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
Answer:
Latisha rode 16 rides
Step-by-step explanation:
Look at picture attached. Have a blessed night!
Just slope intercept formula (y=mx+b). X represents the number of rides so you will need to myltiply that by $2.50. $9 is a one time fee so that would be of the b of the equation. That all equals the total Latisha piaid, $49. Then solve like so.
Make x=0 in the second equation, giving you y=-3. If you plug that in to the top equation you would get x-3(-3)=-19, x+9=-19,x=-28.
If you want the slope it's 1/-8
The 2 because I am goated
Step-by-step explanation:
