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Alex777 [14]
3 years ago
7

Nearing the end of his holiday preparations Richard has only one piece of Wrapping paper left for three remaining gifts. The rem

aining paper measures 25” x 45”. For gift a he needs to fi wrapping paper left for three remaining gifts. The remaining paper measures 25“ x 45“. For gift A he needs two-fifths of the wrapping paper. For gift B he needs one-third of wrapping paper.

Mathematics
1 answer:
Ivanshal [37]3 years ago
6 0
<h2>Hello!</h2>

The answer is:

The dimensions of the paper for Gift C, are: 25" x 12"

and its area is:

GiftCArea=300inch^{2}

<h2>Why?</h2>

To solve the problem, we need to calculate the total area of the remaining paper, and then, subtract it from the paper used for the gift A and B.

We know that:

GiftA=TotalPaperArea*\frac{2}{5}\\\\GiftB=TotalPaperArea*\frac{1}{3}

Now, the paper for Gift C will be:

GiftCArea=(TotalPaperArea)-(PaperArea*\frac{2}{5}+PaperArea*\frac{1}{3})

From the statement we know that the dimenstions of the remaining paper are 25" x 45", so calculating the area we have:

TotalArea=25inch*45inch=1125inch^{2}

Now, calculating the area of the paper for Gift A and B, we have:

GiftA=1125inch^{2}*\frac{2}{5}=450inch^{2}\\\\GiftB=1125inch^{2}*\frac{1}{3}=375inch^{2}

Then, calculating the paper for Gift C, we have:

GiftCArea=(TotalPaperArea)-(PaperArea*\frac{2}{5}+PaperArea*\frac{1}{3})

GiftCArea=1125inch^{2}-(450inch^{2}+375inch^{2}+)

GiftCArea=1125inch^{2}-825inch^{2}=300inch^{2}

GiftCArea=300inch^{2}

Therefore, calculating the dimensions of the paper for Gift C, knowing the height of the paper (25inches), we have::

GiftCArea=Height*Width\\\\Width=\frac{GiftCArea}{25inches}=\frac{300inches^{2} }{25inches}=12inches

Hence, the dimensions of the paper for Gift C, are: 25" x 12".

Have a nice day!

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