- #1

SBains88

- 6

- 0

Im not sure how to go about this problem. I am just beginning in physics and am kind of confused. If anyone can help much would be appreciated!

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- Thread starter SBains88
- Start date

- #1

SBains88

- 6

- 0

Im not sure how to go about this problem. I am just beginning in physics and am kind of confused. If anyone can help much would be appreciated!

- #2

konthelion

- 238

- 0

First, ask yourself what shape is the poster and what is the area of that shape?

- #3

SBains88

- 6

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- #4

dynamicsolo

Homework Helper

- 1,648

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Now if you reduced the length and the width

- #5

SBains88

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- #6

konthelion

- 238

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Ah, you can not do it like that. Example, Let's suppose length = 100. Then 100-15%=100-.15= 99.85 which is not equal to 85%(i.e. 15% off) of 100 which is 85.

If you do it like that then It is L-.15*(blank). What is blank? From then, you can simplify it further.

- #7

SBains88

- 6

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Im not sure what you mean. What is the blank suppose to indicate

- #8

konthelion

- 238

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If you are taking 15% off, then L-.15*x. What is x? Then simplify this expression.

- #9

konthelion

- 238

- 0

- #10

SBains88

- 6

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Is x the percentage that comes off of 100%? So would it be 85%?

- #11

konthelion

- 238

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i.e. number - .15*number which is the same as .85* number

- #12

SBains88

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- #13

konthelion

- 238

- 0

No, you can not assume that L=W. You don't need to know the numerical value... for the area nor the width or length. You are figuring out the percentage change in area.

Old Area = LxW.

Okay, good. Since you've now know that the New Area=(.85L)(.85W), simplify this equation. Use basic algebra. Then, use the definition of a percentage change and find it.

i.e. [tex]change_{area} = \frac{Area_{new}-Area_{old}}{Area_{old}}*100[/tex]

I'm sleepy. You can figure this out by yourself. It's straightforward.

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