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8_murik_8 [283]
3 years ago
12

Please help me find the Measure of the arc!!

Mathematics
1 answer:
noname [10]3 years ago
3 0
124 is the answer....
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Solve the equation for Y<br> 2x + y = -9
Westkost [7]

Alright, so first we'll have to get y on the other side

So

2x+y=-9

   -y     -y

  +9     +9

2x+9=-y

Then divide is all by -1 so y is positive

-2x-9=y

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Graph the solution set to this inequality.<br> 35 - 1175 + 9
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Answer:

−1131

Step-by-step explanation:

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Find the price of a $50 item after a 20% discount
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20% = 0.20
50 * 0.20 = $10 discount
50 - 10 = 40

The price after 20% discount is $40
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3 years ago
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Find the value of tan(sin^-1(1/2))
suter [353]

If you know that \sin\dfrac\pi3=\dfrac12, then you know right away

\tan\left(\sin^{-1}\dfrac12\right)=\tan\dfrac\pi3=\dfrac1{\sqrt}3=\dfrac{\sqrt3}3

###

Otherwise, you can derive the same result. Let \theta=\sin^{-1}\dfrac12, so that \sin\theta=\dfrac12. \sin^{-1} is bounded, so we know -\dfrac\pi2\le\theta\le\dfrac\pi2. For these values of \theta, we always have \cos\theta\ge0.

So, recalling the Pythagorean theorem, we find

\cos^2\theta+\sin^2\theta=1\implies\cos\theta=\sqrt{1-\sin^2\theta}=\sqrt{1-\left(\dfrac12\right)^2}=\dfrac{\sqrt3}2

Then

\tan\theta=\tan\left(\sin^{-1}\dfrac12\right)=\dfrac{\sin\theta}{\cos\theta}=\dfrac{\frac12}{\frac{\sqrt3}2}=\dfrac1{\sqrt3}=\dfrac{\sqrt3}3

as expected.

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3 years ago
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Explain what a Euler diagram is...
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