1. tootsi123

2. tootsi123

@jim_thompson5910

3. misty1212

HI!!

4. misty1212

5. tootsi123

yes

6. misty1212

first the $$-2$$ outside means a) flip everything, then b) square it

7. tootsi123

Okay

8. misty1212

so first $\huge\frac{1}{\left(-3u^2v^3\right)^2}$

9. misty1212

then we need so square everything that means square $$-3$$ to get $$9$$ and also double each exponent

10. tootsi123

Okay but where did the 2 come from in the parenthesis

11. misty1212

$\huge\frac{1}{\left(-3u^3v^3\right)^2}=\frac{1}{9u^6v^6}$

12. misty1212

oh oops that was a typo

13. tootsi123

Oh okay :) i'm following

14. misty1212

should have been $\huge\frac{1}{\left(-3u^3v^3\right)^2}$

15. misty1212

then square all answer above is right though, looks like your choice A

16. tootsi123

thank you :) okay there is another question like it... can i post it and tell you how i would do it and then you tell me if i am doing it right or not??

17. misty1212

ok sure $\color\magenta\heartsuit$

18. tootsi123

So i would flip it first right

19. misty1212

no not here

20. misty1212

the reason we flipped before was because there was a $$-2$$ outside the parentheses

21. tootsi123

Oh okay so then i wouldn't flip it i would leave it the same

22. misty1212

there is only one choice here that makes sense take a look you have $\huge\frac{x^4}{x^{-5}}$ right?

23. misty1212

the $$-5$$ in the deominator means bring it up to the numerator as $$+5$$ so $\frac{x^4}{x^{-5}}=x^4\times x^5=x^{4+5}=x^9$

24. misty1212

only one choice as $$x^9$$ in it so we dont really need to do the rest

25. tootsi123

so i take the two common ones and add them

26. misty1212

yeah you want to do it all?

27. tootsi123

Yes please cause i have a lot more questions like this and i want to make sure i understand them

28. misty1212

ok lets take it slow

29. misty1212

first off, unlike the last one there is no parentheses anywheres, so it is somewhat easier

30. misty1212

you have a minus sign out front that stays there

31. misty1212

you also have $\frac{2}{4}$ which is the same as $$\frac{1}{2}$$ so there will be a 2 in the denominator

32. tootsi123

Okay

33. misty1212

as for the x terms, you have $\frac{x^4}{x^{-5}}$ the $$-5$$ has a minus sign, so that comes upstairs as $$x^5$$ which is why you get $\frac{x^4}{x^{-5}}=x^4\times x^5=x^9$

34. tootsi123

Alright

35. misty1212

and for $\frac{y^2}{y^5}$ the 5 is bigger than the 2, so subtract 2 from 5 in the denominator $\frac{y^2}{y^5}=\frac{1}{y^{5-2}}=\frac{1}{y^3}$

36. misty1212

so in total, a) there is a - sign out front b) there is a 2 in the denominator c) there is a $$x^9$$ in the numerator and d) a $$y^3$$ in the denominator

37. tootsi123

Okay so which ever number is bigger you use that to decide whether or not you add or subtract and whether or not it says on the top or the bottom, am i right?

38. misty1212

yes more or less

39. misty1212

if the exponent is negative a) if it is up bring it down b) if it is down bring it up

40. misty1212

if both terms have positive exponents, subtract the smaller one from the bigger one

41. misty1212

here is an example $\frac{x^7}{x^{-3}}=x^{10}$ wheras $\frac{x^4}{x^{10}}=\frac{1}{x^6}$

42. tootsi123

Okay that makes a little more since now

43. misty1212

whew

44. tootsi123

Thanks