Based on the given variation, y does not vary directly with x and the constant of variation are 8, 3.2 and 1.25 respectively.
<h3>Variation</h3>
y = k × x
where,
k = constant of proportionality
y = -40
x = -5
y = k × x
-40 = k × -5
-40 = -5k
k = -40/-5
k = 8
when,
y = 8 and x = 2.5
y = k × x
8 = k × 2.5
8 = 2.5k
k = 8/2.5
k = 3.2
when,
y = 5 and x = 4
y = k × x
5 = k × 4
5 = 4k
k = 5/4
k = 1.25
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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
I believe it’s c. I hope that this is helpful!
