Simple grade 12 equation

[Sixth_Rule]

i need a quick awnser to this

(y^7/4)(1/4y)^1/2(1/8(y^10/8)

anyone remeber grade 12 algebra??

[jthegreat]

Write it in a way that's clear and maybe someone could help. Your use of parentheses is very unclear.

[eOXevious]

he probably just forgot the last parenthese, but I wish I could help, I forgot how to do that stuff

(y^7/4) * ((1/4y)^1/2) * (1/8(y^10/8))

1/4*1/8* y^7/4 *y^1/2 * y ^10/8

1/32 * y^(28/8)

[ZeeJustin]

[ QUOTE ]

(y^7/4) * ((1/4y)^1/2) * (1/8(y^10/8))

1/4*1/8* y^7/4 *y^1/2 * y ^10/8

1/32 * y^(28/8)

[/ QUOTE ]

Reduuuuuce!

You can't factor out the 1/4 because it is also ^1/2. When you factor out the 1/4 it becomes 1/2 when taken outside of the ^1/2.

Answer is:
1/16*((y)^(7/2))

[BruceZ]

[ QUOTE ]
i need a quick awnser to this

(y^7/4)(1/4y)^1/2(1/8(y^10/8)

anyone remeber grade 12 algebra??

[/ QUOTE ]

You probably don't mean what you wrote, but if we add a ')' to the end, and take 1/4y = 1/(4y), then this has a clear meaning by the heirarchy of operations (^,/,*,+,-) which is:

(y^7)/4 * 1/(4y)/2 * (1/8)*(y^10)/8

= (y^16)/2048.


But if you mean

y^(7/4) * [1/(4y)]^(1/2) * (1/8)*y^(10/8)

Then that equals (1/16)*y^(5/2)


Or if the 1/4y means 1/4 * y, then this is

y^(7/4) * (y/4)^(1/2) * (1/8)*y^(10/8)

and that equals (1/16)*y^(7/2).


In any case, just add the powers of y, and multiply the divisors (after taking a power if necessary).

[mmmmmbrother]

i bet this is for a free bigmac or somthing