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kimye21 Group Title

The length of a rectangle is 7 inches more than the width. The perimeter is 42 inches. Find the length and width

  • one year ago
  • one year ago

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  1. kimye21 Group Title
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    |dw:1348706471681:dw|

    • one year ago
  2. dpflan Group Title
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    Cool, so for that rectangle, perimeter is 42. How do you find the perimeter?

    • one year ago
  3. dpflan Group Title
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    Do you know the formula for finding the perimeter of an object?

    • one year ago
  4. kimye21 Group Title
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    i think its p=2l+2w

    • one year ago
  5. dpflan Group Title
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    yeah, in this case for the rectangle, P = 2*length + 2*width. So you have values for the length and the width, right? L = x + 7 W = x

    • one year ago
  6. dpflan Group Title
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    Plug those values into the perimeter function, what does the equation look like after doing that?

    • one year ago
  7. kimye21 Group Title
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    my equation looks like this 42=2(7)+2w 42-14=28 which makes w = 14 then for L 14+7=21 am i right

    • one year ago
  8. dpflan Group Title
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    Getting there, so the values you have for each dimension are actually variables - (x+7), and x Therefore P = 2*L + 2*W becomes L = x + 7 W = x P = 2*(x+7) + 2*(x) 42 = 2*(x+7) + 2*(x)

    • one year ago
  9. dpflan Group Title
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    Because there is only one variable, one unknown, that being 'x', we can easily then solve for it from the adjusted equation.

    • one year ago
  10. dpflan Group Title
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    Does that make sense? Give that equation a shot, and solve for x.

    • one year ago
  11. kimye21 Group Title
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    x=7

    • one year ago
  12. dpflan Group Title
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    Definitely! that's it, so L = ? and W = ?

    • one year ago
  13. dpflan Group Title
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    Just have to do that last step and you're donw

    • one year ago
  14. dpflan Group Title
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    Quite easy after finding x...

    • one year ago
  15. kimye21 Group Title
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    so w = 7 and l=14

    • one year ago
  16. dpflan Group Title
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    Nice work

    • one year ago
  17. kimye21 Group Title
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    thanks

    • one year ago
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