math
posted by Alyssa Pacheco on .
Theses are just some problems i did not understand how to do so if you could please explain and write the answer(:
anything helps
thank you(:
2. Simplify
(3x^11z^5)^3
b. (3x^5)^ 3 / 2x5(3y)^4
c.3x^5)^ 3 / 2x ^ 5(3y)^4 (same as b but  5 is the exponent)
3a.x^2 / 5 + x5 / 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)

2.
(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 + (x5)/6 = 0
then
x^2/5 = (5x)/6
crossmultiply
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 
this is how 2b and 2c is written:
(3x^5)^3

2x^5(3y)^4
Does that help?
(3x^5)^3

2x5(3y)^4 
that is what think alyssa is trying to say?