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Bookmark or Share- How do you calculate 6!? - Socratic6! = 6*5*4*3*2*1 = 720 For any natural number n, n! = n(n-1)(n-2)...3*2*1 Then, for n=6 we have 6! = 6*5*4*3*2*1 = 720
- How do you evaluate #sin(pi/6) - Socraticsin(pi/6) = 1/2 Start with an equilateral triangle of side 2. The interior angle at each vertex must be pi/3 since 6 such angles make up a complete 2pi circle. Then bisect the triangle through a vertex and the middle of the opposite side, dividing it into two right angled triangles. These will have sides of length 2, 1 and sqrt(2^2-1^2) = sqrt(3). The interior angles of each right angled ...
- How do you factor x^3+6x^2+11x+6? - Socraticx^3+6x^2+11x+6=color(red)((x+1)(x+2)(x+3)) There are several ways to approach this. One of the most reliable is to hope that the expression has rational roots and apply the Rational Root Theorem. In this case, the Rational Root Theorem tells us that (if the expression has rational roots) those roots are integer factors of 6 (the constant term of the expression). We can build a table ...
- How do you find the value of cos (pi)/6? - Socraticsqrt3/2 There are 2 ways, that don't need calculator a. Trig table of special arc --> cos (pi/6) = sqrt3/2 b. Use triangle trigonometry Consider a right triangle ABH that is half of an equilateral triangle ABC Angle A = pi/6 = 30^@, Angle B = 60^@, Angle H = 90^@ Leg AH = sqrt3; Leg BH = 1, Hypotenuse AB = 2. We have cos A = cos (pi/6) = (AH)/(AB) = sqrt3/2 === Edit from r3ce: I believe ...
- Planck's constant - Chemistry - SocraticPlanck's constant is denoted h and is 6.26 x 10(^ -34) J x s where J is Joules and s is seconds. It can be used to relate the energy of a photon (E) and the frequency of its related electromagnetic wave (v) through the relationship E=hv.
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