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
<span>potential; kinetic is the answer.</span>
Answer
4000
900
0
7
Step-by-step explanation:
Not so sure about the answer though
Answer:
12 30
Step-by-step explanation:
Answer:
C
Step-by-step explanation:
Using Pythagoras' identity in the right triangle
The square on the hypotenuse is equal to the sum of the squares on the 2 other sides, that is
x² + 6² = 17²
x² + 36 = 289 ( subtract 36 from both sides )
x² = 253 ( take the square root of both sides )
x = 
→ C
