Please detailed steps and answers, step answers are detailed

1.It can be known that the coordinates of the center of the circle are o(-1,2), and the radius is 2y (x-4), which means that the slope of the line from the point on the circle to the point of e(4,0), then you can know that the connection between any point and e on the circle falls between the two tangents of be and de, then the smallest slope is the tangent of de, and then the slope of de is found: let the equation of de be = k ( x - 4), that is, kx - 4k - y = 0, then the distance from o to de is equal to the radius, that is, (-k - 4k - 2) (k 2 +1) = 2, and k = 0 or -20 21, i.e., its minimum value is -20 21.

2.The maximum value from the fixed point (1,0) to the point on the circle is the maximum when the line passes through the origin o, that is, AC in the figure (in the same way, AC is the smallest when the diameter is removed), AC = R + OC, OC = [1 - 1) 2 + 2 - 0) 2] = 2 2, so AC = 2 + 2 2.

Hope my answer is helpful to you

The first question is that if y (x-4)=k, then y = k(x-4), and k is the slope of the straight line, which passes the fixed point (4,0), and has a maximum and minimum value if and only if the line is tangent to the circle.

The linear equation is connected to the circular equation and subtracts y

x+1）^2+[k(x-4)-2]^2=4

That is, (k 2 + 1) * x 2 + (2-4k-8 * k 2) * x + 16 * k 2 + 16k + 1 = 0

Then k=0 or k=-20 21 is solved by the tangent condition =(2-4k-8*k 2) 2-4*(k 2+1)*(16*k 2+16k+1)=0

Therefore, k = -20 21, i.e. the minimum value of y (x-4) is -20 21

The second question is that the point that intersects the farthest point between the straight line and the circle (-1,0) and the center of the circle (-1,2) is the required point.

The distance is the distance from the fixed point (1,0) to the center of the circle (-1,2) 2 2 plus 2 (radius).

d=2+2√2

I didn't understand the first question.

For the second question, you can first draw a circle with coordinates (-1,2).

The maximum value is the distance from the point (1,0) to the center of the circle plus a radius.

The center of the circle is (-1,2) and the radius is 2, so the x range is (-3,1) and the y range is (0,4). In the first problem, y (x-4) can be seen as a negative value of the tangent of the line connecting the point on the circle with the point (4,0) (because it is x-4, not 4-x), and it is a negative number. When the absolute value of this negative value (i.e., the tangent value) is the maximum value in the problem, then the point on the circle and the line of (4,0) should be tangent to the circle, and the relationship between the side length of the right triangle can be obtained (x-4) +y-0) 2+4=(-1-4) 2+(2-0) 2, after sorting out and standing with the circle equation, x,y is obtained as (11 29,100 29) and (-1,0), remove the latter answer, so (11 29,100 29) point is the point sought, and the final result y ( x-4)=-20 21。

As for the second question, the maximum value is the distance from the point to the center of the circle plus a radius, resulting in 2+2 2.

1, 3, 5, 7, 9, 11, 13, 15 all the numbers appear, this is the correct solution.

There is no solution to this problem according to general mathematics: because the sum of 3 odd numbers is still an odd number.

If you change your thinking, you will have a solution, that is, you can change the decimal system to the decimal system for calculations:

1 all, 1 +2 +3 +....2002³=（1+2+3+..2002）²=[n（n+1）/2]²=4020037030009

4020037030009 7=574291004287 is just divisible So it's Sunday.

Fold 1 in half once

Fold in half twice 3

Fold in half three times 7

Fold in half four times 15

The rule is 2 n+1

Fold in half ten times 2 10-1=1023

2＜8＜3

So the two decimal places of x are closer than 8. ]

x is kept to three decimal places.

Let the width of the iron sheet be xcm, then the length of the bottom of the box is 24cm, and the width is (x-8)cm, and the volume column equation: 4x(x-8)=787, and then find x.