Mathematics: Nearing the end of his holiday preparations richard has only one piece of wrapping paper left for three remaining gifts. the remaining 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.

the answer is 1Step-by-step explanation:...

Nearing the end of his holiday preparations richard

gtjermaine3249

17.01.2022

Hello!

The answer is:

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

and its area is:

$GiftCArea=300inch^{2}$

Why?

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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