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moneybird

  • 4 years ago

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  1. moneybird
    • 4 years ago
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    What are all the values of "x" between 0 and 2 pi that satisfy the equation? \[(5+2\sqrt{6})^{\sin x} +(5-2\sqrt{6})^{\sin x} = 2 \sqrt{3}\]

  2. Ishaan94
    • 4 years ago
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    oh wow

  3. Walleye
    • 4 years ago
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    my exact thoughts. I have never seen anything like this

  4. Ishaan94
    • 4 years ago
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    Hmm \[(5 + 2 \sqrt6)(5 - 2\sqrt6) = 25 - 24 =1\]

  5. Ishaan94
    • 4 years ago
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    Okay I think maybe I got it!

  6. Ishaan94
    • 4 years ago
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    Hmm \[(5+2\sqrt{6})^{\sin x} +(\frac{1}{5 + 2\sqrt{6}})^{\sin x} = 2 \sqrt{3}\] How about this? :-D

  7. moneybird
    • 4 years ago
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    o...

  8. Ishaan94
    • 4 years ago
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    \[\frac{(5 + 2\sqrt{6})^{2\sin{x}} +1}{(5 + 2\sqrt{6})^{\sin{x}}} = 2\sqrt{3}\] Are we getting somewhere? lol

  9. moneybird
    • 4 years ago
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    I think I got it

  10. Walleye
    • 4 years ago
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    awesome ishaan!

  11. moneybird
    • 4 years ago
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    Let (5 + 2 sqrt (6)) be a Let sinx be b \[a^b + \frac{1}{a^b} = 2 \sqrt{3}\] \[\frac{a^{2b} + 1}{a^b} = 2 \sqrt{3}\] \[a^{2b} + 1 - 2 a^b \sqrt{3}\]

  12. moneybird
    • 4 years ago
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    Let a^b be c \[c^2 -2c \sqrt{3} + 1 = 0 \]

  13. Ishaan94
    • 4 years ago
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    \[(5 + 2\sqrt{6})^{2\sin{x}} + 1 = 2\sqrt{3}\times (5 + 2\sqrt{6})^{\sin{x}} \] \[ 1 = 2\sqrt{3}\times (5 + 2\sqrt{6})^{\sin{x}} - (5 + 2\sqrt{6})^{2\sin{x}}\] Hmm....

  14. Ishaan94
    • 4 years ago
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    Oh i see what you did.... Hmm nice

  15. moneybird
    • 4 years ago
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    Thanks for the hint :3

  16. Ishaan94
    • 4 years ago
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    No Problem :-)

  17. moneybird
    • 4 years ago
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    x = 60, 150, 210, 330 degrees

  18. Ishaan94
    • 4 years ago
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    Cool! :-D If only there was an easier way to get this

  19. ali.athar
    • 4 years ago
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    \[\lim_{x \rightarrow 1} x ^{3}-1/x ^{2}-1\]

  20. ali.athar
    • 4 years ago
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    is someone answer this limit question please

  21. ali.athar
    • 4 years ago
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    2 ali.athar 0 limx→1x3−1/x2−1 someone solve this question

  22. robtobey
    • 4 years ago
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    From Mathematica:\[\text{Limit}\left[x^3-\frac{1}{x^2}-1,x\to 1\right]=-1 \]

  23. ali.athar
    • 4 years ago
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    0 2 ali.athar 0 limx→1 x3−1/x2−1

  24. ali.athar
    • 4 years ago
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    \[\lim_{x \rightarrow 2} x ^{3}-8/x ^{2}+x ^{}-6\]

  25. moneybird
    • 4 years ago
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    x^3 - 8 = (x-2) (x^2 + 2x + 4) x^2 + x - 6 = (x+3)(x-2) x-2 cancels out (x^2 + 2x + 4)/(x+3) = (4 + 4 + 4 )/ 5 = 12/5

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