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IceJOKER [234]
3 years ago
6

A rhombus with vertices P(2, 8), Q(3, 11), R(4, 8), and S(3, 5) is rotated 90° counterclockwise about the origin and then dilate

d by a factor of 2 with the origin as the center of dilation to obtain rhombus P′Q′R′S′. Which statement about the transformed rhombus P′Q′R′S′ is true?
Mathematics
1 answer:
jenyasd209 [6]3 years ago
7 0
The third choice is the correct answer if you're on plato
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2 years ago
Rewrite with only sin x and cos x.
Annette [7]

Option A

\cos 3 x=\cos x-4 \cos x \sin ^{2} x

<em><u>Solution:</u></em>

Given that we have to rewrite with only sin x and cos x

Given is cos 3x

cos 3x = cos(x + 2x)

We know that,

\cos (a+b)=\cos a \cos b-\sin a \sin b

Therefore,

\cos (x+2 x)=\cos x \cos 2 x-\sin x \sin 2 x  ---- eqn 1

We know that,

\sin 2 x=2 \sin x \cos x

\cos 2 x=\cos ^{2} x-\sin ^{2} x

Substituting these values in eqn 1

\cos (x+2 x)=\cos x\left(\cos ^{2} x-\sin ^{2} x\right)-\sin x(2 \sin x \cos x)  -------- eqn 2

We know that,

\cos ^{2} x-\sin ^{2} x=1-2 \sin ^{2} x

Applying this in above eqn 2, we get

\cos (x+2 x)=\cos x\left(1-2 \sin ^{2} x\right)-\sin x(2 \sin x \cos x)

\begin{aligned}&\cos (x+2 x)=\cos x-2 \sin ^{2} x \cos x-2 \sin ^{2} x \cos x\\\\&\cos (x+2 x)=\cos x-4 \sin ^{2} x \cos x\end{aligned}

\cos (x+2 x)=\cos x-4 \cos x \sin ^{2} x

Therefore,

\cos 3 x=\cos x-4 \cos x \sin ^{2} x

Option A is correct

7 0
2 years ago
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lesya692 [45]

We know that the sum of the interior angles of a triangle equals 180, then,in this case we have the following equation:

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then, solving for b, we get:

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now that we have that b = 41, we can find the measure of each angle:

\begin{gathered} b=41 \\ 2b=2(41)=82 \\ b+16=41+16=57 \end{gathered}

7 0
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Answer:

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Step-by-step explanation:

Took FLVS test and got it right

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