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:
the answer will be 1 consistent
Answer:
Step-by-step explanation:
add 9 to both sides.
then divide by 1.5
so p > 8
number your number line: 2,4,6,8,10
draw a circle at 8 and the line to the right since p>8
In order to answer the question, we simply substitute the value of p and q to the given expression and solve. We do as follows:
<span>P^2q^2+pq–q^3–p^3
</span>0.5^2(-0.5)^2+0.5(-0.5)–(-0.5)^3–(0.5)^3
-3/16 or -0.1875
Hope this answers the question. Have a nice day.
