sqrt{3}}}{ \frac{3+1}{3}}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{\frac{2}{\sqrt{3}}}{ \frac{4}{3}}$

Express the fraction as:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{2}{\sqrt 3} / \frac{4}{3}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{2}{\sqrt 3} * \frac{3}{4}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{1}{\sqrt 3} * \frac{3}{2}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{3}{2\sqrt 3}$

Rationalize

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{3}{2\sqrt 3} * \frac{\sqrt{3}}{\sqrt{3}}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{3\sqrt{3}}{2* 3}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{\sqrt{3}}{2}$

In trigonometry:

$sin(60^{\circ}) = \frac{\sqrt{3}}{2}$

Hence:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= sin(60^{\circ})$

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3 years ago

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AlladinOne [14]

Answer:

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Step-by-step explanation:

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2 years ago

Factor-3/4 out of 3/2x-9/4y

S_A_V [24]

For this case we must factor $- \frac {3} {4}$ of the following expression:

$\frac {3} {2} x- \frac {9} {4}y$

By law of signs of multiplication it is fulfilled that:

$- * - = +\\- * + = -$

So we have to:

$- \frac {3} {4} * (- 2) = + \frac {6} {4} = + \frac {3} {2}\\- \frac {3} {4} * 3 = - \frac {9} {4}$

Thus, by factoring we have:

$- \frac {3} {4} (- 2x + 3y)$

Answer:

$- \frac {3} {4} (- 2x + 3y)$

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2 years ago