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77julia77 [94]
3 years ago
13

Evaluate the limit if it exists

Mathematics
1 answer:
raketka [301]3 years ago
5 0
Hello  here is a solution : 

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3 years ago
5<br> 0 = - in<br> =<br> in quadrant II<br> Given cos =<br> Find sin<br> 3
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Answer:

sinΘ = \frac{2}{3}

Step-by-step explanation:

using the identity

sin²x + cos²x = 1  ( subtract cos²x from both sides )

sin²x = 1 - cos²x ( take square root of both sides )

sinx = ± \sqrt{1-cos^2x}

given

cosΘ = - \frac{\sqrt{5} }{3} , then

sinΘ = ± \sqrt{1-(-\frac{\sqrt{5} }{3})^2 }

        = ± \sqrt{1-\frac{5}{9} }

        = ± \sqrt{\frac{4}{9} }

        = ± \frac{2}{3}

since Θ is in quadrant II where sinΘ > 0 , then

sinΘ = \frac{2}{3}

7 0
3 years ago
Which triangles can not be proved congruent?#37
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