A community for students. Sign up today

Here's the question you clicked on:

anonymous one year ago Which of the following expressions are equivalent? Justify your reasoning. OPTIONS LISTED BELOW

• This Question is Closed
1. anonymous

A. $\sqrt[4]x{3}$ B. $\frac{ 1 }{ x^{-1}}$ C.$\sqrt[10]{x}^5 \times x^4 \times x^2$ D. $x^{1/3} \times x ^{1/3} \times x^{1/3}$

2. triciaal

review exponent rules see which applies solve and choice what matches

3. triciaal

|dw:1439938113254:dw|

4. anonymous

|dw:1439938334626:dw|

5. triciaal

|dw:1439938289435:dw|

6. anonymous

so what do i do first for the first one to try and simply it

7. triciaal

|dw:1439938437676:dw|

8. anonymous

would the first one be x 3/4?

9. triciaal

shift the number 6 key for the exponent symbol yes

10. anonymous

okay and for the second one would it just be 1^-1?

11. triciaal

no look again 1 is not x hint x^0 = 1

12. anonymous

so it would be 0^-1 right? or just x^-1?

13. anonymous

would B and D be matches?

14. triciaal

|dw:1439938786538:dw|

15. anonymous

ok

16. anonymous

@mathway

17. triciaal

|dw:1439938986593:dw|

18. anonymous

@AliceCullen @arindameducationusc @angel12310 @ayee_ciera @beckyg111 @BioHazard9064 @bubbleslove1234 @Butterfield1215 @BAdhi @CrazyTurtle483 @Crazyandbeautiful @clairvoyant1 @Dragonwiz88 @DenishaH please help :-)

19. anonymous

im not good at this stuff sorry!

20. anonymous

ok thanks anyways

21. anonymous

@KKLL99 @KarlaKalurky please help me

22. anonymous

what confuses you? :)

23. anonymous

how to compare them all @KarlaKalurky

24. anonymous

and how to tell which ones are equal @KarlaKalurky

25. anonymous

$1.) x ^{3/4}$ $2.) \frac{ 1 }{ x ^{-1} } = x ^{-1(-1)}= x ^{1}= x$ $3.) x ^{5/10}* x ^{4} * x ^{3} = x ^{1/2}*x ^{4}*x ^{3} = x ^{(1/2)+(4)+(3)} = x ^{15/2}$ $4.) x ^{1/3} * x ^{1/3} * x ^{1/3} = x ^{(1/3)+(1/3)+(1/3)}= x ^{3/3}= x ^{1}=x$

26. anonymous

is that how to simplify all of them? now how do I compare to see which ones are equal?

27. anonymous

so, 2 and 4 would be equivalent right?

28. anonymous

you should practice different problems about this :) this should be easy :) yes it is :)

29. anonymous

okay, thank you! I just started learning this

30. anonymous

goodluck! :D

31. anonymous

thank you!

32. anonymous

you are welcome :)

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy