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astraxan [27]
3 years ago
12

A line that is perpendicular to y=-2/7x+9y and that passes through the point (4, -6)

Mathematics
1 answer:
IrinaK [193]3 years ago
8 0
--------------------------------------------------
Find Slope :
--------------------------------------------------
y = -2/7x + 9
Slope = -2/7
Perpendicular slope= 7/2

--------------------------------------------------
Find y - intercept :
--------------------------------------------------
y = mx + b
y = 7/2 x + b
At (4, -6)
-6 = 7/2 (4) + b
-6 = 14 + b
b = -6 - 14
b = -20

--------------------------------------------------
Form equation :
--------------------------------------------------
y = mx + b
y = 7/2 x - 20
2y = 7x - 40

--------------------------------------------------
Answer: 2y = 7x - 40
--------------------------------------------------

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What is the answer please
Elena L [17]

Hello from MrBillDoesMath!

Answer:

The third choice,  99


Discussion:

I'm not sure what that strange cutout in the parallelogram has to do with the problem but ..... the area of parallelogram is given by (base)(height). In our case, assuming the length ("height') ED =   11 and length  AB ("base") =  5 + 4 = 9, the area = 11 * 9 = 99 square centimeter. This is the third choice.


Thank you,

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antoniya [11.8K]
687,500 is the answer
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Answer:

The volume is:

\displaystyle\frac{37\pi}{10}

Step-by-step explanation:

See the sketch of the region in the attached graph.

We set the integral using washer method:

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Notice here the radius of the washer is the difference of the given curves:

x-\sqrt{x}

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\displaystyle\int_1^4\pi(x-\sqrt{x})^2dx

We solve it:

Factor \pi out and distribute the exponent (you can use FOIL):

\displaystyle\pi\int_1^4x^2-2x\sqrt{x}+x\,dx

Notice: x\sqrt{x}=x\cdot x^{1/2}=x^{3/2}

So the integral becomes:

\displaystyle\pi\int_1^4x^2-2x^{3/2}+x\,dx

Then using the basic rule to evaluate the integral:

\displaystyle\pi\left[\frac{x^3}{3}-\frac{2x^{5/2}}{5/2}+\frac{x^2}{2}\right|_1^4

Simplifying a bit:

\displaystyle\pi\left[\frac{x^3}{3}-\frac{4x^{5/2}}{5}+\frac{x^2}{2}\right|_1^4

Then plugging the limits of the integral:

\displaystyle\pi\left[\frac{4^3}{3}-\frac{4(4)^{5/2}}{5}+\frac{4^2}{2}-\left(\frac{1}{3}-\frac{4}{5}+\frac{1}{2}\right)\right]

Taking the root (rational exponents):

\displaystyle\pi\left[\frac{4^3}{3}-\frac{4(2)^{5}}{5}+\frac{4^2}{2}-\left(\frac{1}{3}-\frac{4}{5}+\frac{1}{2}\right)\right]

Then doing those arithmetic computations we get:

\displaystyle\frac{37\pi}{10}

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