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anonymous

  • 5 years ago

off subject..but can yall help me figure this problem out PLEASE?! 24v^2 + 5v - 36 i have to do the reverse FOIL method

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  1. anonymous
    • 5 years ago
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    (8x - 9)(3x + 4)

  2. anonymous
    • 5 years ago
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    Recognize that\[24v^2+5v-36=8v \times3v+8v \times4-9\times3v-9\times4\]

  3. anonymous
    • 5 years ago
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    Take out the common factors,\[8v(3v+4)-9(3v+4)\]

  4. anonymous
    • 5 years ago
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    i.e.\[(3v+4)(8v-9)\]

  5. anonymous
    • 5 years ago
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    It's the reverse way of going about FOIL.

  6. anonymous
    • 5 years ago
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    thank you i find these hard to figure out

  7. anonymous
    • 5 years ago
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    Have you been taught the quadratic formula yet?

  8. anonymous
    • 5 years ago
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    If not, don't worry, but I was just asking because you can use it to give you a clue as to what factors to use...

  9. anonymous
    • 5 years ago
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    umm i dont know...

  10. anonymous
    • 5 years ago
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    It's okay. For now, you just have to find the right combination of factors by trial and error.

  11. anonymous
    • 5 years ago
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    okay....math is very hard

  12. anonymous
    • 5 years ago
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    Yeah, you just have to practice.

  13. anonymous
    • 5 years ago
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    what about this one 16x^2 +32xy +15y^2

  14. anonymous
    • 5 years ago
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    \[16x^2+32xy+15y^2=4x \times 4x -4x \times 5y-3y \times 4x +2y \times 5y\]

  15. anonymous
    • 5 years ago
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    \[=4x(4x-5y)-3y(4x-5y)\]\[=(4x-3y)(4x-5y)\]

  16. anonymous
    • 5 years ago
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    Wait up, I stuffed one of the signs...long day

  17. anonymous
    • 5 years ago
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    ok

  18. anonymous
    • 5 years ago
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    should be

  19. anonymous
    • 5 years ago
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    thank you

  20. anonymous
    • 5 years ago
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    \[16x^2+32xy+15y^2=4x \times 4x +4x \times 5y+3y \times 4x + 3y \times 5y\]

  21. anonymous
    • 5 years ago
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    \[=4x(4x+5y)+3y(4x+5y)\]\[=(4x+3y)(4x+5y)\]

  22. anonymous
    • 5 years ago
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    Does your teacher expect you to eyeball solutions to these?

  23. anonymous
    • 5 years ago
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    no we have to show work,

  24. anonymous
    • 5 years ago
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    that was the last one on that lesson. Now its - Diffrence of Squares and Perfect Square trinomials

  25. anonymous
    • 5 years ago
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    How many questions are there?

  26. anonymous
    • 5 years ago
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    30

  27. anonymous
    • 5 years ago
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    I mightn't be around for much longer. Do you want me to see if there's someone else that can help?

  28. anonymous
    • 5 years ago
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    yes please

  29. anonymous
    • 5 years ago
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    I'll try to help out, yet I won't be here for long though

  30. anonymous
    • 5 years ago
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    oh ok thank you

  31. anonymous
    • 5 years ago
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    16a^2- 25b^2

  32. anonymous
    • 5 years ago
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    the book says the answer is " no" how is that

  33. anonymous
    • 5 years ago
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    you can write it like this too :\[4^2a^2 -5^2b^2\] all you have to do is simplify it into 2 brackets, give it a try :)

  34. anonymous
    • 5 years ago
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    Post your assignment questions at www.aceyourcollegeclasses.com you can earn money answering questions there too!

  35. anonymous
    • 5 years ago
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    so you find the square of the 16 and 25

  36. anonymous
    • 5 years ago
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    No...

  37. anonymous
    • 5 years ago
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    why not?

  38. anonymous
    • 5 years ago
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    wait, isn't the question = factor?

  39. anonymous
    • 5 years ago
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    You have something in the form,\[a^2-b^2\]These things are called 'difference of two squares' and it factors as\[a^2-b^2=(a+b)(a-b)\]

  40. anonymous
    • 5 years ago
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    factor right?

