Question

help pleaze

Answer 1

@GoldPhenoix

Answer 2

I don't know how to do this, sorry.

Answer 3

ok

Answer 4

@Jhannybean

Answer 5

Woo! If there was a counter for how many times this question has shown up xD Ok.

Answer 6

We've got \[\large \sqrt[3]{625x^5}\] We can split this up and evaluate this function like so.\[\large \sqrt[3]{625}\cdot \sqrt[3]{x^5}\]

Answer 7

ok

Answer 8

Now,there's one of two ways of solving this,factoring, and powers. I'll show you both methods so you can use whichever one makes you feel more comfortable.

Answer 9

a cubic root, compared to a square root, means for a pair of 3 like terms, 1 of those terms comes out of the cube root.

Answer 10

ok

Answer 11

Therefore, We expand out function out first to look for pairs of 3 like terms. \[\large \left( {\sqrt[3]{\color{blue}{5 \cdot 5\cdot 5 \cdot 5}}}\right) \cdot \left( \sqrt[3]{\color{red}{x \cdot x \cdot x \cdot x}}\right) \]

Answer 12

Now looking at these two expansions, you pull out 1 term for every pair of 3 like terms. Pulling out 3 terms from 5 leaves you with 1 inside = \(\large 5\sqrt[3]{5}\)

and pulling out 3 terms from the other leaves you with \(\large x\sqrt[3]{x}\)

Answer 13

You see how that works so far?

Answer 14

Sorry i missed an x in the expansion.

Answer 15

yes

Answer 16

It should have said \[\large \sqrt[3]{x \cdot x \cdot x \cdot x \cdot x}= x\sqrt[3]{x^2} \]

Answer 17

So we now have \(\large 5\sqrt[3]{5}\) and \(\large x\sqrt[3]{x^2}\) Right? with me so far?

Answer 18

yes im folloowing

Answer 19

alright. These two are being multiplied. So you'll have \[\large 5\sqrt[3]{5} \cdot x\sqrt[3]{x^2}\] When you have two things multiplying, you can multiply the coefficients separately,and the stuff inside the square root separately. Doing so,we have \[\large (5 \cdot x) \cdot \left( \sqrt[3]{5}\cdot \sqrt[3]{x^2}\right) \] We can simplify this to \[\large 5x \cdot \left( \sqrt[3]{5 \cdot x^2}\right) \] You can finish this up :)

Answer 20

ummmmmmmm i think its \[\sqrt[3]{25x^2}\]

Answer 21

@johnweldon1993 please help

Answer 22

No.. you're going to keep the \(\large 5x\) and \(\large \sqrt[3]{5x^2}\) Separate.

Answer 23

oh so what do i do

Answer 24

Take a look at your answer choices again,it might make more sense :)

Answer 25

If we're multiplying both \(\large 5x\) and \(\large\sqrt[3]{5x^2}\) together but keeping them separate,which answer choices can we eliminate?

Answer 26

i see thanks

Answer 27

Awesome:) Do you kind of understand how it works now?

Answer 28

yes thanks can u help me witha few more please

Answer 29

Use the same method of expanding this function like we id above, but this time since it's a square root,we're pulling out 1 term for every pair of 2 terms.

Answer 30

ok so i do\[64\sqrt{64n^6w^4}\]

Answer 31

is that right????

Answer 32

No, breakdown 64 first. what is it? hint:refer to how we broke down 625.

Answer 33

8

Answer 34

so its \[8\sqrt{8n^6w^4}\]

Answer 35

8 can be broken down further. This method of breaking down a large number is called prime factorization. Were factoring out the large number into its LCM.

64 = \(2^6\) or 2 x 2 x 2 x 2 x 2 x 2

Answer 36

4

Answer 37

or 2?????

Answer 38

Use the same method of expanding this function like we id above, but this time since it's a square root,we're pulling out 1 term for every pair of 2 terms.

Answer 39

so it \[4\sqrt{4n^3w^2}?????\]

Answer 40

So ifyou ahve 6 2's.and we make pairs of twos ,how many 2's will we have? 6/2 =?

Answer 41

3

Answer 42

Good. we're pulling out 3 2's.

(2 x 2) x (2 x 2) x (2 x 2) = 2 x 2 x 2 = 6.

Answer 43

since it's a square root,were pulling out 1 term for every 2 like-terms.

Answer 44

ok so it \[2\sqrt{2n^2w^3}\]

Answer 45

We can solve this question using an easier method too.
\[\large \sqrt{64n^6w^4} = \sqrt{64} \cdot \sqrt{n^6} \cdot \sqrt{w^4} = 8 \cdot n^{6/2} \cdot w^{4/2} \] Just simplify this and you will have your answer.Imade it a fraction because another way of writing a square root is \(\large (\text{term})^{1/2} \)

Answer 46

Did that make things esier?

Answer 47

easier*

Answer 48

yes

Answer 49

um i belive it 4w^3n^2

Answer 50

Where did you get 4 though?

Answer 51

\(\large\sqrt{64}\) = ?

Answer 52

8???????????

Answer 53

Yes :D

Answer 54

That's why I was wondering how you got 4 in your answer....

Answer 55

oh i thought i was suppose to simplify 8 lol myb

Answer 56

Nah, we already simplified it as much as as we could by breaking down \(\sqrt{64}\)

Answer 57

I've got to go though, good luck with your other problems!

Answer 58

ok bye

Answer 59

Bye bye :)