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lisabon 2012 [21]

3 years ago

11

I need help badly again....

2 answers:

Shkiper50 [21]3 years ago

5 0

Hello!

First of all we need to find the area of the circle (A= $\pi$ r²). Our radius is one. We square it which leaves us with $\pi$ . Therefore we subtract pi from the triangle, which is 4√3.

Therefore, our answer is C) 4√3- $\pi$ square units.

I hope this helps!

taurus [48]3 years ago

4 0

The second answer is correct.

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(2pm^-1q^0)^-4 • 2m ^-1 p^3 / 2pq^2

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$\dfrac{m^3}{16p^2q^2}$

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Given:

$(2pm^{-1}q^0)^{-4}\cdot \dfrac{ 2m^{-1} p^3}{2pq^2}$

1.

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2.

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3.

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5.

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7.

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8.

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