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koban [17]
3 years ago
14

How many zeros are there in -5x(x-3)(x^2-16)

Mathematics
1 answer:
34kurt3 years ago
4 0
Simplify:

(( - 5x)(x - 3))(x^2 - 16)

= (( -5x)(x - 3))(x^2 + - 16)

= (( -5x)(x - 3))(x^2) + (( -5x)(x - 3))( - 16)

= - 5x^4 + 15x^3 + 80^2 - 240x

= - 5x^4 + 15x^3 + 80x^2 - 240x


There are 4 Zeros...

Hope it helps!!!
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Answer:

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Step-by-step explanation:

To solve this, we are using the Cylinder's volume formula:

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where

V is the volume of the cylinder

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Remember that radius is half the diameter, so we can divide the diameter of our cylinder by 2 to find its radius: r=\frac{d}{2} =\frac{8in}{2} =4in. Now, 9 1/2 inches is 9 and a half inches, so h=9.5in.

Replacing values:

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