Best Answer

[The sqrt(3)+sqrt(2)]/sqrt(6)

can be solved by multiply the numerator and denominator by sqrt(6)

This gives sqrt(18)+sqrt(12)/6

This can be simplified to

[3(sqrt(2)+2(sqrt(3)]/6

Q: What is sqrt3 plus sqrt2 divided by sqrt 6?

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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

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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 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)

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.