The third question of mathematics, the process, how to do the third question of mathematics?

It's the distance between the two straight lines of OA and DB, and when x is the same, the top minus the bottom is equal to 1000

You already know the coordinates of the four points, and you just have to come up with the equation for two straight lines, and the absolute value of the subtraction magnitude is equal to 1000

Solve the equation and you can figure it out.

If the distance from the parabolic point to the x-axis is 4, the point of y 4 is found. There should be 2 points with a distance of 4 from the x-axis, and it can be seen from the graph that point A is to the left of point B.

Discuss:The first case: when y 4

4 -x 2+2x+3.

x＝1The second case: when y -4

4＝-x＾2+2x＋3

x＝1±2√2

x 1 2 2, x segment hood 1-2 2

It can be known that when 1 m-1 1-2 2, that is, 2 m 2-2 2

There are two points on the image g (1 2 2,-4) and (1,4) and the distance from the x-axis is 4.

From the parabolic equation, we can see that the coordinates of the parabolic vertices are (1,4). When m-1 4, the straight line y m-1 has no intersection point with the parabola.

When m-1 4 is ignited, synact y x with the parabolic equation.

Substitute the points (m-1, m-1) into the parabolic equation.

m-1＝-（m-1）＾2+2（m-1）＋3

Solution, m 1 2 (3 13).

When 1 2(3-13) m 1 2(3 13), the parabola is above y m-1 and y m-1 has 1 intersection with image g.

When m 1 2 (imaginary shed 3-13), the parabola is below y m-1 and y m-1 has 2 intersections with image g.

Figure:

x x ——— is a wall, and the other 3 sides are fences.

y1, because the fence is 30 meters long, one side is against the wall, and the other three sides are surrounded by the fence, so the other three sides add up to 30 meters, i.e.

y is the side parallel to the wall, i.e. y 2x 30, simplify y 30 2x, and then find the defined area, and because the wall is only 18 meters long, y cannot exceed 18 meters, i.e. 30 2x< 18, x> 6, and y cannot be less than 0 (nonsense), i.e. 30 2x>0, x<15, so the final answer is y 30 2x(6<=x<15).

2, y 30 2x, to y maximum, x take the smallest, because 6< = x<15, so x takes 6.

or maximum when y 18 meters, at this time 18 30 2x, x 6

3, (the area of the rectangle is long times wide, the length is y 30 2x, and the width is x).

If the area of the nursery garden is s, then s xy (30 2x) x 2x 2+30x (x 2 represents the square of x).

To make s >=88, i.e. -2x 2+30x>=88

2x^2+30x-88>=0

x^2+15-44>=0

x^2-15+44<=0

Factoring yields (x-11)(x-4)<=0

4<=x<=11

Because the domain 6<=x<15 is defined

Combined, we get 6<=x<=11

I think it's written in enough detail, and if you don't understand anything, you can continue to ask.

Solution:1 y＝30－2x (0＜x＜15)2.The length of the wall is only 18 meters.

The maximum can only be 18 meters.

i.e. 30 2x 18

x＝63. x（30－2x ）≧88

2x²－30x＋88≦0

4≦x≦11

0 is up to 18, and it has nothing to do with x.

3.Area s=xy=x(30-2x) 88

x-11）(x-4)≤0

4≤x≤11

Again 6 x 15

6≤x≤11

an=2n-1 bn= bn+1=n+ bn+1 is greater than n to prove that (3) is true.