<h2>
Maximum area is 25 m²</h2>
Explanation:
Let L be the length and W be the width.
Aidan has 20 ft of fence with which to build a rectangular dog run.
Fencing = 2L + 2W = 20 ft
L + W = 10
W = 10 - L
We need to find what is the largest area that can be enclosed.
Area = Length x Width
A = LW
A = L x (10-L) = 10 L - L²
For maximum area differential is zero
So we have
dA = 0
10 - 2 L = 0
L = 5 m
W = 10 - 5 = 5 m
Area = 5 x 5 = 25 m²
Maximum area is 25 m²
The first one is b=75the second one is c= 15Hope this helps!
Move the variables to one side and the constants to the other
50q - 43 = 52q - 81
50q (-52q) - 43 (+43) = 52q (-52q) - 81 (+43)
50q - 52q = -81 + 43
-2q = -38
isolate the q, divide -2 from both sides
-2q/-2 = -38/-2
q = -38/-2
q = 19
hope this helps
