Question

http://prntscr.com/4bxn1r

Answer 1

@Hero

Answer 2

Two hints:

\[\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}\]

\(\sqrt{24} = \sqrt{(4)(6)}\)

Answer 3

B?

Answer 4

sorry i kinda took long, I was looking at your questions :)

Answer 5

Sounds like a guess. You should try showing your work

Answer 6

and i didn't get notified again :P

Answer 7

O: i didn't guess lol

Answer 8

umm

Answer 9

did i get it right?

Answer 10

@dannyrod2000, if you show the work you did to attempt to solve this, then I'll be able to help you better.

Answer 11

xD ok np

Answer 12

ok let me see

Answer 13

A? because it will be\[6*x^2/x*24 then~\it~will~be~ 6x^2/24x \]

Answer 14

\[6x^2/24x=x/4?\]

Answer 15

i hope i kinda doing it right :P

Answer 16

wait I wrong.

Answer 17

What happened to the square roots?

Answer 18

I don't know how to do that :P

Answer 19

where do i go to do square root?

Answer 20

The way I started out is by separating the numbers from the variables:

\(\dfrac{\sqrt{6}}{\sqrt{24}} \dot\ \dfrac{\sqrt{x^2}}{\sqrt{x}}\)

Answer 21

how do you do square root?

Answer 22

and how do you do a fraction. I never knew how :p

Answer 23

Since \(\sqrt{24} = \sqrt{(6)(4)}\) we have:

\(\dfrac{\sqrt{6}}{\sqrt{(4)(6)}} \dot\ \dfrac{\sqrt{x^2}}{\sqrt{x}}\)

And since \(\sqrt{ab} = \sqrt{a}\sqrt{b}\) we have:
\(\dfrac{\sqrt{6}}{\sqrt{4}\sqrt{6}} \dot\ \dfrac{\sqrt{x^2}}{\sqrt{x}}\)

Answer 24

Notice we end up with \(\sqrt{6}\) on top and bottom of the numerical fraction, so the \(\sqrt{6}\)'s cancel out leaving just:
\(\dfrac{1}{\sqrt{4}} \dot\ \dfrac{\sqrt{x^2}}{\sqrt{x}}\)

Answer 25

Furthermore, \(\sqrt{4} = 2\). I'm sure you already knew that which simplies the fraction to:
\(\dfrac{1}{2} \dot\ \dfrac{\sqrt{x^2}}{\sqrt{x}}\)

Answer 26

xD I got it right?

Answer 27

Now to deal with the variables

Answer 28

ohh nvm

Answer 29

so Its D

Answer 30

or C lol

Answer 31

would it have a square root?

Answer 32

@dannyrod2000, you've guessed just about every letter. Just let this play out.

Answer 33

sorry :P

Answer 34

Notice that we can immediately re-write the variable fraction as

\(\dfrac{1}{2} \dot\ \sqrt{\dfrac{x^2}{x}}\)

Answer 35

so do you keep the square root on?

Answer 36

B,C, and D are the same except for the square roots

Answer 37

And \(\dfrac{x^2}{x}\) simplifies to just \(x\) so we end up having:
\(\dfrac{1}{2} \dot\ \sqrt{x}\)

Answer 38

At this point, it should be clear which choice is correct.

Answer 39

B

Answer 40

Notice, the square root is only applied to x

Answer 41

i mean yea

Answer 42

You could probably use a personal tutor.

Answer 43

?

Answer 44

These concepts may be a bit confusing for you.

Answer 45

?

Answer 46

1/2*x square root would be B

Answer 47

Square root is only applied to x after simplification

Answer 48

\(\dfrac{1}{2} \dot\ \sqrt{x} = \dfrac{\sqrt{x}}{2}\)

Answer 49

:p

Answer 50

oh

Answer 51

but aren't you multiplying both 1 and 2?

Answer 52

with x?

Answer 53

NO, you're multiplying fractions:

\(\dfrac{1}{2} \dot\ \dfrac{\sqrt{x}}{1} = \dfrac{\sqrt{x}}{2}\)

Answer 54

oh ok you should of did that before because that makes a lot more sense to me :)

Answer 55

There are some things that go without saying. \(a = \dfrac{a}{1}\) That's a basic algebraic identity that you should already know.

Answer 56

do you mind helping me with another question? @Hero

Answer 57

and i do know :P

Answer 58

It's getting kinda late.

Answer 59

what time is it?