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hodyreva [135]
4 years ago
5

40 POINTS !!!!!PLEASE HELPPP!!!!!!

Mathematics
2 answers:
TEA [102]4 years ago
7 0

Answer:

sin(a) = \frac{7\sqrt{5} }{20}

Step-by-step explanation:

cos(2a) = cos²(a)-sin²(a) =  [1-sin²(a)]-sin²(a)= 1 - 2sin²(a)

1 - 2sin²(a)  = -20/29

- 2sin²(a) = -49/40

sin²(a) = 49/80

sin(a) = \sqrt{\frac{49}{80} }

sin(a) = \frac{\sqrt{49} \sqrt{80} }{80} = \frac{7*4\sqrt{5} }{80}

sin(a) = \frac{7\sqrt{5} }{20}

zlopas [31]4 years ago
3 0

Answer:

cos(theta) = 3sqrt(58)/58

sin(theta) = 7sqrt(58)/58

Step-by-step explanation:

Cos(2theta) = 2cos²(theta) - 1

-20/29 = 2cos²(theta) - 1

2cos²(theta) = 9/29

cos²(theta) = 9/58

cos(theta) = sqrt(9/58)

cos(theta) = 3sqrt(58)/58

sin²(theta) = 1 - cos²(theta)

sin²(theta) = 1 - 9/58

sin²(theta) = 49/58

sin(theta) = sqrt(49/58)

sin(theta) = 7sqrt(58)/58

sqrt: square root

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Answer:

Clockwise vertices of q' ( -3 ,4) →→ q( 4 , 3 ).

Counter clock wise rule : q' (-3 ,4 ) →→ q( -4 , -3 ).

Step-by-step explanation:

Given  : rectangle has vertices located at r'(–4, 4), s'(–4, 1), p'(–3, 1), and q'(–3, 4)

To find :  transformed according to the rule 90º , what is the location of q?

Solution : we have given that

vertices located at r'(–4, 4), s'(–4, 1), p'(–3, 1), and q'(–3, 4).

By the rule of 90º rotation clock wise rule : (x ,y ) →→ ( y , -x )

90º rotation counter clock wise rule : (x ,y ) →→ ( -y , x ).

Then   Clockwise vertices of q' ( -3 ,4) →→ q( 4 , 3 ).

counter clock wise rule : q' (-3 ,4 ) →→ q( -4 , -3 ).

Therefore, Clockwise vertices of q' ( -3 ,4) →→ q( 4 , 3 ).

counter clock wise rule : q' (-3 ,4 ) →→ q( -4 , -3 ).

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3 years ago
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Answer:

see explanation

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