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suter [353]
2 years ago
11

Can someone tell me if i did this right lol

Mathematics
2 answers:
Yanka [14]2 years ago
8 0

Answer:

I'm 96% sure its right, and those are pretty good odds ^^

baherus [9]2 years ago
6 0

Answer:

Your calculation is correct, but I'd answer 33.0 in.

Step-by-step explanation:

You are correct. Good job.

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Answer:

x = 4

Step-by-step explanation:

ΔTRQ is an isosceles right triangle, so if we find the value of RT then the value of 'x' will be the same

We can find RT by creating a proportion based on the ratio of sides in a 30-60-90° triangle which, respectively, is 1 : \sqrt{3} : 2

2\sqrt{3}/RT = \sqrt{3}/2

cross-multiply:

\sqrt{3}· RT = 4\sqrt{3}

RT = 4

Therefore, x = 4

3 0
2 years ago
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Bogdan [553]
E= 116 joules
That is what I think the answer is
7 0
3 years ago
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Answer:

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Step-by-step explanation:

7 0
3 years ago
The area of circle K when c=24x^2 pi
Gala2k [10]

Answer:\frac{\sqrt{6}\sqrt{\pi }\sqrt{Kc}}{12\pi },\:x=-\frac{\sqrt{6}\sqrt{\pi }\sqrt{Kc}}{12\pi }

Step-by-step explanation:

c=24x^2 pi

Switch sides:

24x^2\pi =Kc

divide both sides by 24\pi

\frac{24x^2\pi }{24\pi }=\frac{Kc}{24\pi }

simplify:

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x^2=\frac{Kc}{24\pi }x=\sqrt{\frac{Kc}{24\pi }},\:x=-\sqrt{\frac{Kc}{24\pi }}

answer:x=\frac{\sqrt{6}\sqrt{\pi }\sqrt{Kc}}{12\pi },\:x=-\frac{\sqrt{6}\sqrt{\pi }\sqrt{Kc}}{12\pi }

i hope it's help!

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Vanyuwa [196]
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