If 36x^3a+45xa^2=9a() then 9a is a factor the other factor, (), is the left side divided by 9a

(36x^3a+45xa^2)/(9a)

4x^3+5xa

So the () term is (4x^3+5xa)

7 0

3 years ago

Matthew spent 2/3 of his birthday on books and 1/4 of his birthday money on snacks. He says he spent 3/7 of his birthday money.

Solnce55 [7]

Answer:

Step-by-step explanation:

$\frac{2}{3}$ + $\frac{1}{4}$ = $\frac{8+3}{12}$ = $\frac{11}{12}$ ≈ 1

The correct answer is the last one (C)

7 0

3 years ago

Find tanθ if sinθ=−25√5 is in the third quadrant.

irinina [24]

Using a trigonometric identity, it is found that the tangent of the angle is of:

$\tan{\theta} = \frac{2}{11}$

<h3>How to find the sine of an angle given the cosine, or vice versa?</h3>

The sine and the cosine of the angle are related according to the following identity:

$\sin^2{\theta} + \cos^2{\theta} = 1$

For this problem, the sine is given by:

$\sin{\theta} = -\frac{2}{5\sqrt{5}}$

Hence the cosine is found as follows:

$\left(-\frac{2}{5\sqrt{5}}\right)^2 + \cos^2{\theta} = 1$

$\frac{4}{125} + \cos^2{\theta} = 1$

$\cos^2{\theta} = \frac{121}{125}$

$\cos{\theta} = \pm \sqrt{\frac{121}{125}}$

On the third quadrant the cosine is negative, hence:

$\cos{\theta} = -\frac{11}{5\sqrt{5}}$

<h3>What is the tangent of an angle?</h3>

The tangent of an angle is given by the sine divided by the cosine, hence:

$\tan{\theta} = \frac{-\frac{2}{5\sqrt{5}}}{-\frac{11}{5\sqrt{5}}} = \frac{2}{11}$

More can be learned about trigonometric identities at brainly.com/question/26676095

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4 0

2 years ago