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
P^4, p^3,p^2,p is descending powers of p. Hope this helps!
The values on the interval [0, 2π) that makes f (θ) equal to g(θ) is; A: 0, π
<h3>How to solve Trigonometric Functions?</h3>
We are given the trigonometric functions;
f(θ) = 4sin θ + 1
g(θ) = cos 2θ
We know from trigonometric identities that;
cos 2θ = 1 - 2sin²θ
At f(θ) = g(θ), we have;
4sin θ + 1 = 1 - 2sin²θ
2sin²θ + 4sin θ = 0
2sinθ(sin θ + 2) = 0
2sinθ = 0
θ = sin⁻¹(0/2)
θ = 0, π
Read more about Trigonometric Functions at; brainly.com/question/6904750
#SPJ1
Answer: The written expression for this equation is <u>one - fourth multiplied by nine minus a number.</u> The variable is "a number" and you use the pronunciation of the other numbers.
