Answer:
Dimensions :
x (the longer side, only one side with fence ) = 90 ft
y ( the shorter side two sides with fence ) = 45 ft
Total fence used 45 * 2 + 90 = 180 ft
A(max) =
Step-by-step explanation: If a farmer has 180 ft of fencing to encloses a rectangular area with fence in three sides and the river on one side, the farmer surely wants to have a maximum enclosed area.
Lets call "x" one the longer side ( only one of the longer side of the rectangle will have fence, the other will be along the river and won´t need fence. "y" will be the shorter side
Then we have:
P = perimeter = 180 = 2y + x ⇒ y = ( 180 - x ) / 2 (1)
And A (r) = x * y
A(x) = x * ( 180 - x ) /2 ⇒ A(x) = (180/2) *x - x² / 2
Taking derivatives on both sides of the equation :
A´(x) = 90 - x
Then if A´(x) = 0 ⇒ 90 - x = 0 ⇒ x = 90 ft
and from : y = ( 180 - x ) / 2 ⇒ y = 90/2
y = 45 ft
And
A(max) = 90 * 45 = 4050 ft²
Answer:
The scale factor is 2.
Step-by-step explanation:
Divide 2 by 1 then you get 2. I got the 2 and the one from the shapes. So don’t get confused ;)
Answer:
Simplify the expression: 
Use the exponent power:
Apply the given rules;
Simplify:
then;
Therefore, the simplified expression of is, 
(50 yards/minute) x (3 feet/yard) x (1 minute / 60 seconds)
= (50 · 3 · 1 / 60) (yard-feet-minute) / (minute-yard-second)
= 2.5 feet/second .
Answer:
y_c = 2 + 10*x
Step-by-step explanation:
Given:
y'' = 0
Find:
- The solution to ODE such that y(0) = 2, y'(0) = 10
Solution:
- Assuming a solution y = Ce^(mt)
So, y' = C*me^(mt)
y'' = C*m^2e^(mt)
- Back substitute into given ODE, we get:
y'' = C*m^2e^(mt) = 0
e^(mt) can not be equal to zero
- Hence, m^2 = 0
m = 0 , 0 - (repeated roots)
- The complimentary function for repeated roots is:
y_c = (C1 + C2*x)*e^(m*t)
y_c = C1 + C2*x
- Evaluate @ y(0) = 2
2 = C1 + C2*0
C1 = 2
-Evaluate @ y'(0) = 10
y'(t) = C2 = 10
Hence, y_c = 2 + 10*x
