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anonymous

  • 5 years ago

3x/2 + 1/x = 3x/4

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  1. mathteacher1729
    • 5 years ago
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    Is this the problem? \[\frac{3}{2}x+\frac{1}{x}=\frac{3}{4}x\]

  2. anonymous
    • 5 years ago
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    yes

  3. anonymous
    • 5 years ago
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    oh, that...ok hang on!

  4. anonymous
    • 5 years ago
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    and this problem: 5+ radical 7 divided by 5- radical 7. simplify the expression.

  5. anonymous
    • 5 years ago
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    I answered it, it's (32+ (10) times (Rad. 7) the whole thing divided by 12)

  6. anonymous
    • 5 years ago
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    i got a different answer though

  7. anonymous
    • 5 years ago
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    The problem have no real solutions....Are you learning about imagery?

  8. anonymous
    • 5 years ago
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    \[\sum_{n=1}^{3}\rightarrow(1^n-1)\]

  9. mathteacher1729
    • 5 years ago
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    First -- please post each problem separately on the left. Now, to talk about the first problem: \[\frac{3}{2}x+\frac{1}{x}=\frac{3}{4}x\] Let's subtract (3/2)x from both sides: \[\frac{3}{2}x-\frac{3}{2}x+\frac{1}{x}=\frac{3}{4}x-\frac{3}{2}x\] \[\frac{1}{x}=\frac{3}{4}x-\frac{3}{2}x\] \[\frac{1}{x}=x(\frac{3}{4}-\frac{3}{2})\] So now we have to do a bit of algebra with fractions here. Find the common demoniator (4) and re-write as: \[\frac{1}{x}=x(\frac{3}{4}-\frac{3}{2}*\frac{2}{2})\] \[\frac{1}{x}=x(\frac{3}{4}-\frac{6}{4})\] This is now gets us another step further \[\frac{1}{x}=x(\frac{3-6}{4})\] and now we have \[\frac{1}{x}=\frac{-3x}{4}\] From here you must cross multiply: \[4 = -3x^2\] and we will have an imaginary answer... Are you sure you copied the problem correctly?

  10. anonymous
    • 5 years ago
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    \[\sum_{n=1}^{3}i^n-1\]

  11. mathteacher1729
    • 5 years ago
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    Brii, this is asking: \[(i^1-1)+(i^2-1)+(i^3-1)\] (just let n = 1 , then let n = 2, and let n = 3 and add.

  12. anonymous
    • 5 years ago
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    yea i got i is that correct?

  13. mathteacher1729
    • 5 years ago
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    How did you get your answer? Can you show me the steps?

  14. anonymous
    • 5 years ago
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    i plugged it in the calculator for example: i^1-1 = 1... and so on

  15. mathteacher1729
    • 5 years ago
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    i^1-1 is not 1... :( \[i^1=i\] \[i^2=-1\] \[i^3=-i\] \[i^4=1\] So the sum \[(i^1-1)+(i^2-1)+(i^3-1)\] \[=(i-1)+(-1-1)+(-i-1)\] \[=i-1-1-1-i-1\] I'll let you finish the rest. :)

  16. anonymous
    • 5 years ago
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    but wouldn't you put i to the 1-1 so its i to the 0 which equals 1?

  17. mathteacher1729
    • 5 years ago
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    This is where the math type comes into play. Did you mean the sequence is: \[\sum_{n=0}^{3}i^{n-1}\]

  18. anonymous
    • 5 years ago
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    no the bottom of the sigma should read n=1

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