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anonymous

  • one year ago

PLEASE HELP I REALLY NEED TO PASS! Find The Product. (t-sa)(9t+6sa)

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  1. Mehek14
    • one year ago
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    use FOIL

  2. calculusxy
    • one year ago
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    |dw:1444596888643:dw|

  3. calculusxy
    • one year ago
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    I am not sure if this is correct ...

  4. calculusxy
    • one year ago
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    sorry. one moment

  5. Mehek14
    • one year ago
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    \(\tt{(t-sa)(9t+6sa)\\t*9t=9t^2\\-sa*9t=-9tsa\\t*6sa=6tsa\\-sa*6sa=-6sa^2\\9t^2-9tsa+6tsa-6sa^2}\)

  6. Mehek14
    • one year ago
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    you have to combine like terms

  7. Mehek14
    • one year ago
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    \(\tt{-9tsa+6tsa=-3tsa}\)

  8. calculusxy
    • one year ago
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    yes that's correct

  9. freckles
    • one year ago
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    |dw:1444597193897:dw|

  10. calculusxy
    • one year ago
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    @Mehek14 How did you change the font? I only know /rm and /scr

  11. freckles
    • one year ago
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    -sa*6sa=-6s^2a^2

  12. Mehek14
    • one year ago
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    i used \tt

  13. freckles
    • one year ago
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    |dw:1444597235822:dw|

  14. calculusxy
    • one year ago
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    @freckles YEAH!

  15. Mehek14
    • one year ago
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    \(\tt{9t^2-3tsa-6s^2a^2}\)

  16. anonymous
    • one year ago
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    So 9t^2-3tsa-6sa?

  17. calculusxy
    • one year ago
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    no use @Mehek14 's one.

  18. anonymous
    • one year ago
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    Thanks Guys your real life savers

  19. anonymous
    • one year ago
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    (4a-ts)(2a+2ts)? Its so confusing

  20. anonymous
    • one year ago
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    @Mehek14

  21. Mehek14
    • one year ago
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    \(\tt{(4a-ts)(2a+2ts)\\4a*2a=8a^2\\4a*2ts=8ats\\-ts*2a=-2ats\\-ts*2ts=-2t^2s^2}\)

  22. Mehek14
    • one year ago
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    combine like terms

  23. Mehek14
    • one year ago
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    \(\tt{8ats-2ats=6ats\\8a^2+6ats-2t^2s^2}\)

  24. calculusxy
    • one year ago
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    |dw:1444597636184:dw| Answer would be something like:

  25. anonymous
    • one year ago
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    So ever many coefficients are the number there powered to?

  26. calculusxy
    • one year ago
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    \[\large \rm -2t^2s^2+2ats + 8a^2\]

  27. Mehek14
    • one year ago
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    if the same variables are multiplied to each other, then you get exponents ex \(\tt{t*t=t^2}\)

  28. anonymous
    • one year ago
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    Oh Ok thank you

  29. anonymous
    • one year ago
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    Wait you guys have two different answers

  30. Mehek14
    • one year ago
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    it's the same answers order doesn't matter

  31. anonymous
    • one year ago
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    Oh my teacher said biggest number first

  32. anonymous
    • one year ago
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    i see

  33. calculusxy
    • one year ago
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    degree of polynomials..

  34. calculusxy
    • one year ago
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    that's what i tried to do

  35. Mehek14
    • one year ago
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    \(\tt{8ats-2ats=6ats\\8a^2+6ats-2t^2s^2}\) I keep them in the order I multiply

  36. calculusxy
    • one year ago
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    |dw:1444598189445:dw|

  37. calculusxy
    • one year ago
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    @Mehek14 's right

  38. anonymous
    • one year ago
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    What is it?

  39. Mehek14
    • one year ago
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    what is what?

  40. anonymous
    • one year ago
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    Which is the answer his has 8 yours has 6

  41. calculusxy
    • one year ago
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    \(\color{#0cbb34}{\text{Originally Posted by}}\) @Mehek14 \(\tt{8ats-2ats=6ats\\8a^2+6ats-2t^2s^2}\) \(\color{#0cbb34}{\text{End of Quote}}\) you can order with degree of polynomials if you want

  42. calculusxy
    • one year ago
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    i made a silly mistake.

  43. Mehek14
    • one year ago
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    \[\large \rm -2t^2s^2+2ats + 8a^2\] \(\tt{8a^2+6ats-2t^2s^2}\) they are the same

  44. Mehek14
    • one year ago
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    oh wait you put 2ats

  45. calculusxy
    • one year ago
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    \(\color{#0cbb34}{\text{Originally Posted by}}\) @Mehek14 \[\large \rm -2t^2s^2+2ats + 8a^2\] \(\tt{8a^2+6ats-2t^2s^2}\) they are the same \(\color{#0cbb34}{\text{End of Quote}}\) not really...

  46. anonymous
    • one year ago
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    Thanks again Guys

  47. anonymous
    • one year ago
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    (2n^2+3by)(3n^2-by)?

  48. anonymous
    • one year ago
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    How to i do them with exponents? @mehek14 @calculusxy

  49. anonymous
    • one year ago
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    -3b^y^2+7bn^2y+6n^4?

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