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
I need a point to this question in order to answer it
Answer:
the second one
Step-by-step explanation:
in order for it to be a function the x values CANNOT repeat.
in answer a the x value -6 repeats so that cannot be a function
in answer b none of the x values repeat so its probably this one but lets check the other ones just in case
in answer c the x value 8 repeats so it cannot be this one
in answer d the x value -2 repeats so it cant be this one \
b is the only one that does have repeating x values therefore it is the second one
To solve the problem, we need to get the formula for area of a circle -
Area = pi * r^2; r= radius
= 3.14 * (13cm)^2
= 3.14 * 169cm^2 or 3.14 * 13^2cm^2
= 530.66cm^2
Therefore, the area of the circular plate is 530.66cm^2
