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Tati_Lee

  • 2 years ago

Simplify completely

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  1. Tati_Lee
    • 2 years ago
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    \[\frac{ 5x ^{2}+9x-2 }{ x ^{2}+12x+20 } * \frac{ x ^{2}+17x+70 }{ 15x-3 }\]

  2. ghass1978
    • 2 years ago
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    can you factorize the respective numerators and denominators?

  3. Tati_Lee
    • 2 years ago
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    no, thats all i need help with & i can do the rest

  4. ghass1978
    • 2 years ago
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    look for example at 5x^2+9x-2 look for 2 numbers whose product is -2*5=-10 and whose sum is 9

  5. AidanNims
    • 2 years ago
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    ((x+2)(5x-1))/(x+10)(x+2)* ((x+7)(x+10)/3(5x-1) the way u apporach them is what numbers multiply to give lets say 70... well 7 and 10 . do they add to form the 17x yes cause 10 +7+17.

  6. Compassionate
    • 2 years ago
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    \[\frac{ (x+2)(5x+1) }{ (x+10)(x+2)} * \frac{ (x+10)(x+7) }{ 3(5x+1) }\] (5x+1) cancels (x + 10) cancels \[\frac{ x+2 }{ x+2 } * \frac{ x+7 }{ 3} = 1 * \frac{ x + 7 }{ 3 } = \frac{ x+7 }{ 3 }\] This, I cannot factor anymore.

  7. Tati_Lee
    • 2 years ago
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    thank you @AidanNims & @Compassionate

  8. Tati_Lee
    • 2 years ago
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    can you help me with one more @Compassionate ?

  9. Compassionate
    • 2 years ago
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    Yea.

  10. Tati_Lee
    • 2 years ago
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    great! .. hold on let me write it out

  11. Tati_Lee
    • 2 years ago
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    \[\frac{ x ^{2}+x-12 }{ x ^{2}-x-20 }\div \frac{ 3x ^{2}-24x+45 }{ 12x ^{2}-48x-60}\]

  12. Compassionate
    • 2 years ago
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    One sec, putting on music then I'll solve it. c:

  13. Tati_Lee
    • 2 years ago
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    okie dokie

  14. Compassionate
    • 2 years ago
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    Do you want me too factor it, or just give you the answer?

  15. Tati_Lee
    • 2 years ago
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    you can just give me the answer lol

  16. Compassionate
    • 2 years ago
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    Damn, this expression is walking all over me.

  17. Compassionate
    • 2 years ago
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    \[\frac{ (x+4)(x-3) }{ (x+4)(x-5) } - \frac{ 3(x-3)(x-5) }{ 4(3x^2-x-5) }\]

  18. Compassionate
    • 2 years ago
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    After this part I know what too do, I just can't figure out of to construct a common denominator.

  19. Tati_Lee
    • 2 years ago
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    okay thank you so much!

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