Find the volume of the solid generated by revolving the region enclosed by the triangle with vertices left parenthesis 2 comma 0 right parenthesis , left parenthesis 2 comma 5 right parenthesis , and left parenthesis 6 comma 5 right parenthesis about the y-axis.

The triangle is a right triangle with sides aligned with the x- and y-axes. The desired volume can be found several ways. Perhaps the easiest is to make use of the fact that the volume is the product of the area of the triangle and the length of the path of revolution of the centroid of the triangle.

The centroid of a triangle is located at the intersection point of the medians, one-third of the distance from any side toward the opposite vertex. Here, we want the radius to the centroid from the y-axis, so the side of interest is the one parallel to the y-axis.

In the x-direction, the altitude of the triangle is 6-2=4, so the centroid is located at 4/3 units from the left side, which is x=2. The x-coordinate of the centroid is then 2+4/3 = 10/3, and this is the radius of revolution for the area of the triangle.

The sides of the right triangle are of length 5 and 4, so the area is

... area = (1/2)bh = (1/2)·5·4 = 10 . . . . . square units

Then the volume of interest is

... V = 2π·(radius of revolution)·(area)

... V = 2π·(10/3)·10

... V = 200π/3 ≈ 209.44 . . . units³

2 x^2 -3x+p

sum of roots = 3/2

let other root be x

3+ x = 3/2

x= 3/2 - 3 = -3/2

product of roots = p/2

3× -3/2 = p/2

-9 = p

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step-by-step explanation: