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keana

  • one year ago

help! (picture in comments)

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  1. keana
    • one year ago
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  2. zepdrix
    • one year ago
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    Ok let's see what we know so far...\[\large\rm \color{green}{P}=2\color{royalblue}{L}+2\color{orangered}{W}\] They told us that the \(\large\rm \color{green}{\text{perimeter is 212}}\). So let's plug that in.\[\large\rm \color{green}{212}=2\color{royalblue}{L}+2\color{orangered}{W}\]

  3. zepdrix
    • one year ago
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    They told us that the \(\large\rm \color{royalblue}{\text{Length is two more than the Width}}\). We can write that relationship like this: \(\large\rm \color{royalblue}{L=W+2}\)

  4. keana
    • one year ago
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    yes

  5. zepdrix
    • one year ago
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    \[\large\rm \color{green}{212}=2\color{royalblue}{(L)}+2\color{orangered}{W}\]We plug this information into our equation, replacing L.

  6. keana
    • one year ago
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    right

  7. zepdrix
    • one year ago
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    \[\large\rm \color{green}{212}=2\color{royalblue}{(W+2)}+2\color{orangered}{W}\]

  8. zepdrix
    • one year ago
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    And from there, you have an equation involving only ONE VARIABLE! :) So you can proceed to solve for W.

  9. zepdrix
    • one year ago
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    What do you think? :O

  10. keana
    • one year ago
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    lemme work it out

  11. keana
    • one year ago
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    212=4w+4?

  12. zepdrix
    • one year ago
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    So you ditributed and then combined like-terms? Ya that's a good start! :)

  13. keana
    • one year ago
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    52=w?

  14. zepdrix
    • one year ago
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    Looks good! And then don't forget about the fact that the \(\large\rm \color{royalblue}{\text{Length is two more than the Width}}\).

  15. keana
    • one year ago
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    so 54?

  16. zepdrix
    • one year ago
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    yay good job \c:/

  17. keana
    • one year ago
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    thanks!

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