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There are infinitely many of them. In fact, there are more Irrational Numbers between sqrt(2) and sqrt(3) than there are rational numbers in total.

Q: How many irrational numbers are there between sqrt2 and sqrt3?

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tan3A-sqrt3=0 tan3A=sqrt3 3A=tan^-1(sqrt3) 3A= pi/3+npi A=pi/9+npi/3 n=any integer

the formula for the area of equilateral triangle with side length a is (a^2)(sqrt(3)/4 ) So draw an equilateral triangle with sides a,a,a. Now divide it into two triangles by bisecting the top angle and extending that line down so it makes a 90 angle with the base. (any angle will do since it is symmetric, but I am trying to help you draw it) Now the new half triangle you have is a 30 60 90 triangle since the top angle is half the original which was 60 and the lower left angle is still 60. The 90 degree angle is the third angle and we are drawing it this way so we have a that the side between the 30 and 90 degree angles is the height of the triangle and we can use 1/bh which is the area formula. The new base has length a and the height is sqrt3(a) because it is a 30-60-90 triangle so 1/2 bxh= 1/2(1/2 )a (sqrt3)a=(1/4)a^2( sqrt3 )since 1/2 base = a/4 and height is sqrt3 x a

Half the length of one side multiplied by (square root of 3)/(2). in triangle ABC with height H: (A•sqrt3)/2 or A (sqrt3)/2 This is because of the Pythagorean theorem. You draw a line from the top vertex of the triangle vertically to the bottom side. This line is perpendicular to the bottom side and it will bisect that side. Now you want to know the length of your new line. On each side of it, you have a smaller triangle, one side with the length of the side of the original triangle (let's call it s) and one side with length half that, (1/2)s. Since the side with length s will be the hypotenuse of the triangle, we know s2 = (s/2)2 + h2 by the Pythagorean theorem. (h stands for height.) s2 = s2/4 + h2 s2 (1-1/4) =h2 s2(3/4) =h2 (sqrt3)s/2 = h

Related questions

1, 2+sqrt3, sqrt2+sqrt6

In general, no. It is possible though. (2pi)/pi is rational. pi2/pi is irrational. The ratio of two rationals numbers is always rational and the ratio of a rational and an irrational is always irrational.

Two different irrationals can't make a rational...

[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)/6This can be simplified to[3(sqrt(2)+2(sqrt(3)]/6

-(sqrt3)/2

tan3A-sqrt3=0 tan3A=sqrt3 3A=tan^-1(sqrt3) 3A= pi/3+npi A=pi/9+npi/3 n=any integer

Yes. sqrt(3)/3 = sqrt(3)/[sqrt(3)*sqrt(3)] = 1/sqrt(3)

4 + 2sqrt3

The geometric mean of any two numbers is the square root of their product, in this case sqrt 12 or sqrt4 x sqrt3 ie 2 root 3 or 3.4641

Do you mean which 2 integers the square root of 27 falls between? If so, then the square root of 27 is 3*sqrt3, or about 5.2. So between 5 and 6.

12

The square root of 147 is simplest as 7*sqrt3.