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mart [117]
3 years ago
9

What is the rate of change of the volume of the cylinder at that instance (in cubic kilometers per second)?

Mathematics
1 answer:
Kay [80]3 years ago
3 0

Answer:

hello your question is incomplete attached below is the complete question

answer : -2400 π

Step-by-step explanation:

dr/dt = -12 km/s

h = 2.5 km

r = 40 km at some instant

hence the rate of change of Volume of cylinder at that instance

= dv / dt = d(\pi r^h ) / dt  

              = \pi h\frac{d(r^2)}{dt}  = -\pi h *2r *dr/dt

hence ; dv / dt = -2400 π

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Answer:

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Step-by-step explanation:

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the double negative on the bottom becomes an addition=> (12-18)/(-4+8)

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this fraction can be reduced to -3/2, which is the slope of the graph.

Now, use point slope form (y-y1=m(x-x1)) to find the equation of the graph. Plug any coordinate on the graph in for x1 and y1 here. It should be correct as long as it is a point on the graph, but I am using the point (-8, 18) here.

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