Answer:
HAVE A NICE DAY :)))
Step-by-step explanation:
thank
Answer:
Step-by-step explanation:
Given
The attached triangle
Required
Find y
The attached triangle is isosceles; so:
Also, we have:
--- angles in a triangle
Substitute:
Collect like terms
Answer:
D
Step-by-step explanation:
to find a term in the sequence multiply the previous term by r , then
a₄ = - 8 , so
a₅ = - 8 × 0.5 = - 4
a₆ = - 4 × 0.5 = - 2
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
A graphing calculator shows the ball reaches its maximum height of 69 feet at
t = 2 seconds.
