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anonymous

  • 5 years ago

I've got a math exam on friday. I have some of the previos exams here and there are a whole bunch of questions which I cant solve. I really needs some help with them. I'll be posting them here now. Thank you very much for all who help me out, I really appreciate it :)

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  1. anonymous
    • 5 years ago
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    Limits problems, find the limit: a) \[\lim_{x \rightarrow 5}x/(x-5)\] b) \[\lim_{x \rightarrow \infty}(\sqrt{x+1}-\sqrt{x})\]

  2. anonymous
    • 5 years ago
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    a) add and subtract 5 frm d numerator that gives u 1 + 1/(x-5) when u put in infinity it is 1 + 1/inf = 1

  3. anonymous
    • 5 years ago
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    as for b) multiply and divide by \[\sqrt{x+1}+\sqrt{x}\]

  4. anonymous
    • 5 years ago
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    that gives u 1 / √x+1+√x

  5. anonymous
    • 5 years ago
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    when u put in infinity it gives u 1/inf = 0

  6. anonymous
    • 5 years ago
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    \[a) \lim_{x \rightarrow 5} { x \over x-5} =\infty\] That's because when you substitute 5 in the top you get 5, and in the bottom you get 0. A number over 0 is infinity.

  7. anonymous
    • 5 years ago
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    Next up, find the derivative: a) \[f(x) =x^{2}\times e^{x ^{2}}\] b) \[f(x,y)=(y ^{3}x+e ^{2y})^{3}\]

  8. anonymous
    • 5 years ago
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    a) dy/dx= 2xe^x3 + 3x^4e^x^3

  9. anonymous
    • 5 years ago
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    b) him1618 already got the right answer.

  10. anonymous
    • 5 years ago
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    was a wrong?

  11. anonymous
    • 5 years ago
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    I guess :)

  12. anonymous
    • 5 years ago
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    By the way, in a) if the limit approaches 5 from the left, it would be -infinity.

  13. anonymous
    • 5 years ago
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    so ur saying it doesnt exist?

  14. anonymous
    • 5 years ago
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    Yep.

  15. anonymous
    • 5 years ago
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    yeah neway its a sort of hyperbola centred at 5

  16. anonymous
    • 5 years ago
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    It has a vertical asymptote at x=5.

  17. anonymous
    • 5 years ago
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    Question 3: Define the min/max of \[f(x)=x^{3}-12x ^{2}+21x+25\] in [-1, infinity[

  18. anonymous
    • 5 years ago
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    differentiate it f'(x) = 3x^2 - 24x + 21

  19. anonymous
    • 5 years ago
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    the roots are 7, 1

  20. anonymous
    • 5 years ago
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    Question 4: Define the local min/max of \[f(x,y)=x ^{3}-y ^{3}+3xy\]

  21. anonymous
    • 5 years ago
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    Sick how good you are :) Two more questions...

  22. anonymous
    • 5 years ago
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    yeah go on

  23. anonymous
    • 5 years ago
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    Question 5: Solve the following inequalities a) \[x^{3}+2x ^{2}+2x \le0\] b) \[\log_{2} x <2+\log_{4} x\]

  24. anonymous
    • 5 years ago
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    Question 6: Solve the following equations a) \[\left| x-3 \right|+\left| x+2 \right|=10\] b) \[2^{3x}=3^{2x}\]

  25. anonymous
    • 5 years ago
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    a) x<=0

  26. anonymous
    • 5 years ago
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    is it right

  27. anonymous
    • 5 years ago
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    I believe it's right since it's factored to x(x^2+2x+2). But the part x^2+2x+2 is always >0. So, the sign just depends on x. Hence the whole expression is <=0 when x<=0.

  28. myininaya
    • 5 years ago
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    anwar is like totatally right!

  29. anonymous
    • 5 years ago
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    so am i man..dont demean me..lol

  30. anonymous
    • 5 years ago
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    Why is the part x^2+2x+2 always >0?

  31. anonymous
    • 5 years ago
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    because its determinant is negative...

  32. anonymous
    • 5 years ago
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    Yeah. You're the one who got the answer. I was just following your foot steps :)

  33. anonymous
    • 5 years ago
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    it is an upward parabola where the min value is greater than 0

  34. anonymous
    • 5 years ago
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    Ah I understand :)

  35. anonymous
    • 5 years ago
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    You deserve a medal him1618.

  36. anonymous
    • 5 years ago
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    thnx a lot

  37. anonymous
    • 5 years ago
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    I gave one too :)

  38. anonymous
    • 5 years ago
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    You're welcome :)

  39. anonymous
    • 5 years ago
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    so kind

  40. anonymous
    • 5 years ago
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    I want to solve part b) in inequalities.

