X=4
4(4) + 1 =
16 + 1= 17
the answer is 17
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:
- Least Value = 2
- Median = 6
- Greatest Value = 10
- Third Quartile = 8
- Range = 8 (10 - 2 = 8)
- First Quartile = 5
Step-by-step explanation:
Answer:
41
Step-by-step explanation:
an = a1 + (n-1)d
a_n = the nᵗʰ term in the sequence = 10
a_1 = the first term in the sequence = 5
d = the common difference between terms = 4
a10 = 5 + (10-1)4
a10 = 5 + (9)4
a10 = 5 + 36
a10 = 41
