Answer:
Dh/dt = 0.082 ft/min
Step-by-step explanation:
As a perpendicular cross section of the trough is in the shape of an isosceles triangle the trough has a circular cone shape wit base of 1 feet and height h = 2 feet.
The volume of a circular cone is:
V(c) = 1/3 * π*r²*h
Then differentiating on both sides of the equation we get:
DV(c)/dt = 1/3* π*r² * Dh/dt (1)
We know that DV(c) / dt is 1 ft³ / 5 min or 1/5 ft³/min
and we are were asked how fast is the water rising when the water is 1/2 foot deep. We need to know what is the value of r at that moment
By proportion we know
r/h ( at the top of the cone 0,5/ 2) is equal to r/0.5 when water is 1/2 foot deep
Then r/h = 0,5/2 = r/0.5
r = (0,5)*( 0.5) / 2 ⇒ r = 0,125 ft
Then in equation (1) we got
(1/5) / 1/3* π*r² = Dh/dt
Dh/dt = 1/ 5*0.01635
Dh/dt = 0.082 ft/min
-5/3 can be turned to a mixed number =-1 2/3
So can 18/11 = 1 7/11
10/4 = 2 2/4 or 2 1/2 simplified
And -3/3 as 1
This may make it easier to graph own the number line
False.
Mixed numbers are a combination of whole numbers and fractions, not decimals.
You answered correctly for part a. just simplify it. so for part a it would be x^4 + 8x^3 + 24x^2 + 32x + 16. repeat what you did in part a with part b (add & simplify). btw, the missing side in part c is 2x-x = x
