If x = 111980 + 11981 +11997, then the integral part of X is______ ．

If x = 111980 + 11981 +11997, then the integral part of X is______ ．

181997＜11980+11981+… +11997 < 181980, so 198018 < 111980 + 11981 + +11997 ＜ 199718, that is 110 ＜ 111980 + 11981 + & nbsp &So the integral part of X is 110.95

Given the coordinates of three points of quadratic function ABC, spherical analytic formula
A (0,1) B (- 1,3) C (2,3) requires detailed process and uses two formulas for calculation,

Because the parabola passes B (- 1,3), C (2,3),
Therefore, let y = a (x + 1) (X-2) + 3,
Substituting x = 0, y = 1, we can get 1 = - 2A + 3,
The solution is a = 1,
So the analytic expression of the function is y = (x + 1) (X-2) + 3 = x ^ 2-x + 1

What is the quotient of 3.24 divided by 0.6 over 5.7 times 1.1?

Results = 3.24 / (0.6 * 5.7 * 1.1)
=0.86124402

Given the set a = {x 4 ≤ x ＜ 7} B = {x 2 ＜ x ＜ 9} find Cr (AUB), Cr (a ∩ b), (CRA) ∩ B

∵A∩B={x|4≤x＜7}
A∪B={x|2

Simple calculation of 202 × 29-58

202×29－58
＝101×58－58
＝﹙101－1﹚×58
＝100×58
＝5800.

If x ^ 2-4x + y ^ 2 + 6y + 13 = 0, then x = y=

x^2-4x+y^2+6y+13=(x-2)^2+(y+3)^2-13+13=0
(x-2)^2+(y+3)^2=0
x=2,y=-3

Cos (Wu - 1) - sin (1 - Wu / 2) = what

The original formula = - cos1 - (- cos1)
=0

1: Write a program to find the root of the equation 2x ^ 3-4x ^ 2 + 3x-6 = 0 between [- 10,10] by dichotomy. The ^ sign represents the power, such as x ^ 3 represents the cube of X, so that

#include
#include
void main()
{double f (double x); / * declare functions about X*/
void Erfen (double a, double B, double h, double EP); / * declare the subroutine of dichotomy*/
erfen (- 10,10.0,0.100,0.00001); / * calls the subroutine of dichotomy*/
}
Double f (double x) / * defines a function about X*/
{double y;
Y = 2 * x * x * x-4 * x * x + 3 * X-6; / * assign the function value to y*/
Return y;} / * returns the function value*/
Void Erfen (double a, double B, double h, double EP) / * subroutine defining dichotomy*/
{double x [10], A0, B0, C0, A1, B1; / * store the root in x [] and define the endpoint of the interval as floating-point variable*/
int k,i,j;
k = 0; / * k is the control variable of the cycle*/
a0 = a; / * assign a to the starting point of the first interval*/
while(a0

0.027 ＋ 9.9 × 0.27 should be calculated simply by recursion equation

0.027＋9.9×0.27
=0.27x0.1+0.27x9.9
=0.27x(0.1+9.9)
=0.27x10
=2.7

Finding the limit limx tends to 0 x ^ 2 · e ^ (1 / x ^ 2)

Let u = 1 / x ^ 2, then
Original formula = LIM (U → + ∞) (e ^ u) / u
=lim(u→+∞)(e^u)
=+∞
The lobida rule is applied here