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tester [92]
2 years ago
8

HELP ASAP!! GEOMETRY QUESTION!

Mathematics
1 answer:
levacccp [35]2 years ago
4 0

Answer:

The answer is 6y - 5x = 18

Step-by-step explanation:

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Hey guys, I am about to start studying for a math test happening in two weeks about statistics and I am wondering if you guys ha
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Answer more questions on every topic you’ve been thought. Answer research questions and questions likely to come out in the exam. Dwell on what you are trying to know more that what you already know. A planner could help as well and the Good God would help you
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2 years ago
Kathy uses 7.8 pints of white paint and blue paint to paint her bedroom walls. 3/4
mel-nik [20]

Answer:

1.95

Step-by-step explanation:

8 0
3 years ago
Same question bc i cant find the one i posted earlier
lora16 [44]

Answer:

224 pi

703.36  either might be in your answer list.

Step-by-step explanation:

Givens

r = 8 inches

rotations = 14

Formula

D = 14*C where C = 2 pi r

Solution

D = 14 * (2 pi * 8)

D = 224 * pi                                    One possible answer

D = 224 * 3.14  = 703.36 inches    Another possible Answer

8 0
3 years ago
What is the square root of -2i?
Studentka2010 [4]

Answer:

1-i and -1+i

Step-by-step explanation:

We are to find the square roots of z=0-2i. First, convert from Cartesian to polar form:

r=\sqrt{a^2+b^2}\\r=\sqrt{0^2+(-2)^2}\\r=\sqrt{0+4}\\r=\sqrt{4}\\r=2

\theta=tan^{-1}(\frac{b}{a})\\ \theta=tan^{-1}(\frac{-2}{0})\\\theta=\frac{3\pi}{2}

z=2(\cos\frac{3\pi}{2}+i\sin\frac{3\pi}{2})

Next, use the formula \displaystyle \sqrt[n]{r}\biggr[\cis\biggr(\frac{\theta+2\pi k}{n}\biggr)\biggr] where \displaystyle k=0,1,2,...\:,n-1 to find the square roots:

<u>When k=1</u>

<u />\displaystyle \sqrt[2]{2}\biggr[cis\biggr(\frac{\frac{3\pi}{2}+2\pi(1)}{2}\biggr)\biggr]

\displaystyle \sqrt{2}\biggr[cis\biggr(\frac{3\pi}{4}+\pi\biggr)\biggr]

\sqrt{2}\biggr(cis\frac{7\pi}{4}\biggr)

\sqrt{2}(\cos\frac{7\pi}{4}+i\sin\frac{7\pi}{4})\\ \\\sqrt{2}(\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i)\\ \\1-i

<u>When k=0</u>

<u />\displaystyle \sqrt[2]{2}\biggr[cis\biggr(\frac{\frac{3\pi}{2}+2\pi(0)}{2}\biggr)\biggr]

\sqrt{2}\biggr(cis\frac{3\pi}{4}\biggr)

\sqrt{2}(\cos\frac{3\pi}{4}+i\sin\frac{3\pi}{4})\\ \\\sqrt{2}(-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i)\\ \\-1+i

Thus, the square roots of -2i are 1-i and -1+i

4 0
2 years ago
What conjecture can you make about the sum of the first 30 odd numbers
adoni [48]
It will be an even number!
5 0
3 years ago
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