Question

Simplified form of..

Answer 1

\[\frac{ 8a^2 bc ^{-1}}{ 12ab^3c }\]

Answer 2

Look at each bit individually. \[\frac{8}{12} \times \frac{a^2}{a} \times \frac{b}{b^3} \times \frac{c^-1}{c}\]
Remember also that \(c^-1 = 1/c\)
Let me know if you need more help on this one!

Answer 3

i understand what to do with the first ones but not the c?

Answer 4

No worries, \[c^-1 = \frac{1}{c}\]
That means that \[\frac{c^-1}{c} = \frac{1}{c} \times \frac{1}{c}\]
Does that make sense?

Answer 5

oh yes

Answer 6

so is the final answer \[\frac{ 2a }{ 3bc }\]

Answer 7

@hea

Answer 8

Nope.

You got the numbers right and the a's but \(b/b^3 = 1/b^2\) (if I read the question right)
And with the c's\[\frac{1}{c} \times\frac{1}{c} = \frac{1}{c^2}\]

Answer 9

so is it \[\frac{ 2a }{ 3b^2c^2 }\]

Answer 10

yep!

Answer 11

thanks!