Yes. sqrt(3)/3 = sqrt(3)/[sqrt(3)*sqrt(3)] = 1/sqrt(3)
[-1+sqrt(3)]1/4
It's about 34 and a third
2 sqrt 2 x 2 sqrt2 = 4 x 2 = 8
(5 + sqrt(5)) / 2 = 3.61803399
Yes. sqrt(3)/3 = sqrt(3)/[sqrt(3)*sqrt(3)] = 1/sqrt(3)
[-1+sqrt(3)]1/4
Infinitely many. In fact, the number of irrationals in any interval - no matter how small - is infinite. Furthermore, the cardinality of the set of irrationals in any interval is greater than the cardinality of rational numbers in total!
tan3A-sqrt3=0 tan3A=sqrt3 3A=tan^-1(sqrt3) 3A= pi/3+npi A=pi/9+npi/3 n=any integer
The arithmetic mean is simply (7+21)sqrt(3)/2 = 14sqrt(3) = 24.25 (to 2 d.p.) The geometric mean is sqrt [ 7sqrt(3) x 21sqrt(3) ] = sqrt [ 7 x 21 x 3 ] = sqrt [21 x 21] = 21
1/(5+sqrt(6)) is equal to about .134237
sqrt 3/sqrt3 = 1
It's about 34 and a third
2 sqrt 2 x 2 sqrt2 = 4 x 2 = 8
(5 + sqrt(5)) / 2 = 3.61803399
The square root of 72 is 6x(sqrt2). You can find this by breaking 72 down into its prime factors (2x6x6 or 2x62). You can then take the square root of each part which means the 62 will become 6 and then because the sqrt2 is an irrational number it is left under the sqrt sign.
The question is ambiguous because it could refer to [sqrt(A2B) + sqrt(AB2)]/sqrt(AB) = [A*sqrt(B) + B*sqrt(A)]/[sqrt(A)*sqrt(B)] = A/sqrt(A) + B/sqrt(B) = sqrt(A) + sqrt(B) or sqrt(A2B) + sqrt(AB2)/sqrt(AB) = A*sqrt(B) + B*sqrt(A)/[sqrt(A)*sqrt(B)] = A*sqrt(B) + B/sqrt(B) = A*sqrt(B) + sqrt(B) = sqrt(B)*(1 + A)