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777dan777 [17]
3 years ago
5

A-line includes the points (8,9) and (9,1). What is its equation in point-slope form?

Mathematics
1 answer:
sergij07 [2.7K]3 years ago
4 0

Answer:

y-(9)=(-8)(x-8)\\

Step-by-step explanation:

(x_1, y_1) = (8, 9)\\(x_2, y_2) = (9,1)\\

Step one: Find the slope (m)

slope=m=\frac{rise}{run} =\frac{y_2-y_1}{x_2-x_1} = \frac{1-9}{9-8} =\frac{-8}{1} =-8

Step two: Plug your known values into the point-slope equation

m=-8\\y_1=9\\x_1=8\\

Point-slope form is y-y_1=m(x-x_1)

y-(9)=(-8)(x-8)\\

If you want to convert point-slope form to general form, isolate for y.

y-9=-8x+64\\y=-8x+73

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Answer:

b<-7 or b>-1

Step-by-step explanation:

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for the other inequality, you  take 4b+6>2 and then subtract6 from both sides, from there you will get 4b>-4 and that will be b>-1

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Write an explicit formula
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3 years ago
How do you do this problem?
Vadim26 [7]

Answer:

Your answer is absolutely correct

Step-by-step explanation:

The work would be as follows:

\int _0^{\sqrt{\pi }}4x^3\cos \left(x^2\right)dx,\\\\\mathrm{Take\:the\:constant\:out}:\quad \int a\cdot f\left(x\right)dx=a\cdot \int f\left(x\right)dx\\=> 4\cdot \int _0^{\sqrt{\pi }}x^3\cos \left(x^2\right)dx\\\\\mathrm{Apply\:u-substitution:}\:u=x^2\\=> 4\cdot \int _0^{\pi }\frac{u\cos \left(u\right)}{2}du\\\\\mathrm{Apply\:Integration\:By\:Parts:}\:u=u,\:v'=\cos \left(u\right)\\=> 4\cdot \frac{1}{2}\left[u\sin \left(u\right)-\int \sin \left(u\right)du\right]^{\pi }_0\\\\

\int \sin \left(u\right)du=-\cos \left(u\right)\\=> 4\cdot \frac{1}{2}\left[u\sin \left(u\right)-\left(-\cos \left(u\right)\right)\right]^{\pi }_0\\\\\mathrm{Simplify\:}4\cdot \frac{1}{2}\left[u\sin \left(u\right)-\left(-\cos \left(u\right)\right)\right]^{\pi }_0:\quad 2\left[u\sin \left(u\right)+\cos \left(u\right)\right]^{\pi }_0\\\\\mathrm{Compute\:the\:boundaries}:\quad \left[u\sin \left(u\right)+\cos \left(u\right)\right]^{\pi }_0=-2\\=> 2(-2) = - 4

Hence proved that your solution is accurate.

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4 years ago
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