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November 28, 2014

November 28, 2014

Posted by **Alyssa Pacheco** on Sunday, March 14, 2010 at 10:56pm.

anything helps

thank you(:

2. Simplify

---(3x^11z^5)^3

b. (3x^5)^ -3 / 2x-5(3y)^4

c.3x^5)^ -3 / 2x ^ -5(3y)^4 (same as b but - 5 is the exponent)

3a.x^2 / 5 + x-5 / 6=0 (proportion, and then i got to the quadratic equation and after i did that i got 5/3 and -2.5 correct?)

3b.18=2(3)^x+3 (x+3 are exponents)

c. 128=4^3x (whats x)

- math -
**Reiny**, Monday, March 15, 2010 at 12:11am2.

(3x^11z^5)^3 = 27 x^33 z^15

b) and c) are not clear since you don't use proper brackets

3a) again, your lack of brackets make it ambiguous

I will assume you mean

x^2/5 + (x-5)/6 = 0

then

x^2/5 = (5-x)/6

cross-multiply

6x^2 = 25 - 5x

6x^2 + 5x - 25 = 0

your answer is correct

3b)

18 = 2(3)^(x+3)

9 = 3^(x+3)

3^2 = 3^(x+2)

so x+2 = 2

x = 0

c)

128 = 4^(3x)

2^7 = (2^2)^(3x)

2^7 = 2^(6x)

6x = 7

x = 7/6

- math -
**nevaeh.**, Monday, March 15, 2010 at 12:47amthis is how 2b and 2c is written:

(3x^5)^-3

-----------

2x^-5(3y)^4

Does that help?

(3x^5)^-3

-----------

2x-5(3y)^4

- math -
**nevaeh.**, Monday, March 15, 2010 at 12:47amthat is what think alyssa is trying to say?

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