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December 20, 2014

December 20, 2014

Posted by **Janae** on Saturday, September 14, 2013 at 6:24pm.

(ab^2c)^-1 / a^8b^2c^-4b^3

- Math -
**Steve**, Saturday, September 14, 2013 at 6:51pm(ab^2c)^-1 = a^-1b^-2c^-1

a^8b^2c^-4b^3 = a^8b^5c^-4

a^-1b^-2c^-1 / a^8b^5c^-4 = a^-9b^-7c^3

= c^3 / a^9b^7

- Math -
**Jackie**, Saturday, September 14, 2013 at 7:08pmok i get it now.

for this question..

3square root of 1/2x+1 +2=5

the +2=5 is out of the square root.

so do i get rid of the square root first or subtract 2 from 5

- Math -
**Steve**, Sunday, September 15, 2013 at 6:22amdo the reverse of PEMDAS

3√1/(2x+1) + 2 = 5

3/√(2x+1) = -3

1/√(2x+1) = -1

1 = -√(2x+1)

squaring yields

1 = 2x+1

0 = 2x

x = 0

check:

3/√1 + 2 = 5

yes

Rather a poor exercise, imho

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