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Morgarella [4.7K]
3 years ago
7

A box has a volume of 308 cm cubed. Its height is 4 cm greater than its length. Its length is 3 cm greater than its width. What

is the width of the box? SHOW YOUR WORK.
Mathematics
1 answer:
kifflom [539]3 years ago
4 0

Answer:

4 cm.

Step-by-step explanation:

Given:

A box has a volume of 308 cm cubed.

Its height is 4 cm greater than its length.

Its length is 3 cm greater than its width.

Question asked:

What is the width of the box?

<u>Solution</u>:

Let width  = x\ cm

Length = x+3\ cm

Height = x+3+4=x+7\ cm

<u>As we know:</u>

<u />Volume\ of\ box=length\times width\times height

308=(x+3)\times x\times(x+7)\\ \\ 308=(x^{2} +3x)(x+7)\\ \\ 308=x^{2} (x+7)+ 3x(x+7)\\ \\ 308=x^{3} +7x^{2} +3x^{2} +21x\\ \\ 308=x^{3} +10x^{2} +21x\\ \\ Subtract\ 308\ from\ both\ sides\\ \\ x^{3} +10x^{2} +21x-308=0\\ \\ factor\ left\ side\ of\ equation\\ \\ (x-4)(x^{2} +14x+77)=0\\ \\  Set\ factors\ equal\ to\ 0.\\ \\ x-4=0 \ or\  x^{2} +14x+77=0\\ \\ \\

As width can never be in negative, x= 4\ cm

By substituting the value:-

Width  = x\ cm = 4 cm

Thus, width of the box is 4 cm.

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\boxed{ \bold{ \huge{ \boxed{ \sf{(5.5 \: , \: -  9.5)}}}}}

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<u>Finding</u><u> </u><u>the</u><u> </u><u>midpoi</u><u>nt</u>

\boxed{ \sf{midpoint = ( \frac{x1 + x2}{2} , \frac{y1 + y2}{2} }}

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