Answer:
Is that a magic spell??
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:
If you cube the numbers in the left column, you get the numbers in the right column! We can figure this out by understanding that 1^3 = 1, 3^3 = 27, and 6^3 = 216. Then it all falls into place!
1 -> 1
2 -> 8
3 -> 27
5 -> 125
6 -> 216
11 -> 1331
8 -> 512
10 -> 1000
7 -> 343
14 -> 2744
p -> 
-> q
Answer:
-24+12(d-3)+22=-24+34(d-3)
10(d-3)
10d=-30
d=-30/10
d=3
