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

1. mathteacher1729

Is this the problem? $\frac{3}{2}x+\frac{1}{x}=\frac{3}{4}x$

2. anonymous

yes

3. anonymous

oh, that...ok hang on!

4. anonymous

and this problem: 5+ radical 7 divided by 5- radical 7. simplify the expression.

5. anonymous

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

6. anonymous

i got a different answer though

7. anonymous

The problem have no real solutions....Are you learning about imagery?

8. anonymous

$\sum_{n=1}^{3}\rightarrow(1^n-1)$

9. 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?

10. anonymous

$\sum_{n=1}^{3}i^n-1$

11. 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.

12. anonymous

yea i got i is that correct?

13. mathteacher1729

How did you get your answer? Can you show me the steps?

14. anonymous

i plugged it in the calculator for example: i^1-1 = 1... and so on

15. 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. :)

16. anonymous

but wouldn't you put i to the 1-1 so its i to the 0 which equals 1?

17. mathteacher1729

This is where the math type comes into play. Did you mean the sequence is: $\sum_{n=0}^{3}i^{n-1}$

18. anonymous

no the bottom of the sigma should read n=1