There is an iterative method for finding the square root X1 of a positive integer. The iterative formula for finding the square root is: x0 = A / 2; X1 = (x0 + A / x0) / 2; #include "math.h" main() { float a; double x0,x1; scanf("%f",&a); if (a1e-5); } printf("%lf",x1); getch(); } How the program is executed; In the loop, x0 = X1; X1 = (x0 + A / x0) / 2; isn't that x0 = X1? How does the do loop?

There is an iterative method for finding the square root X1 of a positive integer. The iterative formula for finding the square root is: x0 = A / 2; X1 = (x0 + A / x0) / 2; #include "math.h" main() { float a; double x0,x1; scanf("%f",&a); if (a1e-5); } printf("%lf",x1); getch(); } How the program is executed; In the loop, x0 = X1; X1 = (x0 + A / x0) / 2; isn't that x0 = X1? How does the do loop?

X0 = X1 is to use x0 to store the value of X1 in the n-1st iteration, which is used to calculate the difference between the n-th iteration and the N-1 iteration. If the difference satisfies Fabs (x0-x1) > 1e-5, the iteration ends. That is, to get the approximate value of X1!

Solution equation: x + 5x + 10 (50-2x) = 240

x+5x+10(50-2x)=240
x+5x+500-20x=240
x+5x-20x=240-500
-14x=-260
x=260/14
x=130/7

It is known that the image of parabola y = - x2 + BX + C passes through points a (m, 0) and B (0, n), where m and N are the two real roots of equation x2-6x + 5 = 0, and m < n. (1) find the analytical expression of parabola; (2) let the other intersection of parabola and X-axis in (1) be C, and the vertex of parabola be D, find the coordinates of points c and D and the area of △ BCD; (3) P is a point on line OC, and pass through point P as pH ⊥ x-axis If the straight line BC divides △ PCH into two parts with equal area, the coordinates of point P can be obtained

(1) By solving the equation x2-6x + 5 = 0, we can get X1 = 5, X2 = 1, from m < n, we can know M = 1, n = 5, a (1, 0), B (0, 5), and ≠ 1 + B + C = 0C = 5, i.e. B = − 4C = 5; the analytical formula of the parabola is y = - x2-4x + 5. (2) from - x2-4x + 5 = 0, we can get X1 = - 5, X2 = 1, so the coordinate of C is (- 5, 0)

Add an integral to the polynomial x2 + 4 to make it a complete square. The integral that can not satisfy the above conditions is ()
A. 4xB. -4xC. 4D. -4

A. X2 + 4 plus 4x can get x2 + 4x + 4, which can be changed into (x + 2) 2; it is a complete square form, so this option is wrong; B, X2 + 4-4x can get x2-4x + 4, which can be changed into (X-2) 2; it is a complete square form, so this option is wrong; C, X2 + 4 plus 4 can get x2 + 8, which is not a complete square form, so this option is correct; D, X2 + 4-4 can get X2, which is a complete square form, so this option is wrong

Let the odd function y = f (x) be a decreasing function in the domain R, and the inequality f (KX ^ 2 + 2K) + F (2x-1) of X less than or equal to 0 hold, and the range of positive number k is obtained

f(kx^2+2k)+f(2x-1)≤0
f(kx^2+2k)≤-f(2x-1)=f(1-2x)
∵ y = f (x) is a decreasing function over the domain R
∴kx^2+2k≥1-2x
It is proved that ｛ KX & ｛ 178; + 2x + 2k-1 ≥ 0
And Δ = 4-4k (2k-1) ≤ 0
(k-1)(2k+1)≥0
K ≥ 1 or K ≤ - 1 / 2
∴k≥1

What changes the statement for next in VB affect the number of loops

For i = a To b
In I, a and B, only changing I can affect the number of cycles
Next

There is a "220v5a" electric energy meter, which can be connected to "220V & nbsp; 40W" electric lamp ()
A. 24 B. 27 C. 29 D. 30

P total = UI = 220V × 5A = 1100W; the number of electric lamps is n = P total P amount = 1100w40w = 27.5 according to the meaning of the title, the number of electric lamps should be 27

Zero point theorem

Let f (x) be continuous on a closed interval [a, b], and f (a) and f (b) have different signs (i.e. f (a) × f (b))

The relationship between the current on the cable and the length of the cable? Please specify,

If it is too long, the resistance will increase, if the resistance is large and the current is large, the voltage drop will also increase, and whether the voltage to the load can meet the requirements is also a condition for selecting the wire diameter. Of course, the size of the current must be taken into account!

When x = x, X and y are equal in the solution of the quadratic equation 5x + 6y = 22
Find x

Then you take the condition of x = y into account~~
5x + 6x = 22, x = 2
So when x = 2, X is equal to y