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
-7n=63
n=-9 is your final answer. Hope it help!
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<span>To subtract 7 from 13 you use the fact that 13 is one ten plus three units, i.e. 13 = 10 + 3. then you can subtract 7 from 10 which is 3, and then add the 3 other 3 units to get 3 + 3 = 6. In this way, you have subtracted 7 from 13 using the fact that 13 is one ten plus 3 units.</span><span>
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Answer:
fully simplified answer: x^2-2x+8
Step-by-step explanation:
