We are given with two equations to find the values of two variables, hence the problem can be solved.

Adding the two equations:

x + y = 12<u>x - y = 10
</u>2x = 22
<u />x =11
y = 1

8 0

3 years ago

Read 2 more answers

What is the volume, in cubic cm, of a rectangular prism with a height of 19cm, a

3241004551 [841]

Answer: 2432

19 x 16 x 8 = 2432

6 0

2 years ago

Read 2 more answers

Has anyone done the segment one practice exam in geometry

konstantin123 [22]

I have done the segment 1 pratice exam

5 0

2 years ago

Help!! Which of the following best represents Z1 • Z2 select all that apply

AURORKA [14]

$z_1=\sqrt{3}\left(cos\:\frac{\pi }{4}+i\:sin\:\frac{\pi }{4}\right)$

$z_2=\sqrt{6}\left(cos\:\frac{3\pi \:}{4}+i\:sin\:\frac{3\pi \:}{4}\right)$

$\cos \left(\frac{\pi }{4}\right)=\frac{\sqrt{2}}{2}$

$\sin \left(\frac{\pi }{4}\right)=\frac{\sqrt{2}}{2}$

$\cos \left(\frac{3\pi }{4}\right)=-\frac{\sqrt{2}}{2}$

$\sin \left(\frac{3\pi }{4}\right)=\frac{\sqrt{2}}{2}\\$

$Z_1*Z_2=\sqrt{3}\left(cos\:\frac{\pi }{4}+i\:sin\:\frac{\pi }{4}\right)\cdot \sqrt{6}\left(cos\:\frac{3\pi \:}{4}+i\:sin\:\frac{3\pi \:}{4}\right)$

$=\sqrt{3}\left(\frac{\sqrt{2}}{2}+\:i\frac{\sqrt{2}}{2}\right)\cdot \sqrt{6}\left(\frac{-\sqrt{2}}{2}+\:i\frac{\sqrt{2}}{2}\right)$

On simplifying, we get

$Z_1* Z_2 =-3\sqrt{2}$

<h2>Therefore, correct option is 1st option.</h2>

3 0

2 years ago