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VikaD [51]
3 years ago
14

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.
Mathematics
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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Consider the equation 5x^2-10x+c=0. What values of c result in the equation having a complex solutions? Represent your answer on
stiv31 [10]

Answer:

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I can not represent this in your number line, but this would be represented with a white dot in the five, that extends infinitely to the right, something like the image below:

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