Answer:
0.90
Step-by-step explanation:
I’m doing the unit test
Answer:
L=21.6in and w=13.6
Step-by-step explanation:
If the length is 8 inches longer than its width, we can write the width as "w" and the length as the width w+ 8
Area is (width)(Length) = (width)(width+8)
W
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| | L = w+8
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A = (w+8)(w)
A = w2 + 8w = 295.
(Problem states that area is 295 sq in)
Need to solve this quadratic equation
w2 +8y - 295 = 0
Factor:
(w - 13.64) (w + 21.64) = 0
So
w - 13.64 = 0. or. w + 21.64= 0
Solve these and get
w = 13.64. or. w = -21.64
Only one that makes sense in real life is the poitive one.
So the dimensions are
Width = 13.64 inches
Length = 21.64 inches
Answer: (7 + d) 4 and 4 (7 + d)
Step-by-step explanation:
<em>4 (d + 7) = 4d + 28</em>
<em />
(7 + d) 4 = 28 + 4d (flipped version of 4d + 28)
4 (7 + d) = 28 + 4d (the same)
<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
