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anonymous

  • one year ago

SOMEONE PLEASE HELP ME WITH THIS ALGEBRA 2 . its hard and im really stressed

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  1. Jhannybean
    • one year ago
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    Well, whats the question? :D

  2. anonymous
    • one year ago
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    Match the complex expression to its correct simplified form. 1. i^22 2. i&7 3. 5 + 8i - 4 + 2i 4. (2 - 3i)(3 + 5i) 5. (2 - i)(2 + i)

  3. Astrophysics
    • one year ago
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    i^22 = -1

  4. UsukiDoll
    • one year ago
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    hmm... well for the first one , we know that \[i^2= -1 \] so we can split the \[i^{22} \] with the exponent rule but that can take long... wow @Astrophysics so fast lol

  5. anonymous
    • one year ago
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    lol . i just dont understand it period

  6. UsukiDoll
    • one year ago
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    well I know why it was done fast.. the exponent was an even number XD it's like writing i^2 11 times

  7. Astrophysics
    • one year ago
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    Main thing is you understand that i^2 = -1

  8. Astrophysics
    • one year ago
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    Which can be proved from \[i^2 = (0+i)^2 \implies (0+i)(0+i) \implies ...-1+0i = -1\]

  9. UsukiDoll
    • one year ago
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    long way... not recommended lol \[(i^2)(i^2)(i^2)(i^2)(i^2)(i^2)(i^2)(i^2)(i^2)(i^2)(i^2)\] \[(-1)(-1)(-1)(-1)(-1)(-1)(-1)(-1)(-1)(-1)(-1)\] \[(1)(1)(1)(1)(1)(-1) \rightarrow (1)(-1)= -1 \]

  10. Astrophysics
    • one year ago
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    lool

  11. Astrophysics
    • one year ago
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    Nice haha

  12. UsukiDoll
    • one year ago
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    i^{100} is gonna hurt by hand XD

  13. UsukiDoll
    • one year ago
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    so next question For 3. 5 + 8i - 4 + 2i we can combine like terms so our answer is in the form of ai+b or a+bi

  14. Astrophysics
    • one year ago
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    \[i^7 \implies (i^2)(i^2)(i^2)i\]

  15. anonymous
    • one year ago
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    so number 2 would be i^7 (-1)(-1)(-1)(-1)(-1)(-1)(-1) = -1

  16. Astrophysics
    • one year ago
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    Naw, \[i = \sqrt{-1}\]

  17. UsukiDoll
    • one year ago
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    oh I read that second problem wrong. I thought it was i and 7 XD

  18. Astrophysics
    • one year ago
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    \[i^7 = i^6 i = -1 \times i = -i\]

  19. anonymous
    • one year ago
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    okay so problem 3 would be i^3 ?

  20. UsukiDoll
    • one year ago
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    problem 3 has 3. 5 + 8i - 4 + 2i we can combine like terms so our answer is in the form of ai+b or a+bi

  21. Astrophysics
    • one year ago
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    Yeah just combine like terms haha

  22. anonymous
    • one year ago
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    -1+10i

  23. Astrophysics
    • one year ago
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    close but should be +1

  24. UsukiDoll
    • one year ago
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    we can rearrange it to make it a bit easier 5-4+8i+2i -> 1+10i

  25. UsukiDoll
    • one year ago
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    well the attempt wasn't that bad.... just 5-4=1

  26. UsukiDoll
    • one year ago
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    the last 2 problems is just using FOIL and using the fact that\[ i^2 = -1 \] again the final form of these questions is a+bi or ai+b

  27. anonymous
    • one year ago
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    (2 - i)(2 + i) 4 + 2i - 2i - i^2 4+i^2

  28. anonymous
    • one year ago
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    did i do that right

  29. Astrophysics
    • one year ago
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    \[(2-i)(2+i) \implies (2)(2)+2i+(2)(-i)-i^2 \implies 4+2i-2i-i^2 = 4-i^2 = 4-(-1) = 4+1 = 5\]

  30. Astrophysics
    • one year ago
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    \[4-i^2 = 4-(-1) \implies 4+1 = 5\]

  31. UsukiDoll
    • one year ago
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    (2-i)(2+i) 4+2i-2i-i^2 4-i^2 since i^2 = -1 4-(-1) = 4+1 = 5

  32. Astrophysics
    • one year ago
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    Sorry it got cut off there

  33. Astrophysics
    • one year ago
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    You're on the right path though, just a sign error :-)

  34. UsukiDoll
    • one year ago
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    it was a good attempt though .. just when we see i^2, we need to use substitution Replace i^2 with -1

  35. anonymous
    • one year ago
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    (2 - 3i)(3 + 5i) 6 + 10i - 9i - 15i 6 - 14i

  36. Astrophysics
    • one year ago
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    Hmm I think not

  37. UsukiDoll
    • one year ago
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    (2-3i)(3+5i) 6+10i-9i-15i^2

  38. UsukiDoll
    • one year ago
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    remember... i^2 = -1

  39. anonymous
    • one year ago
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    how do you apply -15 and i^2

  40. Astrophysics
    • one year ago
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    \[(2-3i)(3+5i) = 6+10i-9i-15i^2 \implies 6+i-15i^2\] \[6+i-15(-1) \implies 6+i+15 \implies 21 +i\]

  41. Astrophysics
    • one year ago
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    Do you see your error now?

  42. UsukiDoll
    • one year ago
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    i^2 = -1 replace the i^2 with -1 6+10i-9i-15i^2 since i^2 = -1 6+10i-9i-15(-1)

  43. UsukiDoll
    • one year ago
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    6+i+15 21+i

  44. Astrophysics
    • one year ago
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    \[\huge \checkmark\]

  45. anonymous
    • one year ago
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    omg thank yall . !!!!!! can you help me with something else ?

