Y = 2x - 3 . . . (1)
y = -2x + 5 . . . (2)
Equating (1) and (2),
2x - 3 = -2x + 5
2x + 2x = 5 + 3
4x = 8
x = 8/4 = 2
x = 2
y = 2(2) - 3 = 4 - 3 = 1
y = 1
Solution = (2, 1)
Answer:
look at the picture i have sent
<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
Answer:
-2 + × = y
Step-by-step explanation:
I think this is the answer because it's asking for an expression which this would be one.
