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MakcuM [25]
3 years ago
12

Which of the following statements describes a correct method for solving this equation?

Mathematics
2 answers:
ANTONII [103]3 years ago
8 0

Answer:

a

Step-by-step explanation:

i think

Klio2033 [76]3 years ago
3 0
Answer: A
Explanation:
3x+5=8x

The three is a positive number so it must be subtracted. This statement eliminates option B. The first thing that should be done in this equations to get rid of an x. There should only be one part of the equation with an x. With this being said, options c and d are also eliminated.
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Each wheel on a car has a diameter of 22 inches.
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I think it will be 69.08
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In the triangle above, what is the measure of A.<br> M B.<br> M C.<br> M D<br> , cannot be determined
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Describe the relationship between the perimeter of a triangle and the perimeter of the triangle that is formed by the midsegment
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3 0
3 years ago
\int\limits^0_\pi {x*sin^{m} (x)} \, dx
Ket [755]

Let

I(m) = \displaystyle \int_0^\pi x\sin^m(x)\,\mathrm dx

Integrate by parts, taking

<em>u</em> = <em>x</em>   ==>   d<em>u</em> = d<em>x</em>

d<em>v</em> = sin<em>ᵐ </em>(<em>x</em>) d<em>x</em>   ==>   <em>v</em> = ∫ sin<em>ᵐ </em>(<em>x</em>) d<em>x</em>

so that

I(m) = \displaystyle uv\bigg|_{x=0}^{x=\pi} - \int_0^\pi v\,\mathrm du = -\int_0^\pi \sin^m(x)\,\mathrm dx

There is a well-known power reduction formula for this integral. If you want to derive it for yourself, consider the cases where <em>m</em> is even or where <em>m</em> is odd.

If <em>m</em> is even, then <em>m</em> = 2<em>k</em> for some integer <em>k</em>, and we have

\sin^m(x) = \sin^{2k}(x) = \left(\sin^2(x)\right)^k = \left(\dfrac{1-\cos(2x)}2\right)^k

Expand the binomial, then use the half-angle identity

\cos^2(x)=\dfrac{1+\cos(2x)}2

as needed. The resulting integral can get messy for large <em>m</em> (or <em>k</em>).

If <em>m</em> is odd, then <em>m</em> = 2<em>k</em> + 1 for some integer <em>k</em>, and so

\sin^m(x) = \sin(x)\sin^{2k}(x) = \sin(x)\left(\sin^2(x)\right)^k = \sin(x)\left(1-\cos^2(x)\right)^k

and then substitute <em>u</em> = cos(<em>x</em>) and d<em>u</em> = -sin(<em>x</em>) d<em>x</em>, so that

I(2k+1) = \displaystyle -\int_0^\pi\sin(x)\left(1-\cos^2(x)\right)^k = \int_1^{-1}(1-u^2)^k\,\mathrm du = -\int_{-1}^1(1-u^2)^k\,\mathrm du

Expand the binomial, and so on.

8 0
3 years ago
I need help with this.
viktelen [127]
Just separate them into there own boxes, which is 38ft
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