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anonymous
 5 years ago
3x/2 + 1/x = 3x/4
anonymous
 5 years ago
3x/2 + 1/x = 3x/4

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mathteacher1729
 5 years ago
Best ResponseYou've already chosen the best response.0Is this the problem? \[\frac{3}{2}x+\frac{1}{x}=\frac{3}{4}x\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh, that...ok hang on!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and this problem: 5+ radical 7 divided by 5 radical 7. simplify the expression.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I answered it, it's (32+ (10) times (Rad. 7) the whole thing divided by 12)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i got a different answer though

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The problem have no real solutions....Are you learning about imagery?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\sum_{n=1}^{3}\rightarrow(1^n1)\]

mathteacher1729
 5 years ago
Best ResponseYou've already chosen the best response.0First  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 rewrite 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 \[\frac{1}{x}=x(\frac{36}{4})\] 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
 5 years ago
Best ResponseYou've already chosen the best response.0\[\sum_{n=1}^{3}i^n1\]

mathteacher1729
 5 years ago
Best ResponseYou've already chosen the best response.0Brii, this is asking: \[(i^11)+(i^21)+(i^31)\] (just let n = 1 , then let n = 2, and let n = 3 and add.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yea i got i is that correct?

mathteacher1729
 5 years ago
Best ResponseYou've already chosen the best response.0How did you get your answer? Can you show me the steps?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i plugged it in the calculator for example: i^11 = 1... and so on

mathteacher1729
 5 years ago
Best ResponseYou've already chosen the best response.0i^11 is not 1... :( \[i^1=i\] \[i^2=1\] \[i^3=i\] \[i^4=1\] So the sum \[(i^11)+(i^21)+(i^31)\] \[=(i1)+(11)+(i1)\] \[=i111i1\] I'll let you finish the rest. :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but wouldn't you put i to the 11 so its i to the 0 which equals 1?

mathteacher1729
 5 years ago
Best ResponseYou've already chosen the best response.0This is where the math type comes into play. Did you mean the sequence is: \[\sum_{n=0}^{3}i^{n1}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no the bottom of the sigma should read n=1
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