3, 4, 5, 6, 7, 8, 9 and 10 are equal to 18 in a square grid

3, 4, 5, 6, 7, 8, 9 and 10 are equal to 18 in a square grid

There should be one less number 2, right? If there is 2, it is arranged as follows:
3,10,5
7,2 ,9
8,6,4
3,10,5
7, ,9
8, 6, 4
How to find the period of odd function f (1 + x) = f (1-x)!
Because f (x) is an odd function whose domain is r, there is f (x) = - f (- x).
Because the image is symmetric with respect to the line x = 1, f (x) = f (2-x),
So f (2-x) = - f (- x),
By replacing - x with X, we can get f (2 + x) = - f (x),
Replace x with 2 + X, so f (4 + x) = - f (2 + x) = f (x), so f (x) is a function with period 4
I don't understand the last step
- f (2 + x) = f (x) why
We can get f (x + 2) = - X,
Replace x with 2 + X, so f (4 + x) = - f (2 + x) = f (x), so f (x) is a function with period 4
The last step of this is not to get f (2 + x) = - f (x), then - f (2 + x) = f (x)
Welcome to ask!
F (x) = - f (- x) from odd function
Let x = Y-1 give f (y) = f (2-y) = - f (Y-2)
Similarly, f (Y-2) = - f [(Y-2) - 2]
That is, f (y) = f (y-4)
The cycle is 4
As shown in the figure, ∠ AOB is an angle placed in the square grid, then the value of cos ∠ AOB is______ .
Suppose that the side length of a square is 12, then Ao = 5, Bo = 5, ab = 2. In △ AOB, according to the cosine theorem, AB2 = AO2 + bo2-2ao · Bo · cos ∠ AOB, ∧ 2 = 5 + 5-2 × 5 × cos ∠ AOB, ∧ cos ∠ AOB = 0.8, so the answer is 0.8
If f (x) is an odd function and f (x + 1) = - f (x), how to find the minimum positive period?
f(x＋1)=－f(x)
f(x＋2)=－f(x＋1)=f(x)
Then the minimum positive period is 2
f(x＋1)=－f(x)
f(x＋2)=－f(x＋1)=f(x)
Then the minimum positive period is 2
F (x) = - f (x + 1) = - [- f (x + 2)] = f (x + 2), so the minimum positive period is two. Generally, in this form, the minimum positive period can be obtained by the formula. I can summarize it
As shown in the figure, in the 3 × 3 square grid, the ∠ 1 and ∠ 2 are marked=______ .
According to Pythagorean theorem, AC = BC = 5, ab = 10. ∵ (5) 2 + (5) 2 = (10) 2, ∵ ACB = 90 degree, ∵ cab = 45 degree. ∵ ad ∥ CF, ad = CF, ∵ quadrilateral ADFC is parallelogram, ∵ AC ∥ DF,