Answer:
it's the first option
Step-by-step explanation:
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
A(n) = a1 + (n - 1) * d
A(n) = -3 + (n - 1) * -2.2
n = term to find
a1 = first term = -3
d = common difference = -2.2
we can already see that the first term is -3 <==
4th term..
A(4) = -3 + (4 - 1) * -2.2
A(4) = -3 + 3(-2.2)
A(4) = -3 - 6.6
A(4) = - 9.6 <== 4th term
10th term...
A(10) = -3 + (10 - 1) * -2.2
A(10) = -3 + 9 * - 2.2
A(10) = -3 - 19.8
A(10) = - 22.8 <===10th term