SOLUTION: Can someone show me how to simplify these 2 questions 1) (2x)^2 * (3x^4) A)6x^8 B)12x^6 C)6x^6 D)324x^6 and 2) (-3x / 2y^3)^-2 A) 4y^3 / 9x^2 B) -4y^6 / 9x^2 C

Algebra ->  Functions -> SOLUTION: Can someone show me how to simplify these 2 questions 1) (2x)^2 * (3x^4) A)6x^8 B)12x^6 C)6x^6 D)324x^6 and 2) (-3x / 2y^3)^-2 A) 4y^3 / 9x^2 B) -4y^6 / 9x^2 C      Log On


   



Question 36403: Can someone show me how to simplify these 2 questions
1) (2x)^2 * (3x^4)
A)6x^8
B)12x^6
C)6x^6
D)324x^6
and
2) (-3x / 2y^3)^-2
A) 4y^3 / 9x^2
B) -4y^6 / 9x^2
C) 4y^6 / 9x^2
D) -4y^6 / 6x^2
Please show me how you worked these out...it confuses me..thanks

Answer by rapaljer(4671) About Me  (Show Source):
You can put this solution on YOUR website!
1) (2x)^2 * (3x^4)
+2%5E2%2Ax%5E2%2A3%2Ax%5E4
+4x%5E2+%2A3x%5E4

Remember when you multiply, you ADD exponents:
12x%5E6
The answer is B.


2) (-3x / 2y^3)^-2
+%28%28-3x%29%2F%282y%5E3%29%29%5E-2

When you raise a fraction to a negative power, you invert the fraction. When you raise a fraction to the 2 power you square the fraction. This means that in raising to the -2 power, you must invert and square:
%28%282y%5E3%29%2F%28-3x%29%29%5E2+
%284y%5E6%29%2F%289x%5E2%29%29

The correct answer is C

R^2 at SCC