  46. anonymous
    • one year ago
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    1. X^2 + 9 = 0 2. 3x^2 + 11 = -181

  47. UsukiDoll
    • one year ago
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    are these questions solve for x?

  48. Astrophysics
    • one year ago
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    You can set x^2 = -9 this has to do with imaginary numbers as well

  49. anonymous
    • one year ago
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    Solve the following: x^2 + 9 = 0 Select one: a. 2 + 4i b. + 3i c. not possible d. + 9i

  50. Astrophysics
    • one year ago
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    Remember as I mentioned before \[i = \sqrt{-1}\]

  51. anonymous
    • one year ago
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    Solve: 3x^2 + 11 = -181 Select one: a. not possible b. + 8i c. 1 + 13i d. + 4i

  52. Astrophysics
    • one year ago
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    \[x^2 = - 9\] what do you get

  53. UsukiDoll
    • one year ago
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    ah yes once we square root both sides on x^2=-9 we're gonna have an imaginary case. since negative signs in radicals aren't allowed, the result is imaginary.. denoted by i

  54. anonymous
    • one year ago
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    3

  55. anonymous
    • one year ago
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    no

  56. anonymous
    • one year ago
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    3i ?

  57. Astrophysics
    • one year ago
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    That's one solution remember when we take a square root we get a positive and negative value :-)

  58. UsukiDoll
    • one year ago
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    hmm I guess for this multiple choice, it's only considering the positive version when taking the square root |dw:1439181642696:dw|

  59. anonymous
    • one year ago
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    so positive and negative 3i

  60. Astrophysics
    • one year ago
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    Yup

  61. UsukiDoll
    • one year ago
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    yeah

  62. Astrophysics
    • one year ago
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    So here is what's going on really

  63. Astrophysics
    • one year ago
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    \[x^2 = -9 \implies x = \pm \sqrt{-9} \implies \pm \sqrt{-1} \sqrt{9} \implies \pm 3i\]

  64. anonymous
    • one year ago
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    the next one is positive or negative 8i

  65. Astrophysics
    • one year ago
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    We just factor out the \[\sqrt{-1}\]

  66. Astrophysics
    • one year ago
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    Yea thats right

  67. UsukiDoll
    • one year ago
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    3x^2 +11=-181 3x^2 = -181-11 3x^2=-192 x^2 = -64 X = \sqrt{-64} x = -8i, 8i

  68. anonymous
    • one year ago
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    (2x^2-8x +10=0) im going to go this one tell me if its correct

  69. Astrophysics
    • one year ago
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    ?

  70. anonymous
    • one year ago
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    |dw:1439181899563:dw|

  71. anonymous
    • one year ago
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    im not sure what to do next

  72. Astrophysics
    • one year ago
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    Why would you do that, just factor

  73. UsukiDoll
    • one year ago
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    I wouldn't do that at all

  74. Astrophysics
    • one year ago
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    or complete the square i guess

  75. UsukiDoll
    • one year ago
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    2x^2-8x+10=0 all of these numbers are even .. .they have one even number in common

  76. Astrophysics
    • one year ago
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    Here lets make it easier on you \[2(x^2-4x+5)=0\]

  77. Astrophysics
    • one year ago
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    Oh sorry Usuki xD

  78. UsukiDoll
    • one year ago
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    2 x 1 = 2 2 x 4 = 8 2 x 5 = 10 all of them have 2 in common . so yank that out and we should have 2(x^2-4x+5) = 0

  79. anonymous
    • one year ago
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    then factor x^2 - 4x + 5 ?

  80. UsukiDoll
    • one year ago
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    then factor (x^2-4x+5) since 5 is a prime number the only combination is 1 x 5 = 5 we need a negative number to produce -4 so 1 -5 = -4 (x+1)(x-5) ... only that produces a different sign patern but I'm wondering if this is written right. by discriminant formula (-4)^2-4(1)(5) = 16-20 = -4 oh great we need quadratics -_-! .. complex roots.. mhm

  81. Astrophysics
    • one year ago
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    Yup :p

  82. UsukiDoll
    • one year ago
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    \[\frac{ -b \pm \sqrt{b^2-4ac}}{2a}\]

  83. Astrophysics
    • one year ago
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    Quadratic is the easiest way to approach this

  84. UsukiDoll
    • one year ago
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    since a = 1, b = -4, and c = 5 but we got our b^2-4ac

  85. Astrophysics
    • one year ago
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    You'll have two complex solutions

  86. UsukiDoll
    • one year ago
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    \[\frac{ 4 \pm \sqrt{-4}}{2}\] \[\frac{ 4 \pm \sqrt{4}i}{2}\]

  87. UsukiDoll
    • one year ago
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    take the square root of 4 and then reduce these fractions.

  88. anonymous
    • one year ago
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    2 plus or minus i

  89. Astrophysics
    • one year ago
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    Nice!

  90. anonymous
    • one year ago
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    Yay !!! :)

  91. UsukiDoll
    • one year ago
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    \[\frac{ 4 \pm 2i}{2} \rightarrow 2 \pm i\] correct :)

  92. anonymous
    • one year ago
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    I have other work to complete if I need help will you all be on ?

  93. UsukiDoll
    • one year ago
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    depends on the time zone... I'm a bit worn out from last night.

  94. Astrophysics
    • one year ago
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    Just open a new question, and someone will get to it eventually :)

  95. anonymous
    • one year ago
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    ok thanks for yall help & eastern

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