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iogann1982 [59]
3 years ago
15

PLEASE PLEASE HELP ME:(( I NEED WORKING AS WELL

Mathematics
1 answer:
mr Goodwill [35]3 years ago
8 0

the equation of the line has form

y=kx+b

\left \{ {{5=k*0+b} \atop {9=k*2+b}} \right. \\b=5\\9=2k+5\\2k=4\\k=2, so

y=2x+5

The second line is parallel to the first and runs 7 units lower, so b=5

0=1k+5\\k=-5\\y=-5x+5

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A rectangle and a square have the same area. The width of the rectangle is 2 in less than the side of the square and the length
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The area of a rectangle is A=LW, the area of a square is A=S^2.

W=S-2 and L=2S-3

And we are told that the areas of each figure are the same.

S^2=LW, using L and W found above we have:

S^2=(2S-3)(S-2)  perform indicated multiplication on right side

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S^2=2S^2-7S+6  subtract S^2 from both sides

S^2-7S+6=0  factor:

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S(S-1)-6(S-1)=0

(S-6)(S-1)=0, since W=S-2, and W>0, S>2 so:

S=6 is the only valid value for S.  Now we can find the dimensions of the rectangle...

W=S-2 and L=2S-3  given that S=6 in

W=4 in and L=9 in

So the width of the rectangle is 4 inches and the length of the rectangle is 9 inches.


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