Good math helps!!! Thank you

There are two cases in this problem, first, ab is on the same side as cd, doing of of perpendicular to cd and intersecting with ab with the point e, with ob and od

Because the diameter of 50 and because of the chord ab cd, from the chord cut theorem, ae is equal to be, df is equal to cf, and the square of 25 minus the square of 24 is equal to 49

The root number 49 is equal to 7, so oe is 7 25 squared minus 20 squared to get 225 225 square root is 15

So the distance between AB and CD is 8 with an OF of 15

When AB and CD are opposite, the distance between AB and CD is 7 plus 15 equals 22

For chord AB, d1 = root number ((50 2) -40 2) ) = root number (225) = 15;

For chord cd, d2 = root number ((50 2) -48 2) ) = root number (49) = 7;

Considering the position of AB and CD, there are:

When they are on the same side of the parallel diameter, d=d1-d2=8cm, and when they are on different sides of the parallel diameter, d=d1+d2=22cm.

The distance from circle o to chord ab is h1 = 25 20 = 15 the distance from circle o to chord is h2 = 25 24 = 7 When ab and cd are on the same side of the center of the circle o, the distance between the two strings is h1 h2 = 8 When ab and cd are not on the same side of the center of the circle o, the distance between the two strings is h1 h2 = 22

The distance between AB and CD is 56I remember that I had heard this question from the teacher before. Additional questions on the test paper, forgetting ...... in the process

I'll draw it on paper, and I'll pass it on later.

This is a matter of probability with a put-back. (It should be the same size of the balls) (1) is to touch two white balls in box A, and one white ball in box B, then the probability of p= c32 c52 * (c21 c32) = 1 5 permutation and combination.

2. There are two cases when two white balls are drawn: two white balls from box A and two boxes, one white ball is touched, and three white balls are drawn.

then p=c32 c52*(c22 c32)+c31c21 c52*(c11c21 c32)+1 5=7 10

3)x=0,1,2

When x=0, the square of p0=c2o*(1-7 10) = x=1 p1=c21*7 10*(1-7 10)x=2 p2=c.

to draw a distribution column.

e(x)=0*p0+1*p1+2*p2

x-a > 0, so, x>a, i.e., a0, so, x<3 2 So, the first integer that a is smaller than 3 2 is 1, followed by 0, so, the 6 integer solutions are 1, 0, -1, -2, -3, -4

Substitute the last -4, a<-4, so -5 a<-4 We assume that a=-4, that is, -4a=, that is, a=-5, that is, -5, the key is to substitute hypothesis=a, and that's it.