Factorial: 2011! How many zeros at the end

Factorial: 2011! How many zeros at the end

[2011/5]+[2011/5^2]+[2011/5^3]+[2011/5^4]+[2011/5^5]+……
=402+80+16+3+0+……
=501
So there are 501 zeros
Where [x] is the largest integer not exceeding X
Why is the factorial of 0 1?
0 means negative, 1 means positive, negative means positive
Let the quadratic function f (x) = - 4x & # 178; + 2 (A-1) x + B be a decreasing function in the interval (1, + ∞)
It is easy to know that the opening of quadratic function is downward, and the axis of symmetry x = (A-1) / 4
When the axis of symmetry is on the left side of the interval, the interval is a decreasing function
Then (A-1) / 4 ≤ 1
So a ≤ 5
Given the function f (x) = sin2x-2sinx (1), find the minimum positive period of function FX. (2) find the maximum, minimum and maximum of function
When f (x) = sin2x-2sin & # 178; X = sin2x + (1-2sin & # 178; x) - 1 = sin2x + cos2x-1 = √ 2Sin (2x + π / 4) - 1. (1) the minimum positive period T = 2 π / 2 = π. (2) sin (2x + π / 4) ∈ [- 1,1], and when sin (2x + π / 4) = 1, the maximum value is: F (x) | max = √ 2-1; sin (2x + π / 4) = - 1
If the quadratic function f (x) = X2 - (A-1) x + 5 is an increasing function in the interval (12,1), the value range of F (2) is obtained
The quadratic function f (x) is an increasing function in the interval (12,1). Because the opening of its image (parabola) is upward, its axis of symmetry x = a − 12 or coincides with the straight line x = 12 or lies on the left side of the straight line x = 12, so a − 12 ≤ 12, the solution is a ≤ 2, so f (2) ≥ - 2 × 2 + 11 = 7, that is, f (2) ≥ 7
Given the function f (x) = (√ 3cosx SiNx) sin2x / 2cosx + 1 / 2. (1) find the value of F (π / 3); (2) find the minimum positive period and monotonicity of F (x)
Decreasing function
Cos α sin β = [sin (α + β) - sin (α - β)] / 2 F (x) = (√ 3cosx SiNx) sin2x / 2cosx + 1 / 2 = (√ 3cosx SiNx) SiNx + 1 / 2 = 2cos (x + π / 4) SiNx + 1 / 2 = sin (x + π / 4 + x) - sin (x + π / 4-x) + 1 / 2 = sin (2x + π / 4) - (√ 2-1) / 2 you will
If the function f (x) = x2 + (a2-4a + 1) x + 2 is a decreasing function in the interval (- ∞, 1], then the value range of a is______ .
Since the equation of the axis of symmetry of the function f (x) = x2 + (a2-4a + 1) x + 2 is x = - A2 − 4A + 12, and the function is a decreasing function in the interval (- ∞, 1], there is - A2 − 4A + 12 ≥ 1, and 1 ≤ a ≤ 3 is obtained, so the answer is: [1, 3],
Given vector a = (sin2x minus 1, cosx), vector b = (1,2cosx), Let f (x) = vector a times vector B, find the minimum positive period of function f (x) and X belong to [...]
Given vector a = (sin2x minus 1, cosx), vector b = (1,2cosx), Let f (x) = vector a times vector B, find the minimum positive period of function f (x) and the maximum value of X when it belongs to [0, Pie / 2]
A = (sin2x-1, cosx), B = (1,2cosx) f (x) = A.B = (sin2x-1, cosx) (1,2cosx) = sin2x-1 + 2cos ^ 2x = √ 2Sin (2x + 0.25 π) minimum positive period T = 2 π / ω = 2 π / 2 = π x ∈ [0, π / 2] 2x + 0.25 π ∈ [0.25 π, 1.25 π] f (x) max = √ 2Sin (2x + 0.25 π) | x = π / 8 = √ 2
Given that f (x) = (3a-1) x + 4a (x = 1) is a decreasing function on R, the range of a is obtained
x=1 y=1 X+Y=3
(1 + sin2x) / (cos2x) = - 3, find TaNx
A: (1 + sin2x) / cos2x = - 3 (Sin & # 178; X + 2sinxcosx + cos & # 178; x) / (COS & # 178; x-sin & # 178; x) = - 3, so: (SiNx + cosx) &# 178; / [(cosx SiNx) (cosx + SiNx)] = - 3, so: (SiNx + cosx) / (cosx SiNx) = - 3, the left molecular denominator of (SiNx + 1) /