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Gennadij [26K]
3 years ago
5

I need help doing the steps

Mathematics
1 answer:
elixir [45]3 years ago
5 0
Http://chestofbooks.com/architecture/Building-Construction-V2/images/Stone-Steps-173.jpg
There ya go
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The tennis racket is $36.75 before tax in the sales tax is 6% what is the finial price
scoundrel [369]
The final price for the racket is $38.96
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PLEASE HELP ASAP WILL GIVE BRAINLIEST!!!!!
olasank [31]

Answer:

A. (0,7)

B. -1.25

C. Negative

D. I started at (0,7), then plotted the second point by moving the point down 5 and right 4.

Step-by-step explanation:

Look at the equation more carefully.

I hope this helps :)

5 0
3 years ago
Read 2 more answers
Convert 5 feet 4 inches to centimeters. (2.54 cm = 1 inch) *<br> PLEASE HELPPPP
hjlf

Answer:

162.56 cm

Step-by-step explanation:

also you can use a conversion calculator for faster results too

can i get brainliest plz

7 0
3 years ago
Four less than twice a number is ten. Find the number
Amanda [17]

the answer is going to be 7

6 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
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