anonymous
  • anonymous
3x/2 + 1/x = 3x/4
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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mathteacher1729
  • mathteacher1729
Is this the problem? \[\frac{3}{2}x+\frac{1}{x}=\frac{3}{4}x\]
anonymous
  • anonymous
yes
anonymous
  • anonymous
oh, that...ok hang on!

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anonymous
  • anonymous
and this problem: 5+ radical 7 divided by 5- radical 7. simplify the expression.
anonymous
  • anonymous
I answered it, it's (32+ (10) times (Rad. 7) the whole thing divided by 12)
anonymous
  • anonymous
i got a different answer though
anonymous
  • anonymous
The problem have no real solutions....Are you learning about imagery?
anonymous
  • anonymous
\[\sum_{n=1}^{3}\rightarrow(1^n-1)\]
mathteacher1729
  • mathteacher1729
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?
anonymous
  • anonymous
\[\sum_{n=1}^{3}i^n-1\]
mathteacher1729
  • mathteacher1729
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.
anonymous
  • anonymous
yea i got i is that correct?
mathteacher1729
  • mathteacher1729
How did you get your answer? Can you show me the steps?
anonymous
  • anonymous
i plugged it in the calculator for example: i^1-1 = 1... and so on
mathteacher1729
  • mathteacher1729
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. :)
anonymous
  • anonymous
but wouldn't you put i to the 1-1 so its i to the 0 which equals 1?
mathteacher1729
  • mathteacher1729
This is where the math type comes into play. Did you mean the sequence is: \[\sum_{n=0}^{3}i^{n-1}\]
anonymous
  • anonymous
no the bottom of the sigma should read n=1

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