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anonymous

  • one year ago

Two sets of equatic expressions are shown below in various forms: Line 1: x^2 + 3x + 2 (x + 1)(x + 2) (x + 1.5)^2 − 0.25 Line 2: x^2 + 5x + 6 (x + 2)(x + 3) (x + 2.5)^2 + 6.25 Which line contains three equivalent expressions?

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  1. anonymous
    • one year ago
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    line 1 only line 2 only line 1 and line 2 neither line 1 or lime 2

  2. anonymous
    • one year ago
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    @myininaya

  3. johnweldon1993
    • one year ago
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    Do you know how to distribute out the parenthesis? e.g) (x+1)(x+2)

  4. anonymous
    • one year ago
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    I think so don't you use FOIL? @johnweldon1993

  5. johnweldon1993
    • one year ago
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    Yes use foil to distribute those out..in both cases

  6. anonymous
    • one year ago
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    line 1 x^2 +2x+x+2 line 2 x^2+3x+2x+6 is that correct @johnweldon1993

  7. johnweldon1993
    • one year ago
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    yes...now just combine like terms...and we have line 1) x^2 + 3x + 2 line 2) x^2 + 5x + 6 So far those match the first part of each line...so we move onto the next part as well So now we need to distribute the line 1) (x + 1.5)^2 - 0.25 line 2) (x + 2.5)^2 + 6.25

  8. anonymous
    • one year ago
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    then it would be x^2 +1.5^2-0.25 x^2+2.5^2+6.25

  9. anonymous
    • one year ago
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    @johnweldon1993

  10. johnweldon1993
    • one year ago
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    Ahh not quite :) So think of something like \[\large (x + a)^2 \rightarrow (x + a)(x + a)\]

  11. anonymous
    • one year ago
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    oo so it would be (x+1.5)(x+1.5)-0.25 (x+2.5)(x+2.5)+6.25 @johnweldon1993

  12. johnweldon1993
    • one year ago
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    Correct....so you foil those parenthesis out and then just simplify :)

  13. anonymous
    • one year ago
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    the first one should be x^2+3x+2

  14. anonymous
    • one year ago
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    the second one should be x^2+5x+12.5

  15. johnweldon1993
    • one year ago
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    Perfect....so as you can see...line 1 is the same...but line 2 is no longer the same..thus the answer is...?

  16. anonymous
    • one year ago
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    line 1 only @johnweldon1993

  17. johnweldon1993
    • one year ago
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    There we go :)

  18. anonymous
    • one year ago
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    thank you so much @johnweldon1993

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