Simplify 4/x-2 + 5/x+3

Mathematics

1 answer:

yulyashka [42]3 years ago

4 0

Answer:

9x+2/(x+3)(x-2)

Step-by-step explanation:

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Gravel is being dumped from a conveyor belt at a rate of 35 ft3/min, and its coarseness is such that it forms a pile in the shap

Luda [366]

Answer:

0.45 ft/min

Step-by-step explanation:

Given:-

- The flow rate of the gravel, $\frac{dV}{dt} = 35 \frac{ft^3}{min}$

- The base diameter ( d ) of cone = x

- The height ( h ) of cone = x

Find:-

How fast is the height of the pile increasing when the pile is 10 ft high?

Solution:-

- The constant flow rate of gravel dumped onto the conveyor belt is given to be 35 ft^3 / min.

- The gravel pile up into a heap of a conical shape such that base diameter ( d ) and the height ( h ) always remain the same. That is these parameter increase at the same rate.

- We develop a function of volume ( V ) of the heap piled up on conveyor belt in a conical shape as follows:

$V = \frac{\pi }{12}*d^2*h\\\\V = \frac{\pi }{12}*x^3$

- Now we know that the volume ( V ) is a function of its base diameter and height ( x ). Where x is an implicit function of time ( t ). We will develop a rate of change expression of the volume of gravel piled as follows Use the chain rule of ordinary derivatives:

$\frac{dV}{dt} = \frac{dV}{dx} * \frac{dx}{dt}\\\\\frac{dV}{dt} = \frac{\pi }{4} x^2 * \frac{dx}{dt}\\\\\frac{dx}{dt} = \frac{\frac{dV}{dt}}{\frac{\pi }{4} x^2}$

- Determine the rate of change of height ( h ) using the relation developed above when height is 10 ft:

$h = x\\\\\frac{dh}{dt} = \frac{dx}{dt} = \frac{35 \frac{ft^3}{min} }{\frac{\pi }{4}*10^2 ft^2 } \\\\\frac{dh}{dt} = \frac{dx}{dt} = 0.45 \frac{ft}{min}$