The answer is 78% that was rounded from 77.78%
Answer:
21 ft by 28 ft
Step-by-step explanation:
To maximize the area, see the attached.
Perimeter will be 4l+3w which is equal to the fencing perimeter, given as 168
4l+3w=168
Making l the subject then
4l=168-3w
l=42-¾w
Area of individual land will be lw and substituting l with l=42-¾w
Then
A=lw=(42-¾w)w=42w-¾w²
A=42w-¾w²
Getting the first derivative of the above with respect to w rhen
42-w6/4=0
w6/4=42
w=42*4/6=28
Since
l=42-¾w=42-¾(28)=21
Therefore, maximum dimensions are 21 for l and 28 for w
[(7**13)**3]*[7**0]
[7**39]*[1]..........> 9.0954 E 32 Strawberries in the field
Answer: C
Step-by-step the table that's flipped the y and x values of the given table
Q = -60 and P ≠ 32 will result in an equation with no solutions. (Both conditions must be met.)
_____
For Q = -60 and P = 32, there will be an infinite number of solutions. For any other values of Q and P, the solution is
.. x = (32 -P)/(Q +60)