Answer:
0.95p and 1 *0.95p
Step-by-step explanation:
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:
B
Step-by-step explanation:
the measure of the chord- chord angle ABC is half the sum of the arcs intercepted by the angle and its vertical angle , that is
∠ ABC = 
(AC + DE) = (75 + 45)° = × 120° = 60°
Answer:
y = 1
Step-by-step explanation:
x + y = 16
x - y = 14 add the equations
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2x = 30
x = 15
x + y = 16
15 + y = 16
y = 1
the numbers are 15 and 1
