Question

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

Answer 1

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

Answer 2

yes

Answer 3

oh, that...ok hang on!

Answer 4

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

Answer 5

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

Answer 6

i got a different answer though

Answer 7

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

Answer 8

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

Answer 9

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?

Answer 10

\[\sum_{n=1}^{3}i^n-1\]

Answer 11

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.

Answer 12

yea i got i is that correct?

Answer 13

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

Answer 14

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

Answer 15

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. :)

Answer 16

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

Answer 17

This is where the math type comes into play.

Did you mean the sequence is:

\[\sum_{n=0}^{3}i^{n-1}\]

Answer 18

no the bottom of the sigma should read n=1