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3 years ago

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a shipping box will be filled with two different machine parts. one type of machine part weights 0.15 ponds and the other weight

s half that amount. which equation used to determine total weight

1 answer:

gulaghasi [49]3 years ago

8 0

$\frac{x}{2} =0.15$ hope so this helps

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The answer to the question is 3/8

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Solve the equation g(x) = 2k(x) algebraically for x, to the nearest tenth. given

Alekssandra [29.7K]

Using the degree of freedom rule, we can solve three unknown variables if and only if the number of independent equations is equal to 3. Thus the number of equations should be equal to the number of variables. We can use substitution to find x.

<span>g(x) = 2k(x) (1)
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k(x)=2x+16 (3)

we substitute 2 to 1 and also 3 to 1. The resulting function hence becomes:
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Simplifying the equation on the right.
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we group then the like terms on one side. That is,
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4 years ago

\int\limits^0_\pi {x*sin^{m} (x)} \, dx

Ket [755]

Let

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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>

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

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.

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3 years ago

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siniylev [52]

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riadik2000 [5.3K]

I think it would be 95,047,816,007,026

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4 years ago

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