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2 years ago

Number c is really difficult

Mathematics

1 answer:

olya-2409 [2.1K]2 years ago

6 0

Multiply each term in the equation by 4/1.
9-2y+2y+4=24

Combine like terms
13=24

Because the result is false, it is no solution.

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Do these side lengths make a triangle 8, 2, 12,

Paul [167]

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3 years ago

Sandy and Maria worked the following weekly hours this summer at a summer camp. Which of the following statements are true?

LiRa [457]

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3 years ago

Write the solution to the given inequality in interval notation.

DIA [1.3K]

Option A: $[4, \infty)$ is the solution

Explanation:

The solution of the given inequality is the set of all the possible values of x.

The graph shows the number line in which the shaded region is from the right of -4 and the arrow of solution goes to infinity.

Also, There is a closed circle at the point -4.

This means that -4 is included in the solution set.

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Thus, the solution is x ≥ -4

Hence, the solution is $[4, \infty)$

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3 years ago

A train traveled for 1.5 hours to the first station, stopped for 30 minutes, then traveled for 4 hours to the final station wher

stealth61 [152]

So are you trying to find the total time as a function or?

6 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

u = x ==> du = dx

dv = sinᵐ (x) dx ==> v = ∫ sinᵐ (x) dx

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 m is even or where m is odd.

If m is even, then m = 2k for some integer k, 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 m (or k).

If m is odd, then m = 2k + 1 for some integer k, 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 u = cos(x) and du = -sin(x) dx, 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

2 years ago