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2 years ago

7

a cylindrical paint roller has a length of 10 inches and a diameter of 1 1/2 inches. As the roller is dragged along a wall, if i

t completes 4 revolutions, what is the approximate area the roller will cover with paint?

1 answer:

sergij07 [2.7K]2 years ago

3 0

I hope this helps you radius= 1 1/2=1.2+1/2=3/2 lateral area= 2pi.r.height area= 2.pi.3/2.10 area =30pi 4.30pi=120pi

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Hatshy [7]

Let

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y-------> the temperature in Fahrenheit degrees

$A(2,150)\\B(3,190)$

we know that

<u>the change in the y-values between two points is equal to</u>

$(y2-y1)$

Substitute the y-values of the point A and B in the formula

$y2=190\\y1=150$

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<u>the change in the x-values between two points is equal to</u>

$(x2-x1)$

Substitute the x-values of the point A and B in the formula

$x2=3\\x1=2$

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therefore

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2 years ago

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tamaranim1 [39]

Well (3-2^2)5 = -5
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2 years ago

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gtnhenbr [62]

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2 years ago