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avanturin [10]
3 years ago
6

Simplify the expression. (−5g5h6)2(g4h2)4

Mathematics
2 answers:
Tanzania [10]3 years ago
7 0
= 25*g^10 h^12 * g^16 h^8

= 25 g^26 h^20
Colt1911 [192]3 years ago
4 0

We have to simplify the expression here. The expression given,

(-5g^5h^6)^2(g^4h^2)^4

Now to find (-5g^5h^6)^2 we will use power of a product property. We have to distribute the power here. We will get,

(-5g^5h^6)^2 = (-5)^2(g^5)^2(h^6)^2

= 25(g^5)^2(h^6)^2

Now we will use power of a power property. If given (a^m)^n we will have to multtiply the powers and we will get a^{mn}. So we will get here,

25(g^5)^2(h^6)^2 = 25 g^{10} h^{12}

Now the next part is (g^4h^2)^4

By using power of a power property we will get,

(g^4)^4(h^2)^4 = g^{16}h^8

Now we have to multiply them.

(25g^{10}h^{12}  )(g^{16}h^8)

Now we have to use product of powers property. If we have same base them we will have to add the exponents there. The formula is (a^m)(a^n) = a^{(m+n)}. So we will get here,

25(g^{10} g^{16})(h^{12}  h^{8})

25g^{(10+16)} h^{(12+8)}

25g^{26}h^{20}

So we have got the required simplified answer here.

The simplified answer is 25g^{26}h^{20}.

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Identify the area of segment MNO to the nearest hundredth. HELP PLEASE!! I don't understand it!
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Answer:

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Step-by-step explanation:

I don't understand it, either.

Point N is part of a "segment" that above and to the right of chord MO. It is the sum of the areas of 3/4 of the circle and a right triangle with 7-inch sides. The larger segment MO to the upper right of chord MO has an area of about 139.95 in², which <u>is not</u> an answer choice.

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The remaining segment, to the lower left of chord MO does not seem to have anything to do with point N. However, its area is 13.98 in², which <u>is</u> an answer choice. Therefore, we think the question is about this segment, and we wonder why it is called MNO.

The area of a segment is given by the formula ...

  A = (1/2)(θ -sin(θ))r² . . . . . . where θ is the central angle in radians.

Here, we have θ = π/2, r = 7 in, so we can compute the area of the smaller segment MO as ...

  A = (1/2)(π/2 -sin(π/2))(7 in)² = 24.5(π/2 -1) in² ≈ 13.9845 in²

Rounded to hundredths, this is ...

  ≈ 13.98 in²

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