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anonymous

  • 4 years ago

i have 2 equations (4x3+24x2+7x+42)/(x+6)and(2x2+6x+a) the first formula has an x<-6 and the 2nd has x</=-6 and i have to find which value would make it continuous. i set them equal to each and plugged in -6 in for the x and solved for a getting -36 which is wrong

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  1. anonymous
    • 4 years ago
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    factor and cancel to get \[4x^2+7\]

  2. anonymous
    • 4 years ago
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    replace x by 6 and get \[4\times 6^2+7=189\]

  3. anonymous
    • 4 years ago
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    so my algebra is wrong but my method is right?

  4. anonymous
    • 4 years ago
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    what you then have to do, now that you know the limit of the first expression is equal to 189, is replace x by 6 in the second expression, set it also equal to 189 and solve for "a"

  5. anonymous
    • 4 years ago
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    there is only 1 a though

  6. anonymous
    • 4 years ago
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    and that is in the 2nd equation

  7. anonymous
    • 4 years ago
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    right. the second expression is \[2x^2+6x+a\] and you need to to match up to the first expression when x =6

  8. anonymous
    • 4 years ago
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    you know the limit of the first expression at 6 is 186 so write \[2\times 6^2+6\times 6+a=189\] and solve for "a"

  9. anonymous
    • 4 years ago
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    sorry i meant "189"

  10. anonymous
    • 4 years ago
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    oh ok that makes sense

  11. anonymous
    • 4 years ago
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    i get \[108+a=189\] and therefore \[a=81\] but don't take my word for it

  12. anonymous
    • 4 years ago
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    it says 81 is in incorrect

  13. anonymous
    • 4 years ago
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    x had to equal -6 not 6

  14. anonymous
    • 4 years ago
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    x=-7, a=147 x=-8, a=183 and so on

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