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:
-2, 3, -0.5 + 0.866i, -0.5 - 0.866i.
Step-by-step explanation:
As the last term is -6 , +/- 2 , +/- 3 are possible zeroes.
Try 2:-
(2)^4 - 6(2)^2 - 7(2) - 6 = -28 so 2 is not a zero.
3:-
(3)^4 - 6(3)^2 - 7(3) - 6 = 0 so 3 is a zero.
(-2)^4 - 6(-2)^2 - 7(-2) - 6 = 0 so -2 is also a zero.
Divide the function by (x +2)(x - 3), that is x^2 - x - 6
gives x^2 + x + 1
x^2 + x + 1
So we have x^2 + x + 1 = 0
x = [-1 +/- √(1^1 - 4*1*1)] / 2
= -1 + √(-3) / 2 , - 1 - √(-3) / 2.
= -0.5 + 0.866i, -0.5 - 0.866i
The units add 20 to it, for example. 20,40,60, and 80.
Answer:
80 cars will maximize revenue
Step-by-step explanation:
The revenue per square meter for parked cars is ...
$2.00/5 = $0.40
The revenue per square meter for buses is ...
$6.00/32 = $0.1875
Thus the available space should be used to park the maximum number of cars.
80 cars should be in the lot to maximize income.
