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3 years ago

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An engine cylinder 13.7 cm deep is being bored such that the radius increases by 0.13 mm divided by min. How fast is the volume

V of the cylinder changing when the diameter is 9.7 cm question mark Use 3.14 for pi. Round to the nearest hundredth.

1 answer:

zzz [600]3 years ago

6 0

Answer:

Step-by-step explanation:

Given that an engine cylinder 13.7 cm deep is being bored such that the radius increases by 0.13 mm divided by min.

We convert everything in mm for uniformity

We have formula for volume as

$V=\pi r^2 h$

Given that $\frac{dr}{dt} =0.13$

h is constant

$V= 3.14 (r^2) (13.7)\\\frac{dV}{dt} =6.28 (137) r dr/dt\\= 6.28(137)\frac{ (97)}{2} 0.13$

= 6954.67 mm/sec^3

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ANSWER

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This implies that

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or

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$x=\frac{1}{2}$ does not belong to $3\leq x$

This means it can never be an intercept for this piece-wise function.

Hence our x-intercept is $(-4,0)$

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Answer:

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Step-by-step explanation:

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