Maths question?

Looks ok, but assuming youre getting assessed on line, the usual form would be to put the y^(-11) on the denom as y^11. Similarly, doing that at the start/throughout the working would be more usual.

Edit - note a simpler/quicker way to do it would simply to consider the x, y and z terms independently and then multiply at the end. So the y's would give
1/y^2 * 1/y^9 = 1/y^11
...

Right, got it thank you!

I put it into that form you said is preferable to get x^3/y^11 z^2

But the answer is x^3/ y^11 z

So no z^2. Can you see anything I may have done wrong?

Where is the z^2? I can't see it.

Ah here unless I’ve made an error when simplifying?

Ah here unless I’ve made an error when simplifying?

In the OP you had just z on the denominator which is correct.

This isn't the same as your second attachment - can you post the whole question in one picture?

I thought to put it into the more desirable format I had to do this stuff I did at the end?

3^-1 * 2 / 5 = 2/15
not 2/75. There is just a single z on the denominator. y^-11 on the numerator is equivalent to y^11 on the denominator, nothing else changes.

Ok thank you for all the replies!

It youre a bit unsure about this stuff, a simple tip is to write it all as multiplications, so division becomes multiplication by a reciprocal, so youre working with
(1/y)^2 (x^3)^2 (1/z^2)^2 (1/x)^3 (z)^3 (1/y^3)^3
then its easy to read off the relevant exponents for each base as per #2.

Yes I’ve got that thank you

Also, what do you think of the following problem?

I know you don’t have to actually work it out/try to work it out but disprove it by giving an example that doesn’t work

Sounds like an interesting problem, youre starting the right way, so good luck.