Question

cube root (8xh^-5xs^-3)/cube root (h^-9xs^9). Please help!

Answer 1

do the x's mean variable x or times?

Answer 2

times

Answer 3

ok. people don't use x in algebra to mean times. too confusing.
(8xh^-5xs^-3)/cube root (h^-9xs^9)
\[ \frac{ \sqrt[3]{8h^{-5}s^{-3}}}{ \sqrt[3]{h^{-9}s^{9}}} \]
is this it?

Answer 4

yes!

Answer 5

I would rewrite that using exponent 1/3 instead of cube root sign
It means the same thing, but easier to manipulate
\[\left( \frac{ 8h^{-5}s^{-3}}{h^{-9}s^{9}}\right)^\frac{1}{3} \]

Answer 6

do you know how to figure out
\[ \frac{h^{-5}}{h^{-9}}\]?

Answer 7

do you make the -9 positive by bring it up to the numerator?

Answer 8

you could. 1/h^-9 means the same as h^9
there are "short cut rules"
when multiplying or dividing the same base with different exponents.

Answer 9

the simplest rule is an extension of this idea:
\[ h \cdot h = h^2\]
if we put in the exponents to make it more obvious
\[ h^1 \cdot h^1 = h^{1+1}= h^2\]
we add the exponents.

another example:
h*h*h * h*h
using the "short version" that is obviously h^5
it is also h^3 * h^2 and we see adding 3+2 gives us the right answer: h^5

Answer 10

okay, so the question wants me to put it into kh^rs^t. I got 2h^4/3s^2

Answer 11

on the other hand
\[ \frac{h\cdot h}{h} = h \]
and that is
\[ \frac{h^2}{h^1} = h^{2-1}= h^1= h \]
we subtract top exponent minus bottom exponent

Answer 12

using the subtract rule h^-5/ h^-9 = h^(-5 - (-9))
-5 - -9 is -5+9 = +4
h^4 is the simplified version

Answer 13

then it'll be 4/3?

Answer 14

now do the s's
s^-3 / s^9
the new exponent will be -3 - 9 = -12
so s^-12

and though not obvious, 8= 2^3
we have so far
\[ \left(2^3 h^4 s^{-12}\right)^\frac{1}{3} \]

Answer 15

oh is it 2h^4/3s^-4

Answer 16

and we can take the cube root of each term inside
notice the s^-4 is up top. (if we put it down below it will become 1/s^4 )
\[ 2 h^\frac{4}{3} s^{-4} \]

Answer 17

so (2h^4/3)/s^4

Answer 18

I think no parens confused me
I would write it as
2 h^(4/3) s^(-4)

Answer 19

because they want it in the form
k h^r s^t

Answer 20

oh I got the question right! thank you!

Answer 21

they wanted me to keep the negative exponent