By the power and chain rules, taking derivatives on both sides with respect to
gives

or

Answer: ∆V for r = 10.1 to 10ft
∆V = 40πft^3 = 125.7ft^3
Approximate the change in the volume of a sphere When r changes from 10 ft to 10.1 ft, ΔV=_________
[v(r)=4/3Ï€r^3].
Step-by-step explanation:
Volume of a sphere is given by;
V = 4/3πr^3
Where r is the radius.
Change in Volume with respect to change in radius of a sphere is given by;
dV/dr = 4πr^2
V'(r) = 4πr^2
V'(10) = 400π
V'(10.1) - V'(10) ~= 0.1(400π) = 40π
Therefore change in Volume from r = 10 to 10.1 is
= 40πft^3
Of by direct substitution
∆V = 4/3π(R^3 - r^3)
Where R = 10.1ft and r = 10ft
∆V = 4/3π(10.1^3 - 10^3)
∆V = 40.4π ~= 40πft^3
And for R = 30ft to r = 10.1ft
∆V = 4/3π(30^3 - 10.1^3)
∆V = 34626.3πft^3
Answer:
1.15
Step-by-step explanation:
(1.8*10-3)/(2*10-7)
(18-3)/(20-7)
15/13
1.15
..................
Answer:
i dont know
Step-by-step explanation:
But can you just like it cause im trying get enough points to be able have a hundred points so i could have help with a college essay
please
<span>2**n. (That's two to the nth power: 2 to the first, 2 to the second (or two squared), 2 to the third (or two cubed), 2 to the fourth, etc.)
2**1 = 2 = 2
2**2 = 2 * 2 = 4
2**3 = 2 * 2 * 2 = 8
2**4 = 2 * 2 * 2 * 2 = 16 </span>