Question

Need help with a algebra problem. Can someone assist?

Answer 1

Yes, I can

Answer 2

What do you need help with?

Answer 3

o.k.

Answer 4

just post your questions

Answer 5

\[\sqrt{}56x^5y^6\]
\[\sqrt{2y^4}\]

Answer 6

Everyone the second problem goes underneath the first problem

Answer 7

\[\frac{\sqrt{56 x^5 y^6}}{\sqrt{2y^4}}\]

Answer 8

yes

Answer 9

ok now what I did is

Answer 10

so far I have 28x^5y^2 This is where I got confused

Answer 11

\[\frac{\sqrt{7*4*2* x^4x y^6}}{\sqrt{2y^4}}\]
Make sense?

Answer 12

o.k. I am following you so far

Answer 13

4 is a perfect square so we may pull out 2

Answer 14

\[\frac{2\sqrt{7*2 x^4x *y^6}}{\sqrt{2y^4}}\]

Answer 15

o.k.

Answer 16

imra nice job

Answer 17

In addition we may pull out x^2 for x^4 and y^3 from y^6
\[\frac{2x^2y^3\sqrt{7*2 x }}{\sqrt{2y^4}}\]

Answer 18

same thing for denominator
\[\frac{2x^2y^3\sqrt{7*2 x }}{y^2\sqrt{2}}\]

Answer 19

o.k. I wijth you

Answer 20

Denominator-numerable cancel for y^3 and y^2

Answer 21

\[\frac{2x^2y\sqrt{7*2 x }}{\sqrt{2}}\]

Answer 22

ok I with you

Answer 23

let's pull out x from radical and saprate the radical\[\frac{2x^2x^{\frac{1}{2}}y\sqrt{7 }\sqrt{2}}{\sqrt{2}}\]

Answer 24

o.k.

Answer 25

add x's exponent and cancel Sqrt[2]
2x^{\frac{5}{2}}y\sqrt{7 }

Answer 26

\[2x^{\frac{5}{2}}y\sqrt{7 }\]

Answer 27

ok

Answer 28

is that the answer you came up with?

Answer 29

Hold on let me check to double check everything

Answer 30

What I have on the answer key is 2x^4y^2\[\sqrt{7xy}\]

Answer 31

my answer is :\[2x^2y \sqrt{7x}\]

Answer 32

\[\sqrt{\frac{56x^5y^6}{2y^4}}=\sqrt{28x^5y^6}=2x^2y \sqrt{7x}\]

Answer 33

His seems to be 2/7 x^3 y sqrt(7xy)

Answer 34

\[\frac{\sqrt{56 x^5 y^6}}{\sqrt{2 y^4}}\]
\[\frac{(56)^{\frac{1}{2}}\left(x^5\right)^{\frac{1}{2}}\left(y^6\right)^{\frac{1}{2}}}{2^{\frac{1}{2}}\left(y^4\right)^{\frac{1}{2}}}\]

\[2(7)^{\frac{1}{2}} x ^{\frac{5}{2}}y\]
\[2 \sqrt{7} x^{5/2} y\]

Answer 35

Nancy this is correct but did you miss a step?

Answer 36

Nancy your answer is correct but how did you get rid of the x. Please write the hold problem out for me so i can see it more clearer.

Answer 37

square 2 , you have xxxxx,than
xx xx x=
x x than 1 x keep in radical ?

Answer 38

xx xx = x^2 bring out side radical
x not engout 2 than keep inside radical

Answer 39

can you write the entire problem out for me?

Answer 40

square 2, you have x^5
xxxxx u ok from here ?

Answer 41

no I am a visual person tht needs to see the problem out at it's entire

Answer 42

each two x = one x
xx = x
xx= x total 4, if 4 =2
\[x^2\sqrt{x}\]the x out side = twice time x inside radical
out side 2 + inside 2=
4 + 1 = 5 (x^5)

Answer 43

\[\sqrt{125}=\sqrt{5*5*5}= 5\sqrt{5}\]

Answer 44

two 5 bring out = one 5

Answer 45

ok

Answer 46

\[\sqrt{3125}=\sqrt{5^5}\]\[\sqrt{5^2*5^2*5}=5*5\sqrt{5}\]\[25\sqrt{5}\]

Answer 47

ok

Answer 48

square is mean 2, bring out each pair, if odd one left keep inside

Answer 49

I am following you

Answer 50

r u understand now?

Answer 51

Yes I got that. I just need too see who you came up with the answer by working the problem out for me. That was a good explaination. If you can assist me step by step on that problem then I will figure it out quicker.

Answer 52

u want go to the problem u post?

Answer 53

56/2=28
\[\frac{y^6}{y^4}=y^{6-4}=y^2\]

Answer 54

o.k. so far so good

Answer 55

u have another problem?

Answer 56

no there are more steps to that proble. I need the hold problem work out

Answer 57

r u ok now?

Answer 58

28= 2*2*7
2*2= 2 ( bring 2 out side)
7 not a pair than keep inside radical

Answer 59

good bye

Answer 60

Thanks sorry I could get you to write out the entire problem, but it was good to see what parts of the problem were identified.