  41. anonymous
    • 5 years ago
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    b) x is in 0 to 16

  42. anonymous
    • 5 years ago
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    Please do I can't solve it :/

  43. anonymous
    • 5 years ago
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    thats d ans

  44. anonymous
    • 5 years ago
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    its damn tedious to write these eqns

  45. anonymous
    • 5 years ago
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    Yeah, that sucks....

  46. anonymous
    • 5 years ago
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    on d comp only..on paper it took a minute

  47. myininaya
    • 5 years ago
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    i think i considered all the cases for 6a

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  48. anonymous
    • 5 years ago
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    2 is \[\log_{4}16 \]

  49. anonymous
    • 5 years ago
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    \[\log_2x-\log_4x<2 \implies \log_2x-{\log_2x \over \log_24}<2 \implies \log_2x-{1 \over 2}\log_2x<2\] \[{1 \over 2}\log_2 x<2 \implies \log_2{\sqrt x}<2 \implies \sqrt x<4 \implies x<16\]

  50. anonymous
    • 5 years ago
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    But x has to be a positive number, So 0<x<16

  51. anonymous
    • 5 years ago
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    right

  52. anonymous
    • 5 years ago
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    Nice answer myininaya helpd me a bunch :)

  53. myininaya
    • 5 years ago
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    here is 6b

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  54. anonymous
    • 5 years ago
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    yeah 6a is bloody tedious..nice wrk

  55. anonymous
    • 5 years ago
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    b is right 2

  56. anonymous
    • 5 years ago
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    Right on the spot AnwarA I'll jot that down to my notebook :)

  57. anonymous
    • 5 years ago
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    Did anyone solve 6a, or I can take it?

  58. anonymous
    • 5 years ago
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    yes its been solved by myininaya

  59. anonymous
    • 5 years ago
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    I hate you man >.<

  60. myininaya
    • 5 years ago
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    woman* lol

  61. anonymous
    • 5 years ago
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    OMG.. I don't hate you at all :P

  62. myininaya
    • 5 years ago
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    lol

  63. myininaya
    • 5 years ago
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    you can look over my work. there might be a mistake who knows

  64. myininaya
    • 5 years ago
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    it looks good to me though

  65. anonymous
    • 5 years ago
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    I can hardly read it :P

  66. myininaya
    • 5 years ago
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    :(

  67. myininaya
    • 5 years ago
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    i can rewrite it

  68. anonymous
    • 5 years ago
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    No, It's right. I got the exact same answers.

  69. myininaya
    • 5 years ago
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    you can also graph it in your calculator if you want to verify your answers

  70. anonymous
    • 5 years ago
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    x=15/2 for x>=3, x=-5/2 for x<=2. No solution for 2<x<3.

  71. myininaya
    • 5 years ago
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    right!

  72. myininaya
    • 5 years ago
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    i can write neat when i'm not trying to beat you guys to answer the questions lol

  73. myininaya
    • 5 years ago
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    i wish in high school there was a math quiz bowl and i participated that would have been so fun

  74. anonymous
    • 5 years ago
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    lol we should divide the questions equally. I wish I had a scanner -.-

  75. anonymous
    • 5 years ago
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    One question, what happens here: \[\log_2x-{\log_2x \over \log_24}<2 \implies \log_2x-{1 \over 2}\log_2x<2\] ?

  76. anonymous
    • 5 years ago
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    log4 to base to is 2.

  77. anonymous
    • 5 years ago
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    to base 2*

  78. myininaya
    • 5 years ago
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    because 2^2=4

  79. anonymous
    • 5 years ago
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    Mm, yes of course! :)

  80. anonymous
    • 5 years ago
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    Do you still have any questions, or that's it?

  81. anonymous
    • 5 years ago
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    I think that is about it :) So quick and dirty. Thanks a million for all of you!

  82. myininaya
    • 5 years ago
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    yes all your questions were being answered so quickly by him hes such a meanie

  83. anonymous
    • 5 years ago
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    no probs mate

  84. anonymous
    • 5 years ago
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    yeah thats kind...

  85. myininaya
    • 5 years ago
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    lol i'm kidding

  86. anonymous
    • 5 years ago
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    dont sweat..i hate studying so much...bt i lov doing this

  87. anonymous
    • 5 years ago
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    I think that is about it :) So quick and dirty. Thanks a million for all of you!

  88. anonymous
    • 5 years ago
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    well thnx

  89. anonymous
    • 5 years ago
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    Good luck in your exam on Friday :)

  90. anonymous
    • 5 years ago
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    yeah good luck...remember me if u feel nervy..lol

  91. anonymous
    • 5 years ago
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    Thanks, I believe I can make it :D

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