  41. anonymous
    • 5 years ago
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    it says for each binomial is the binomial a diffrence of squares

  42. anonymous
    • 5 years ago
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    \[16a^2-25b^2=(4a)^2-(5b)^2=(4a-5b)(4a+5b)\]

  43. anonymous
    • 5 years ago
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    lol, that's what I've meant loki!

  44. anonymous
    • 5 years ago
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    I wanted him to give it a try :)

  45. anonymous
    • 5 years ago
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    oops..sorry :(

  46. anonymous
    • 5 years ago
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    it's alright :)

  47. anonymous
    • 5 years ago
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    dina did you understand the general concept of how to solve the problem?

  48. anonymous
    • 5 years ago
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    so its a NO because one is plus and one is minus?!

  49. anonymous
    • 5 years ago
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    Your answer is YES

  50. anonymous
    • 5 years ago
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    If it factors like it factored above, it is a difference of two squares, so your answer would be YES.

  51. anonymous
    • 5 years ago
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    uhhh ok im getting confused lol

  52. anonymous
    • 5 years ago
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    try the general formula loki showed you which is : (a-b)(a+b) use the FOIL method to check whether it's equal to : \[a^2-b^2\] as long as you understand the general concept of this question, you'll be able to proceed on your own , okay? :)

  53. anonymous
    • 5 years ago
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    Exactly

  54. anonymous
    • 5 years ago
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    clear dina? ^_^

  55. anonymous
    • 5 years ago
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    i think so

  56. anonymous
    • 5 years ago
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    all of the questions are the same, all you have to do is write them in the form of \[a^2-b^2\] and you\'ll be fine ^_^

  57. anonymous
    • 5 years ago
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    then apply the rule which is (a-b)(a+b) and use the FOIL to check your answer, and you'll ace it ! :)

  58. anonymous
    • 5 years ago
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    thanks guys!

  59. anonymous
    • 5 years ago
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    np ^_^, did you undestand?

  60. anonymous
    • 5 years ago
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    So you get it then?

  61. anonymous
    • 5 years ago
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    i think for now at least

  62. anonymous
    • 5 years ago
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    good!

  63. anonymous
    • 5 years ago
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    awesome :) , just practice one to check your understanding

  64. anonymous
    • 5 years ago
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    and you'll get it

  65. anonymous
    • 5 years ago
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    loki, lemmy get you up top real quick pleasee

  66. anonymous
    • 5 years ago
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    ok...16a^2 - 12b^3

  67. anonymous
    • 5 years ago
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    (4 + a)

  68. anonymous
    • 5 years ago
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    for the first part?

  69. anonymous
    • 5 years ago
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    you can take 4 as a common factor , so you'll have:\[4(4a^2-3b^2) = 4(2a -b)(2a+3b)\] correct me if I'm wrong

  70. anonymous
    • 5 years ago
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    That is not correct

  71. anonymous
    • 5 years ago
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    First of all he has b^3 and second of all you can't write 3 as a perfect square. All you can do is take out the 4

  72. anonymous
    • 5 years ago
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    lol, I'm rushing can you take the lead blexting>?

  73. anonymous
    • 5 years ago
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    Sure...

  74. anonymous
    • 5 years ago
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    oh right! i forgot abt b^3

  75. anonymous
    • 5 years ago
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    sorry

  76. anonymous
    • 5 years ago
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    dina are just trying to factor?

  77. anonymous
    • 5 years ago
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    the answers either yes or no

  78. anonymous
    • 5 years ago
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    the answers either yes or no

  79. anonymous
    • 5 years ago
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    to "is it a perfect square" is that the question?

  80. anonymous
    • 5 years ago
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    This last one was NOT a perfect square.

  81. anonymous
    • 5 years ago
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    yes

  82. anonymous
    • 5 years ago
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    yes

  83. anonymous
    • 5 years ago
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    OK.. what is the next one

  84. anonymous
    • 5 years ago
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    thats it for now my daughter just wooke up thanks for all the help! i appreciate it

  85. anonymous
    • 5 years ago
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    OK

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