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Ludmilka [50]
3 years ago
14

Express sin Q as a fraction in simplest terms.​

Mathematics
2 answers:
____ [38]3 years ago
8 0

Answer:

b

Step-by-step explanation:

hgvfigjl

ohaa [14]3 years ago
8 0

Answer:

hypotenuse =  \sqrt{ {27}^{2} +  {36}^{2}  }  \\  =  \sqrt{2025}  = 45 \\  \sin \: Q =  \frac{opposite}{hypotenuse}  \\  \sin \: Q =  \frac{36}{45}  \\  \sin \: Q =  \frac{4}{5}

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maw [93]

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that is the solution to the question

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X + 98<br> x + 98<br> Solve for x
disa [49]

Answer:

I believe your answer is commutative property

4 0
3 years ago
Find angle &lt;QPS in the diagram​
marin [14]

Answer:

40°

Step-by-step explanation:

Because triangle QSR is isosceles ∠SQR=∠SRQ=35°. The sum of the angles in a triangle is 180°, so ∠QSR=180°-35°-35°=110°. The measure of a straight line is 180°, so ∠PSQ=180°-110°=70°. Because triangle PSQ is also isosceles ∠PSQ=∠PQS=70°. Then, ∠QPS=180°-70°-70°=40°.

4 0
3 years ago
Write the standard form of the equation of the circle with the given characteristics.
SOVA2 [1]
<h2><u>Circle Equations</u></h2>

<h3>Write the standard form of the equation of the circle with the given characteristics.</h3><h3>Center: (0, 0); Radius: 2</h3>

To determine the equation of a circle, use the standard form of a circle (x - h)² + (y - k)² = r² where,

  • <u>(h, k)</u> is the center; and
  • <u>r</u> is the radius

Substitute the values of the center and radius to the standard form.

<u>Given:</u>

<u>(0, 0)</u> - <u>center</u>

<u>2</u> - <u>radius</u>

  • (x - h)² + (y - k)² = 2²
  • (x - 0)² + (y - 0)² = 4
  • x² + y² = 4

<u>Answer:</u>

  • The equation of the circle is <u>x² + y² = 4</u>.

Wxndy~~

7 0
2 years ago
PLEASE HELP I DO NOT UNDERSTAND AT ALL ITS PRECALC PLEASE SERIOUS ANSWERS
Elina [12.6K]

You want to end up with A\sin(\omega t+\phi). Expand this using the angle sum identity for sine:

A\sin(\omega t+\phi)=A\sin(\omega t)\cos\phi+A\cos(\omega t)\sin\phi

We want this to line up with 2\sin(4\pi t)+5\cos(4\pi t). Right away, we know \omega=4\pi.

We also need to have

\begin{cases}A\cos\phi=2\\A\sin\phi=5\end{cases}

Recall that \sin^2x+\cos^2x=1 for all x; this means

(A\cos\phi)^2+(A\sin\phi)^2=2^2+5^2\implies A^2=29\implies A=\sqrt{29}

Then

\begin{cases}\cos\phi=\frac2{\sqrt{29}}\\\sin\phi=\frac5{\sqrt{29}}\end{cases}\implies\tan\phi=\dfrac{\sin\phi}{\cos\phi}=\dfrac52\implies\phi=\tan^{-1}\left(\dfrac52\right)

So we end up with

2\sin(4\pi t)+5\cos(4\pi t)=\sqrt{29}\sin\left(4\pi t+\tan^{-1}\left(\dfrac52\right)\right)

8 0
3 years ago
Read 2 more answers
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