3 years ago

Can someone show me how to do this?

Mathematics

1 answer:

andriy [413]3 years ago

6 0

3/4 ( (-50+2)/5)
= 3/4 ( -48/5)
= 3/4 * -48/5
= -36/5

You might be interested in

Write an equation and find the missing angle.

AVprozaik [17]

4+5x+x+2= 180
29

4+5(29)=149degrees

(29)+2= 31 degrees

i hope this helps

3 0

3 years ago

Whats the answer for number 9?

Tasya [4]

<h2>847.8 is the correct answer!</h2><h2></h2><h3>9 x 2 = 18</h3><h3>18 x 3.14 = 56.52</h3><h3>56.52 x 15 = 847.8</h3><h3></h3><h3><em>Please let me know if I am wrong.</em></h3>

3 0

3 years ago

Simplify <img src="https://tex.z-dn.net/?f=7%5E%7B-2log_%7B7%7D3%7D" id="TexFormula1" title="7^{-2log_{7}3}" alt="7^{-2log_{7}3}

jasenka [17]

This expression is equivalent to:

[ 7^(log to the base 7 of 3)]^(-2), or [3]^(-2)], or 1/9.

6 0

3 years ago

Find exact value of cos

mylen [45]

<h3>Answer: Choice B. $\frac{\sqrt{91}}{10}$ </h3>

Work Shown:

$\sin^2 \theta + \cos^2 \theta = 1\\\\\left(\frac{3}{10}\right)^2 + \cos^2 \theta = 1\\\\\frac{9}{100} + \cos^2 \theta = 1\\\\\cos^2 \theta = 1 - \frac{9}{100}\\\\\cos^2 \theta = \frac{100}{100}-\frac{9}{100}\\\\\cos^2 \theta = \frac{91}{100}\\\\\cos \theta = \sqrt{\frac{91}{100}} \ \text{ cosine positive in Q1}\\\\\cos \theta = \frac{\sqrt{91}}{\sqrt{100}}\\\\\cos \theta = \frac{\sqrt{91}}{10}\\\\$

7 0

3 years ago

Read 2 more answers

C. Square root of 3 <br> D. Cannot be determined

diamong [38]

Answer:

Step-by-step explanation:

Angle QSR is 180° - 150°, is 30°

So basically you're getting the answer for tan30° which is $\frac{\sqrt{3} }{3}$

6 0

3 years ago