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Specific steps to solve two problems in senior high school mathematics
1, the first question may be ugly, and the screenshot is much better.

Let the average value be x, then x1+x2+...+x10 =10x.

Disassemble (x1-2) 2 = x12-4x1+4 here.

(x 1-2)2+(X2 2)2+...+(x 10-2)2 = 1 10

So (x12+x22+...+x102)-4 (x1+x2+...+x10)+10× 4 =/kloc-.

(X 12+X22+...+X 102)=40X+70? The same form

Because the variance (x1-x) 2+(x2-x) 2+...+(x10-x) 2 = 2.

(X 12+X22+...+X 102)-2X(X 1+X2+...+X 10)+ 10X2=2

Substitute into the formula? 40X+70-2X2+ 10X2=2,

4X2+20X- 1=0

The solution is X=(-5)/2.

2. As shown in the figure, the vertex of the parabola is (0,0), which is the closest point to m (a a,0) on the parabola. M is on the X axis, and M must be to the left of (0,0), so A.

Personally, remember to adopt!