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December 20, 2014

December 20, 2014

Posted by **JayLynne-I had misprint in previous posting** on Tuesday, October 19, 2010 at 11:37am.

1. Solution to the radical function sqrt(1/2x+1) = 1 is x = 0

True or False I think it is true according to my calculations

2.Solve x = sqrt(8x)

8x = x^2

x^2 - 8x = 0

x(x-8) = 0

x = 0

x-8 =0

x=8

the only solution is x = 0, correct or nobecause x = 8 doesn't work-

3. Simply (3x^2y^-2)^-2/9x^-5y

a.xy^3

b. xy^3/3

xy^3/9

d. xy^3/81

I think it is xy^3/81 but I'm not sure of my calculations

4. Solve 4sqrt(6x-2)>4

a. 1/3<=x<43

b. 1/3<x<43

x>=43

x>43

I think x>43

Thank you

- Algebra-Please check my calculations -
**MathMate**, Tuesday, October 19, 2010 at 11:54am1. true if the equation is

sqrt((1/2)x+1) = 1, or

sqrt(x/2+1) = 1

2. x=√(8x)

"the only solution is x = 0, correct or nobecause x = 8 doesn't work- " ???

x=0 works: 0=√(8*0)

x=8 also works √(8*8)=8

3. Correct.

4. Solve 4sqrt(6x-2)>4

Divide by 4 to get

sqrt(6x-2)>1

For this to happen, 6x-2>1 or x>(1/2)

which does not match any of your answers.

I suspect there is a misprint or misinterpretation of the question.

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