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:
4.25
Step-by-step explanation:
5-3/4=4.25
sry if im wrong hope this helps
brainliest? please?
If i did the math correctly one number is 50
19 has two factors: 1, 19
21 has four factors: 1, 3, 7, 21
23 has two factors: 1, 23
And we need a number that has more than four factors and is greater than 25.
Factors of 50- 1, 2, 5, 10, 25, 50
50has six factors and is greater than 25.
Answer:
sorry i dont do parrallleleleleleleleograms
Step-by-step explanation:
Just finished this for another student.
Length = 9
