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EleoNora [17]
3 years ago
11

PLS help will give brainly points thank you

Mathematics
2 answers:
iris [78.8K]3 years ago
4 0
<h3>Answer : </h3>

a). 20°

<h3>Explanation :</h3>

The angle x is equal to 20° because vertical angle.

frez [133]3 years ago
3 0

Answer:

I think the answer will be 20. It is oppsite sides and the lines look even. The area is small as well. So I think 20˙ will be the best option.

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Which expression is the simplest form of 5(x-3)-3(2x+4)/9?
nataly862011 [7]
-18 0.5 add that the together and u will get your answer
6 0
3 years ago
Can someone help me pls
Andrej [43]

Answer:

x=78

Step-by-step explanation:

q is 90 degrees since it is a right angle,

p=12

90 plus 12 =102

x=180-102

x=78

3 0
3 years ago
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Solve -3 (x + 1) - 2x= -6<br><br> A x=1<br> B x=5<br> C x = -7<br> D x = -9
Sliva [168]

Answer: C

Step-by-step explanation:

4 0
4 years ago
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Find the cube roots of 27(cos 330° + i sin 330°)
Aleksandr-060686 [28]

Answer:

See below for all the cube roots

Step-by-step explanation:

<u>DeMoivre's Theorem</u>

Let z=r(cos\theta+isin\theta) be a complex number in polar form, where n is an integer and n\geq1. If z^n=r^n(cos\theta+isin\theta)^n, then z^n=r^n(cos(n\theta)+isin(n\theta)).

<u>Nth Root of a Complex Number</u>

If n is any positive integer, the nth roots of z=rcis\theta are given by \sqrt[n]{rcis\theta}=(rcis\theta)^{\frac{1}{n}} where the nth roots are found with the formulas:

  • \sqrt[n]{r}\biggr[cis(\frac{\theta+360^\circ k}{n})\biggr] for degrees (the one applicable to this problem)
  • \sqrt[n]{r}\biggr[cis(\frac{\theta+2\pi k}{n})\biggr] for radians

for  k=0,1,2,...\:,n-1

<u>Calculation</u>

<u />z=27(cos330^\circ+isin330^\circ)\\\\\sqrt[3]{z} =\sqrt[3]{27(cos330^\circ+isin330^\circ)}\\\\z^{\frac{1}{3}} =(27(cos330^\circ+isin330^\circ))^{\frac{1}{3}}\\\\z^{\frac{1}{3}} =27^{\frac{1}{3}}(cos(\frac{1}{3}\cdot330^\circ)+isin(\frac{1}{3}\cdot330^\circ))\\\\z^{\frac{1}{3}} =3(cos110^\circ+isin110^\circ)

<u>First cube root where k=2</u>

<u />\sqrt[3]{27}\biggr[cis(\frac{330^\circ+360^\circ(2)}{3})\biggr]\\3\biggr[cis(\frac{330^\circ+720^\circ}{3})\biggr]\\3\biggr[cis(\frac{1050^\circ}{3})\biggr]\\3\biggr[cis(350^\circ)\biggr]\\3\biggr[cos(350^\circ)+isin(350^\circ)\biggr]

<u>Second cube root where k=1</u>

\sqrt[3]{27}\biggr[cis(\frac{330^\circ+360^\circ(1)}{3})\biggr]\\3\biggr[cis(\frac{330^\circ+360^\circ}{3})\biggr]\\3\biggr[cis(\frac{690^\circ}{3})\biggr]\\3\biggr[cis(230^\circ)\biggr]\\3\biggr[cos(230^\circ)+isin(230^\circ)\biggr]

<u>Third cube root where k=0</u>

<u />\sqrt[3]{27}\biggr[cis(\frac{330^\circ+360^\circ(0)}{3})\biggr]\\3\biggr[cis(\frac{330^\circ}{3})\biggr]\\3\biggr[cis(110^\circ)\biggr]\\3\biggr[cos(110^\circ)+isin(110^\circ)\biggr]

4 0
3 years ago
PLEASE HELP ME NEED ASAP!!!!
rosijanka [135]

Answer:

It should be (0, 24)

Step-by-step explanation:

I hope this helped.

4 0
3 years ago
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