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Sagot :
We know that the formula for the area of a rectangle is:
[tex]A=lw[/tex]
We know that the area is 165. Now we need to solve for the dimensions.
Let the width of the rectangle be "[tex]x[/tex]"
Then the length of the rectangle is [tex]3x+2[/tex]
Substitute.
[tex](3x+2)(x)=156[/tex]
Multiply the terms.
[tex]3x^2+2x=156[/tex]
Bring the 156 over.
[tex]3x^2+2x-156=0[/tex]
We have to use the quadratic formula to solve this now. Let us restate it:
[tex]x= \frac{-b± \sqrt{b^2-4ac} }{2a} [/tex]
Now I'm going to fast forward this because all the rest is boring stuff and substitution. The answer is:
[tex] \frac{ \sqrt{469}-1 }{3} [/tex]
Since we cannot have a negative answer, we must cancel out the other answer, which I did not include.
Hope this helped! :)
~Cam943, Junior Moderator
[tex]A=lw[/tex]
We know that the area is 165. Now we need to solve for the dimensions.
Let the width of the rectangle be "[tex]x[/tex]"
Then the length of the rectangle is [tex]3x+2[/tex]
Substitute.
[tex](3x+2)(x)=156[/tex]
Multiply the terms.
[tex]3x^2+2x=156[/tex]
Bring the 156 over.
[tex]3x^2+2x-156=0[/tex]
We have to use the quadratic formula to solve this now. Let us restate it:
[tex]x= \frac{-b± \sqrt{b^2-4ac} }{2a} [/tex]
Now I'm going to fast forward this because all the rest is boring stuff and substitution. The answer is:
[tex] \frac{ \sqrt{469}-1 }{3} [/tex]
Since we cannot have a negative answer, we must cancel out the other answer, which I did not include.
Hope this helped! :)
~Cam943, Junior Moderator
l×b=area
l=2+3b
2+3b ×b=165
3b²=163
b²=54.3
b=√54.3
b=7.37 or 7.4=7
b=7ft
l=2+3(7.4)
l=2+ 22.2
l=24.2=24
l=24ft
l=2+3b
2+3b ×b=165
3b²=163
b²=54.3
b=√54.3
b=7.37 or 7.4=7
b=7ft
l=2+3(7.4)
l=2+ 22.2
l=24.2=24
l=24ft
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