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Vaselesa [24]
3 years ago
14

Please help!! Find the value of x

Mathematics
1 answer:
nevsk [136]3 years ago
5 0

Answer:

x = 62 degrees

Step-by-step explanation:

Since this is a right angle, we can use trig functions

cos theta = adjacent/ hypotenuse

cos x = 3.7/7.9

Taking the inverse cos of each side

cos ^ -1 ( cos x) = cos ^-1 (3.7/7.9)

x =62.07246803

To the nearest degree

x = 62 degrees

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7nadin3 [17]

Answer:

7√3

Step-by-step explanation:

We can use pythagoreans theorem to solve this

Since, we know one side, and the hypotenuse, we can solve for the other side.  

Pythagoreans theorem: a²+b²=c²

Where a and b are two sides, and c is the hypotenuse (the side opposite of the right angle)

In this triangle, 7 is the side, and 14 is the hypotenuse.

I will plug in the values into pythagoreans theorem, and then simplify:

a^{2}+b^{2} =c^{2} \\\\7^{2} +x^{2} =14^{2} \\\\49+x^{2} =196\\\\x^{2} =147\\\\x=\sqrt{147} \\\\x=\sqrt{49*3 }\\\\ x=\sqrt{7^{2} *3 }\\\\x=7\sqrt{3}

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3 years ago
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Strike441 [17]
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2 years ago
What is this answer im being timed so plz help
olchik [2.2K]

Answer:

(Kind of cheating but since you're timed I'm just gonna five you the answer:

3/11

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\mathrm{Cov}(X,Y)=\mathbb E[(X-\mathbb E[X])(Y-\mathbb E[Y])]


\mathrm{Cov}(X,Y)=\mathbb E[XY-\mathbb E[X]Y-X\mathbb E[Y]+\mathbb E[X]\mathbb E[Y]]=\mathbb E[XY]-\mathbb E[X]\mathbb E[Y]


We have


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\mathbb E[aX-b]\mathbb E[cY-d]=ac\mathbb E[X]\mathbb E[Y]-ad\mathbb E[X]-bc\mathbb E[Y]+bd


Putting everything together, we find the covariance reduces to


\mathrm{Cov}(aX-b,cY-d)=ac(\mathbb E[XY]-\mathbb E[X]\mathbb E[Y])=ac\mathrm{Cov}(X,Y)


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2 years ago
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If you times it the answer is going to be 1,200 
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