Question

integral exponents,simplify;
(27p^3q^6r^1)^1/3

Answer 1

Find the distance between the points (6, 5) and (4, 3).

Answer 2

?

Answer 3

\[(27p^3q^6r^1)^{1/3}=27^{1/3}(p^3)^{1/3}(q^6)^{1/3} (r^1)^{1/3}\]

Answer 4

and then i use (ym)n=ymn?

Answer 5

I'm guessing you mean \((y^m)^n=y^{mn}\) , then yes :)

Answer 6

so the answer will be
27^1/3 p^1 q^2 r^1/3?

Answer 7

yes. Although, you can simplify \(27^{1/3}\) even more!

Answer 8

okay then 27^1/3 will be
9?

Answer 9

9 p^1 q^2 r^1/3?

Answer 10

not quite.

\(27=3\times 3 \times 3 = 3^3\)

So: \(27^{1/3}=(3^3)^{1/3}=3^1=3\)

Answer 11

also, you don't need to write \(p^1\), when you have 1's as exponents, you can just leave it as \(p\)

Answer 12

okay.thanks. :)

Answer 13

np