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Hoochie [10]
3 years ago
9

Using point-slope form, write an equation of the line that passes through (2,-1) and has a slope of -5

Mathematics
1 answer:
expeople1 [14]3 years ago
5 0
Y – y1 = m(x – x1)

y-(-1)=-5(x-2)

Y+1=-5(x-2)
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2 years ago
The base of a solid is the region in the first quadrant bounded above by the line y = 5, below by y = sin−1(x), and to the right
Arturiano [62]

Each cross section is a square with side length equal to the vertical distance between the curves y=5 and y=\sin^{-1}x, with 0\le x\le1, or 5-\sin^{-1}x. Then each section has an area of (5-\sin^{-1}x)^2, so the volume of this solid is

\displaystyle\int_0^1(5-\sin^{-1}x)^2\,\mathrm dx=\boxed{33-5\pi+\frac{\pi^2}4}

For computing the integral, consider the substitution

u=5-\sin^{-1}x\implies x=\sin(5-u)\implies\mathrm dx=-\cos(5-u)\,\mathrm du

Then the integral becomes

\displaystyle\int_5^{5-\pi/2}u^2(-\cos(5-u))\,\mathrm du

=\displaystyle\int_{5-\pi/2}^5u^2\cos(u-5)\,\mathrm du

Integrate by parts twice; for the first round, take

f=u^2\implies\mathrm df=2u\,\mathrm du

\mathrm dg=\cos(u-5)\,\mathrm du\implies g=\sin(u-5)

\implies\displaystyle u^2\sin(u-5)\bigg|_{5-\pi/2}^5-2\int_{5-\pi/2}^5u\sin(u-5)\,\mathrm du

For the second round, take

f=u\implies\mathrm df=\mathrm du

\mathrm dg=-\sin(u-5)\,\mathrm du\implies g=\cos(u-5)

\implies\displaystyle\bigg(u^2\sin(u-5)+u\cos(u-5)\bigg)\bigg|_{5-\pi/2}^5-\int_{5-\pi/2}^5\cos(u-5)\,\mathrm du

and the rest is trivial.

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Read 2 more answers
I need help (3/4)x+2=(3/8)x-4
bezimeni [28]

Answer:

x = -16

Step-by-step explantion:

Solve for  x  by simplifying both sides of the equation, then isolating the variable.

Hope this helps!

(・∀・(・∀・(・∀・*)

7 0
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A tissue box shaped like a rectangular prism. The box is 2.3 feet wide, 4 feet long and 6 feet tall. what is the volume of the t
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Multiply 2.3 x 4 x 6 and u shall get yer answer.
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