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The length of a rectangle is twice the width. The area is 338 yd^. Find the length and width.
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1. Area equation 2. equation relating base and width. you got it
I am looking for the answer
338 = 2x + x 338 = 3x \[\frac{338}{3}=x\] \[w = \frac{338}{3}\] \[l = \frac{676}{3}\]
so w=112.666 and l = 225.333, correct?
If you're an engineer, I suppose.
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this is for ALG. II, it is a pain in the a$$
AGL
Well, the fractions are more correct, because they are more exacting.
but, most maths teachers tell you to round, because most math teachers are idiots.
that is so very true. so w=112.0, and l=225.0, lol
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:) if you need to round, yes
I have never been very good at all with word problems, they always screw me up.
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