simplest math journal ever
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16 May 2013, 19:33
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Journals
So you uber geniouses here at CF will laugh at this, but I'm learning myself the basics of math for a future study. I've got this calculation I can't seem to crack, so help would be appreciated.
I'm trying to do the following calculation (the answer is also there). I have no idea how to come to the answer.
Now I need to know how to get from the initial calculation to the 6root6. It's pretty important for me to solve this, thanks in advance!
I'm trying to do the following calculation (the answer is also there). I have no idea how to come to the answer.
Now I need to know how to get from the initial calculation to the 6root6. It's pretty important for me to solve this, thanks in advance!
so it becomes (2^3 * 3^(3/2))/(2^(3/2))
which is 14.69
anyways.
Start by writing out the power to what is actually there, so to [yourthing]*[yourthing]*[yourthing] then start working away the divisions and simplify the resulting root..
(all)².(all)=(all)³
(all)²=(4*3)/2 = 6
so
6.(all) = 12.((sqrt(3).sqrt(2))).sqrt(2) / sqrt(2) = 6.(sqrt(3).sqrt(2))
final - apply exponent law( multiply base give same exponent- exponent = 1/2 for sqrt() )
6.sqrt(6)
its like cf basic etiquete
:PPPPPP
about maths, he said he needed explanation step by step
and no, i did not think of that, its clever that way
(rad(2)*rad(3))^3
(rad(2*3))^3
(rad(6))^3
rad(6)*rad(6)*rad(6)
= 6*rad(6)
=[(2^1)*(3^1/2)*(2^-1/2)]^6
=[(2^1/2)*(3^1/2)]^6 ,because a^n*a*m= a^n+m so 2^1*2^-1/2=2^1/2
=(6^1/2)^3 , because a^m*b^m= (ab)^m
= 6*6^1/2
or just
a= (2*sqrt3)/sqrt2 = (sqrt4*sqrt3)/sqrt2 = sqrt12/sqrt2 = sqrt 6
a^3 = sqrt6*sqrt6*sqrt6= 6*sqrt6
and ((sqrt12)/(sqrt2))^3 is indeed 6 sqrt6
would take a good ten years for me to crack this one :D