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3 years ago

WILL GIVE BRAINLIEST!!! NO UNHELPFUL ANSWERS OR I WILL REPORT!!!

Mathematics

2 answers:

Gemiola [76]3 years ago

8 0

Answer:

B is the answer

Step-by-step explanation:

I figured out that it cant be D because website k is not on the left, Website L is, and Website L is not symmetrical, and B is a better answer then A if you think about it.

matrenka [14]3 years ago

3 0

Answer:

B the range of data for website L is equal to the range of website K

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If tan theta = 10/13 and cos theta > 0 then sin2theta is what ?

svetoff [14.1K]

$sin2\alpha = \frac{260}{269}$

<u>Step-by-step explanation:</u>

We have , $Tan\alpha = \frac{Perpendicular}{Base} = \frac{10}{13}$ ,

We know that $sin\alpha = \frac{Perpendicular}{Hypotenuse} = \frac{Perpendicular}{\sqrt[2]{(Perpendicualr)^{2} + (Base)^{2})} }$

Substituting values of P & B , $sin\alpha = \frac{10}{\sqrt{10^{2} + 13^{2}} } = \frac{10}{\sqrt{269} }$

Now , $sin2\alpha = 2sin\alpha cos\alpha = 2sin\alpha \sqrt{1 - (sin\alpha)^{2} }$

⇒ $sin2\alpha =$ $\frac{10}{\sqrt{269} }$ × $\sqrt{1 - (\frac{10}{\sqrt{269} })^{2} }$ ×2

⇒ $sin2\alpha = \frac{20}{\sqrt{269} }( \sqrt{\frac{269 - 100}{269} } )$

⇒ $sin2\alpha = \frac{20}{\sqrt{269} }( \sqrt{\frac{169 }{269} } )$

⇒ $sin2\alpha = \frac{260}{\sqrt{269} }( \sqrt{\frac{1}{269} } )$

⇒ $sin2\alpha = \frac{260}{269}$

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