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Let's try it this way:

You have 12 children to fill the first group of 3 people -- this is a combinations problem of choosing a group of 3 from a group of 12:  ₁₂C₃  =  220.

We now have only 9 children left to fill the second group of 5 persons -- this is a combinations problem of choosing a group of 5 from a group of 9:  ₉C₅  =  126.

There are 4 children left for the group of 4 -- this can be done in only 1 way.

Combining these answers:  220 x 126 x 1  =  27,720.

4n√(8m⁵n³)  =  4n√(4m⁴n²)√(2mn)  =  4n(2m²n)√(2mn)  =  8m²n²√(2mn)

2m√(18m³n⁵)  =  2m√(9m²n⁴)√(2mn)  =  2m(3mn²)√(2mn)  =  6m²n²√(2mn)

Combining:  8m²n²√(2mn) - 6m²n²√(2mn)  =  2m²n²√(2mn)

X = 1

Y = 2/6

also, as always, youre welcome

correction: kinji fukasaku. and no, ceaser would not chop my head off, cuz i am Ceaser. besides, he is the director for the best movie EVAR!: Battle Royale (2000) and Battle Royale II-Requiem (2003)

thats too confusing.put it in simpler wordsl, pls

2r² - 28r + 38  =  0

Divide all the terms by 2:

r² - 14r + 19  =  0

Since this can't be factored easily, if you use the quadratic formula:

a = 1     b = -14     c = 19

r  =  [-b ± √(b² - 4·a·c)]/(2·a)   --->   r  =  [- -14 ± √( (-14)² - 4·1·19)]/(2·1)

--->   r  =  [14 ± √( 196 - 76)]/(2)    --->   r  =  [14 ± √(120)]/(2)   --->   r  =  [14 ± √4√(30)]/(2)

--->   r  =  [14 ± 2√(30)]/(2)   --->   r  =  7 ± √(30)