Question

Solve each equation. Show all work

1.(1/6x2 ) = ( 1/3x2) - (1/x)

2. (1/n2) + (1/n) = (1/2n2)

3. (1/x) = (6/3x) + 1

4. (1/6x2) = (1/2x) + (7/ 6x2)

5. (1/6b2) + (1/6b) = (1/b2)

Answer 1

help me anyone

Answer 2

help help anyone PLEASE

Answer 3

Please choose ONE question, and then we could focus on that one. Discuss what you already know and what you need to know to solve this question.

Answer 4

1.(1/6x2 ) = ( 1/3x2) - (1/x)

Answer 5

Is it supposed to be this? \(\dfrac{1}{6x^{2}} = \dfrac{1}{3x^{2}} - \dfrac{1}{x}\)

In any case, you would do well to find a common denominator and add things up. Go!!

Answer 6

yes now wat i supose to do

Answer 7

As tkhunny has aptly pointed out, you need to determine what the common denominator is in this particular problem.

Answer 8

aint the common denominator is 6

Answer 9

Hint; IF WE were to add 1/3 and 1/7, we'd identify the lowest common denominator to be 3*7 = 21, and then change each fraction accordingly:

1/3 = 7/21

1/7 = 3/21

1/3 + 1/7 = (3+7)/21 = 10/21.

Think: how could you use this information / example to find the LCD of your posted problem?

Answer 10

6*3=18
1/3=6/18
1/6=3/18
1/3 + 1/6=(3+6)/18=9/18

Answer 11

Please remember that what I typed in most recently was an example. You'll need to apply the same principle(s) towards solving (1/6x2 ) = ( 1/3x2) - (1/x).

\[(1/6x^2 ) = ( 1/3x^2) - (1/x)\] hs common denominator 6x^2. Note that both 3x^2 and x will divide into 6x^2.

\[(1/6x^2 ) = ( 1/3x^2) - (1/x) \rightarrow \frac{ 1 }{ 6x^2 }=\frac{ 1 }{ 3x^2 }-\frac{ 1 }{ x }\]

Answer 12

can be re-written with the common denominator in all terms as:\[\frac{ 1 }{ 6x^2 }=\frac{ 1(2) }{ 2(3x^2) }-\frac{ 1(6x) }{ x(6x) }\]

Answer 13

and then, because the denoms. are now all the same, you can ignore them and concentrate on solving 1=2-6x.

I need to get off OpenStudy now. But I hope to be able to work with you again soon.