math problem! precal / trig?
({ a^2/3 b^3/2}/ a^2b)^6
simplify
if this is confusing, i can explain it.
a has an exponent of 2/3 and b has an exponent of 3/2. Both are over a squared and b ( not squared)
the entire problem is to the power of 6 and it asked me to simplify
- thank you :)
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the answer is b^3/a^8
first u distribute the 6 exponent to the entire problem.....soo after multiplying it...the problem reads: (a^4 b^9)/(a^12 b^6)
now u subtract the expoenents of the same bases...so a^4 - a^12 and b^9 - b^6....which equal a^-8 and b^3....
negative exponents got under the fraction bar as denominators, thus, the answer b^3/a^8
THANK YOU FOR TAKING THE TIME TO EXPLAIN AND WRITE WITH COMPLETE PARENTHESES.
First, let's apply the 6 by multiplying it times every exponent inside:
(a^4b^9) / (a^12b^6) Next subtract exponents of powers with same bases.
a^4/a^1 = a^(4-12) = a^-8. b^9/b6 = b^(9-6) = b^3.
a^-8 •b^3 or rewrite with all positive exponents as b^3 / a^8
= (a^-4/3 b^1/2)^6 = b^3 / a^8
b^3/a^8