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Harrizon [31]
2 years ago
15

A rectangular trough is 8ft long, 2ft across the top, and 4 feet deep. if water flows in at a rate of 2 ft 3 /min, how fast is t

he surface rising when the water is 1 ft deep?
Mathematics
1 answer:
mash [69]2 years ago
5 0
 <span>let V be the current volume of the trough and H the current height of the water 
then we have 
V = 16H 
taking derivative in terms of time we get 
dV/dt = 16 dH/dt 
we know that dV/dt = 2 
thus 
dH/dt = 2/16 = 1/8 
thus the water is rising at 
1/8 ft per minute 
or 
1.5 inches per minute</span>
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Answer:

The true statements about the graph of the function are:

The range of the function is {y : y ≤ 6} ⇒ 2nd

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Step-by-step explanation:

* Lets explain how to solve the problem

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∵ The form of the quadratic function is f(x) = ax² + bx + c

∵ The coordinates of the vertex of the parabola are (h , k) where

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∴ k = -(-2)² - 4(-2) + 2 = -(4) - (-8) + 2

∴ k = -4 + 8 + 2 = 6

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* Look to the attached figure to find the true statements

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∵ The range of the function is the values y-coordinates of the points

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Hello there!

the answer is 8.33

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Answer:

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Step-by-step explanation:

The area of a regular polygon is calculated using the formula;

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