Just want to thank anyone ahead of time for any help!

32)   So, I assume we have  3*3√5y^3* 2*3√50y^4  ???

9√5y^3 * 6√50y^4

9√5y^3 * 6√(25*2)y^4

9√5y^3 * 6*5√(2)y^4

9√5y^3 * 30√(2)y^4

270(y^7)√10

26) 4√64x^3y^6

$$4\sqrt{64}x^3y^6\\\\
=4*8x^3y^6\\\\
=32x^3y^6\\\\$$

How about having a go at the next one Nataszaa and I will help if you get stuck.  

Thanks Melody! I am awful at this...i have spent weeks with my tutor on it but i will try.

for #28 i wasnt sure, but i went off of what you had posted.

so i get the sqrt of 6 and 16? which are 2&4...what next?

28) 3√6* 3√16

So we have

 3√6 * 3*4  =    note that we are just multiplying now, Nataszaa

3*3*4*√6 =

36√6    ...we can't do anything with the square root of 6 !!! 

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30) √8x^5*√3x   =  √(2*2*2)x^5*√(3)x = √(4*2)x^5*√(3)x = 2√(2)x^5*√(3)x =

2* x^5 * x * √(2) * √(3)  =  (2x^6)√(6)

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3√7x/3√5y^2 = (the 3's "cancel) = √7x/√5y^2= (Multiply the numerator and denominator by √5) =  (√35x)/(5y^2)

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Nataszaa, repost 32 again.......are there some parenthesis involved???

inbetween the lines under number 30, is that a continuation?

Cphill

THANKS

32 was weird...posting wise...lol

its a normal 3 then a sqrt sign with another 3 in the opening start of the sign...i cant copy and paste :((

I did   5)  after 30....that's the one between the lines....I look at 32 in a sec !!!

Sorry, CPhill

I messaged you about this equation. Turns ou that that the second three is cubed root.

so 3 cubedroot...

32) \(3\sqrt[3]{5y^3}\times2\sqrt[3]{50y^4}\)

= \(3y \sqrt[3]{5}\times 2\sqrt[3]{5\times5\times2\times y^3\times y}\)

= \(6y\sqrt[3]{5}\times y \sqrt{50y}\)

= \(6y^2 \sqrt[3]{250y}\)

= \(6y^2 \sqrt[3]{5^3\times 2y}\)

= \(6y^2\times 5 \sqrt[3]{2y}\)

= \(30y^2\sqrt[3]{2y}\)