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4 years ago

If the length of a rectangle is 3 feet less then twice its width and the area is 54square feet what are its dimensions

1 answer:

5 0

The length (L) of the rectangle can be written as a function of the width (W) $L = 2W - 3$ :
Now since we know Area = Width*Length, we can write the area as a function of the width:
$A = L*W = (2W-3)*W$
Distributing the W inside the parentheses we have:
$A = 2W^2 - 3W$
We know the area is 54 ft^2, so we can rewrite it as:
$2W^2 - 3W - 54 = 0$
Now solve for W by factoring (or by applying the quadratic formula):
$2W^2 - 12W + 9W - 54 = 0$
Factor out a common 2W from the first two terms and a 9 from the last two terms:
$2W(W-6) + 9(W-6) = 0$
Regroup the terms to get our fully factored equation:
$(2W + 9)(W-6) = 0$
This gives us the roots W = 6 and W = -9/2, but width can't be negative so we have width = 6 ft. Then remember that the length L = 2W - 3, so our length is:
$L = 2W - 3 = 2(6) - 3 = 12 - 3 = 9$
So now we know that our rectangle is 9 feet long and 6 feet wide.

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I need help on number 30.

Naily [24]

Answer:

X=35

Step-by-step explanation:

Angle M= 60 <-- because we know a triangle=180 and so far we have 40 and 80 in the triangle JKM, 60 degrees is needed to complete 180

The other side of angle M (in the triangle KML)= 120 <-- because we know that a straight edge is 180 degrees and to complete the 180, we need another 120 degrees

so, to find X we need to find how many degrees are needed to complete the 180 degrees of triangle KML

120+25= 145

180-145= 35

6 0

4 years ago

PLEASE HELP RIGHT AWAY, <br> THANK U SO MUCH!! <3

vredina [299]

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