Which could be used to solve this equation?
1 answer:
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Answer:
0.28cm/min
Step-by-step explanation:
Given the horizontal trough whose ends are isosceles trapezoid
Volume of the Trough =Base Area X Height
=Area of the Trapezoid X Height of the Trough (H)
The length of the base of the trough is constant but as water leaves the trough, the length of the top of the trough at any height h is 4+2x (See the Diagram)
The Volume of water in the trough at any time
=8h(8+2x)
V=64h+16hx
We are not given a value for x, however we can express x in terms of h from Figure 3 using Similar Triangles
x/h=1/4
4x=h
x=h/4
Substituting x=h/4 into the Volume, V
h=3m,
dV/dt=25cm/min=0.25 m/min
=0.002841m/min =0.28cm/min
The rate is the water being drawn from the trough is 0.28cm/min.
Start by getting the term with the exponent on one side
take the log of both sides
There is a log rule that says we can bring down the exponent
Solve for x
5x=6.64
answer: x=1.33
take the log of both sides
There is a log rule that says we can bring down the exponent
Solve for x
5x=6.64
answer: x=1.33