Answer:
a) see below
b) 40x20 meters
Step-by-step explanation:
Write down what you know:
- The area of the enclosure is length*width, so
- The length of the fencing is 80 meters, so
Now we have to combine these two equations above, and get rid of y in the process.
First rewrite the second as:
Then substitute for y in the first:
b) To maximize A, find the zero of the first derivative:
So y = (80-40)/2 = 20 meters.
Answer: D
Step-by-step explanation:
Answer:
its a
Step-by-step explanation:
its the only one that would be parallel because slope is the same
Answer: 2
Step-by-step explanation:
Every set of point is found if you go up twice and once to the right 2/1 = 2
Equation = Y = 2x + 4
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