Answer:
Step-by-step explanation:
rate of leaking water = 9500 cm³/min
height of tank, H = 6 m = 600 cm
diameter of tank, = 4 m
radius of tank, R = 2 m = 200 cm
dh/dt = 20 cm/min
at any time, h = 2 m Let at that time radius of the tank is r.
According to the diagram
R / H = r / h
200 / 600 = r / h
r = h/3
The volume of tank is given by
Differentiate with respect to t
dV/dt = 279111.11 cm³/min
This is the volume of water increasing per minute in the tank.
Let C be the rate of volume of water pumped into the tank.
So, C = 279111.11 + 9500
C = 288611.11 cm³/min
The answer is (-7, -12).
Attached is the graph to show the answer.
Answer:
m∠QPR = 35° m∠QPM =40° m∠PRS = 30°
Step-by-step explanation:
ΔPRQ is a right triangle with right angle at R. So m∠QPR = 90 - 55 = 35
ΔQPM is a right triangle with right angle at M. So m∠QPM = 90 - 40 = 40
Arc RQ = 2(35) = 70 and arc SR = 2(25) = 50.
So arc PS = 180 - (arc RQ + arc SR) = 180 - (70 + 50) = 180 - 120 = 60
Now arc PS is the intercepted arc for ∠PRS.
Therefore, m∠PRS = 60/2 = 30
I used the fact that an inscribed angle has a measure 1/2 the measure of the intercepted arc several times. Also, I used the fact that the acute angles of a right triangle are complementary. And, finally I used the fact that an inscribed angle in a semicircle is a right angle.
I hope this helped.
Answer:
yeah thats true
Step-by-step explanation:
Answer:
it might be b or a.... im so sorry if its not right.
Step-by-step explanation: