The dimensions of the rectangle can be a length of 2ft and a width of 4ft.
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How to find the dimensions of the garden?</h3>
Remember that for a rectangle of length L and width W, the perimeter is:
P = 2*(L + W)
And the area is:
A = L*W
In this case, we know that the area is 8 square feet and the perimeter is 12 ft, then we have a system of equations:
12ft = 2*(L + W)
8ft² = L*W
To solve this, we first need to isolate one of the variables in one of the equations, I will isolate L on the first one:
12ft/2 = L + W
6ft - W = L
Now we can replace that in the other equation to get:
8ft² = (6ft - W)*W
This is a quadratic equation:
-W^2 + 6ft*W - 8ft² = 0
The solutions are given by Bhaskara's formula:

Then we have two solutions:
W = (-6 - 2)/-2 = 4ft
W = (-6 + 2)/-2 = 2ft
If we take any of these solutions, the length will be equal to the other solution.
So the dimensions of the rectangle can be a length of 2ft and a width of 4ft.
if you want to learn more about rectangles:
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Answer:
39.5 degrees
Step-by-step explanation:
180-101=79
79/2=39.5
The first part of the question is not needed
we only need to know M1 and M2
(y2-y1)^2+(x2-x1)^2=(your answer)^2
(5-16)^2+(-3-2)^2=(your answer)^2
(-11)^2+(-5)^2=(your answer)^2
121+25=(your answer)^2
146=(your answer)^2
√146=12.0830459735945721
nearest tenth=12.1
Answer:
The height as a function of time is
h(t) = -(1/2)gt2 + v0t + h0
h(t) = -16t2 + 32t + 20
The ball reaches the ground when h(t) = 0
-16t2 + 32t + 20 = 0
Solve this quadratic equation for t. You will get a positive solution and a negative solution. Discard the negative solution, since time begins at t=0 at the moment the ball is struck. Can you finish from here?
Answer:
6.7 years (approx)
Step-by-step explanation: