2007-01-31 09:16:04 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
What is the biggest square that divides 48? 16 is
sqrt(48)=sqrt(16*3)
=sqrt(16)*sqrt(3)
=4*sqrt(3)
2007-01-31 09:19:52 · answer #1 · answered by Professor Maddie 4 · 0⤊ 0⤋
you got the right answer.
sqrt 48 = sqrt 16 * sqrt 3
and sqrt 16 can be reduced to 4, but three cannot.
so its 4sqrt3
2007-01-31 09:22:10 · answer #2 · answered by ♥♥♥♥♥ 5 · 0⤊ 0⤋
Do you advise sqrt(40 8*3)*sqrt(27)? Sqrt(27)=sqrt(3^2*3)=sqrt(3^2)*sqrt(3)... Sqrt(40 8*3)=type(3)*sqrt(40 8)=sqrt(3)*sq... word: sqrt(3)*sqrt(3)=3 Altogether, sqrt(40 8*3)*sqrt(27)= 3*2*sqrt(3)*sqrt(3)*sqrt(6) = 2*3*3*sqrt(6)=18*sqrt(6)
2016-11-23 18:21:26 · answer #3 · answered by ? 4 · 0⤊ 0⤋
prime factor 48
so you get
48 = 2*24
24=2*12
12=2*6
6=2*3
so that's 48=2*2*2*2*3
Then, since the index of a sqrt is 2 (and the index of cube root is 3) we take out pairs of numbers from inside.
circling helps, like
, where 3 is the odd one out.
each pair needs only one of it's numbers to represent it out side the sqrt.
So sqrt[(2*2)(2*2)*3] gets the pairs outside with only one of each: 2*2sqrt(3) = 4sqrt(3)
Yeah, if it was 96, and had 5 2's inside, we'd leave that odd two inside and get 4 sqrt(6)
2007-01-31 10:05:49 · answer #4 · answered by Anonymous · 0⤊ 0⤋
sqr(48)= sqr(16*3)=sqr(16)*sqr(3)=4*sqr(3)=4sqr3
so yes 4sqrt(3) is correct
2007-01-31 09:22:15 · answer #5 · answered by Bill F 6 · 0⤊ 0⤋