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Agata [3.3K]
4 years ago
11

A metal worker makes a box from a 15inch by 20inch piece of tin by cutting a square with a side length x out of each corner and

folding up the sides
a) write expressions for length width and height of the box
b) write a polynomial for the volume of the box
Mathematics
1 answer:
Nikolay [14]4 years ago
3 0

Answer:

a) Length = (20 - 2x) inches

Width = (15 - 2x) inches

b) Volume = (20 - 2x)(15 - 2x)x cubic inches.

Step-by-step explanation:

A metalworker makes a box from 15 inches by 20 inches piece of tin by cutting a square with a side length x out of each corner and folding up the sides.

a) So, the length of the box will become (20 - 2x) inches.

And, the width of the box will become (15 - 2x) inches.

b) The volume of the box will be = (Length × Width × Height)

= (20 - 2x)(15 - 2x)x cubic inches. (Answer)

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NEED HELP ASAP<br> Given G (-6,5) and H (2, -7), what is the length of GH
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\bold{\huge{\green{\underline{ Solution }}}}

\bold{\underline{ Given}}

  • <u>We </u><u>have </u><u>given </u><u>that </u><u>the </u><u>coordinates </u><u>of </u><u>the </u><u>end </u><u>point </u><u>G </u><u>and </u><u>H </u><u>are </u><u>(</u><u> </u><u>-</u><u>6</u><u>,</u><u>5</u><u>)</u><u> </u><u>and </u><u>(</u><u> </u><u>2</u><u>,</u><u> </u><u>-</u><u>7</u><u> </u><u>)</u>

\bold{\underline{To  \: Find :- }}

  • <u>We </u><u>have </u><u>to </u><u>find </u><u>the </u><u>length </u><u>of </u><u>GH </u>

\bold{\underline{Let's \:  Begin :- }}

The coordinates of G = ( -6 , 5 )

The coordinates of H = ( 2 , - 7 )

<u>According </u><u>to </u><u>the </u><u>distance </u><u>formula</u><u>, </u><u> </u><u>we </u><u>get </u><u>:</u><u>-</u><u> </u>

\purple{\bigstar}\boxed{\sf{Distance\:\:GH=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2\;}}}

  • <u>Here</u><u>, </u><u> </u><u>x1</u><u> </u><u>=</u><u> </u><u>-</u><u>6</u><u> </u><u>,</u><u> </u><u>x2</u><u> </u><u>=</u><u> </u><u>2</u><u> </u><u>and </u><u>y1</u><u> </u><u>=</u><u> </u><u>5</u><u> </u><u>,</u><u> </u><u>y2</u><u> </u><u>=</u><u> </u><u>-</u><u>7</u>

<u>Subsitute </u><u>the </u><u>required </u><u>values </u><u>in </u><u>the </u><u>above </u><u>formula </u>

\sf{ Distance \:GH = √( -6 - 2)² + ( 5 -(-7))²}

\sf{Distance\:GH= √(-8)^2+(5 + 7)^2}

\sf{Distance\:GH=√(-8)^2+(12)^2}

\sf{ Distance \:GH =√64 + 144 }

\sf{ Distance \:GH =√ 208}

\sf{ Distance\: GH = √ 2 × 2 × 2 × 2 × 13}

\sf{ Distance \:GH = 4√13}

\sf{\red{ Hence\: the \: length\:of \: GH \: is \: 4√13 \: units}}

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