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3 years ago

Give the equation of the line passing through the point (20,2)that is perpendicular to y=−4x.

Mathematics

1 answer:

Cerrena [4.2K]3 years ago

5 0

Answer:

y-2 = 1/4(x-20) point slope form

y = 1/4x -3 slope intercept form

Step-by-step explanation:

y = -4x

The slope is -4

The slope that is perpendicular is the negative reciprocal

- (1/-4) = 1/4

The slope of our new line is 1/4

We have a point and the slope, so we can use point slope form

y-y1 = m(x-x1)

y-2 = 1/4(x-20)

If we want the line in slope intercept form, distribute

y-2 = 1/4x -5

Add 2 to each side

y-2+2 = 1/4x -5+2

y = 1/4x -3

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3 years ago

Sa = 2(pi)(r)(h) + 2(pi)(r^2) v = (pi)(r^2)(h) find the surface area and the volume.

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If Sa=2πrh+2π $r^{2}$ v=π $r^{2}h$ then the surface area is π $r^{2} h$ and volume is

(rh-2h)/2r.

Given Sa=2πrh+2π $r^{2} v$ =π $r^{2}h$ .

We have to find surface area and volume from the given expression.

Volume is basically amount of substance a container can hold in its capacity.

First we will find the value of v from the expression. Because they are in equal to each other, we can easily find the value of v.

2πrh+2π $r^{2}$ v=π $r^{2}$ h

Keeping the term containing v at left side and take all other to right side.

2π $r^{2}$ v=π $r^{2}h$ -2πrh

v=(π $r^{2}$ h-2πrh)/2π $r^{2}$

v=π $r^{2} h$ /2π $r^{2}$ -2πrh/2π $r^{2}$

v=h/2-h/r

v=h(r-2)/2r

Put the value of v in Sa=2πrh+2π $r^{2} v$

Sa=2πrh+2π $r^{2}$ *h(r-2)/2r

=2πrh+2πrh(r-2)/2

=2πrh+πrh(r-2)

=2πrh+π $r^{2}$ h-2πrh

=π $r^{2}$ h

Hence surface area is π $r^{2}$ h and volume is h(r-2)/2.

Learn more about surface area at brainly.com/question/16519513

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