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tatyana61 [14]
3 years ago
11

Can someone solve this?

Mathematics
1 answer:
ladessa [460]3 years ago
6 0
  • First question:

Recall that \cos^2x+\sin^2x=1 and \sqrt{x^2}=|x| for all x. So

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

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

For 0, we expect both \cos x>0 and \sin x>0 (i.e. the sine and cosine of any angle that lies in the first quadrant must be positive). By definition of absolute value, |x|=x if x>0.

So we have

\dfrac{\sqrt{1-\cos^2x}}{\sin x}+\dfrac{\sqrt{1-\sin^2x}}{\cos x}=\dfrac{\sin x}{\sin x}+\dfrac{\cos x}{\cos x}=1+1=\boxed{2}

making H the answer.

  • Second question:

C is always true, because the inequality reduces to x > y.

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Help please 40 points
MrMuchimi

Part 1:

Step 1:

Find the value of G(16)

Using the equation for g(x) =√x, replace x with 16 and solve:

g(16) = √16 = 4

Step 2:

Now using the answer for g(16) replace x in the equation fir f(x) with 4 and solve:

f(x) = x +2

f(4) = 4+2 = 6

Part 2:

following the same steps as part 1:

f(g(x)) would be replacing x in the equation ofr f(x)with the equation for g(x)

f(g(x)) = √x +2

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3 years ago
Find the equation for the circle with a diameter whose endpoints are (-5, -3) and (3,2).
densk [106]

Answer:

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Step-by-step explanation:

The center of the circle is the midpoint of the diameter, so is ...

  (h, k) = ((-5, -3) +(3, 2))/2 = (-5+3, -3+2)/2 = (-1, -1/2)

The circle formula is then ...

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We can find r^2 by substituting one of the points for (x, y).

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Then the circle equation is ...

  (x +1)^2 +(y +0.5)^2 = 22.25

3 0
3 years ago
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neonofarm [45]

Here's link to the answer:

tinyurl.com/wpazsebu

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