2。 If the first person is a white ball (probability is 1/2), there are three balls left, and the probability that the second person touches the white ball is 1/3. Similarly, when the first person touches the black ball, the probability that the second person touches the white ball is 2/3. The total probability is1/2 *1/3+1/2 * 2/3 =1/2.
3。 Draw a picture or calculate it layer by layer.
4。 Connecting OB, we can see that OA=OB, OP=OP, ∠ OAP = ∠ OBP = 90, and the proved △OBP is equal to △OAP, while the tangent of DE to the circle and point C indicates that OP (or OC) is the middle vertical line of △DEP. Then use ∠OAP=∠DCP=90 degrees, ∠APO=∠APO to prove that △AOP is similar to △CDP, while CP=OP-OC, DP, CD and circumference can be obtained.
5。 The length of the vertical line of an isosceles triangle with its base angle as the waist is calculated, that is, the circular tube with the smallest diameter that the isosceles triangle can pass through. It can be found that the oblique waist length of a right-angled trapezoid passing through an angle of 60 degrees is 16. If it is longer than the bottom, it is not the shortest. If the vertical line with oblique waist is taken from the lower right corner, the shortest diameter can be found to be 7 through similar triangles. The root number is 5 times that of 3, so they can all pass through a circle with a diameter of 14CM.
6。 First draw a circle o and make an o point. Take any point on the circle 1 as C, connect CO, and make ∠COE=36 degrees according to CO, then make the vertical line of 0E from point C, intersect with OE at point D, intersect with the circle at point A, and the four points of 0, C, D and A will come out.
Next △ABC may be obtuse, acute or right angle.
Let's talk about right angles, and directly deduce ∠B=36 by similar triangles's principle.
Acute angle and obtuse angle, first make the extension line of CO intersect at point F and connect AF, then you can know that △ACF is a right triangle.
If it is an acute angle, then according to similar triangles, it can be obtained that ∠AFC is ∠AFC=∠COD=36. According to the principle that two different angles of a triangle in a circle correspond to the same arc, then the two angles are the same, then ∠B = 36.
If it is an obtuse angle, we can know that the combination of △ABC and point F can form a circle inscribed quadrilateral. According to the quadrilateral theorem of inscribed circle, if the sum of relative angles is 180 degrees, then ∠B= 180-∠CFA= 144.
7. What I want is to use x to represent y, which is a classification expression. I divide X into three sections to represent Y. Y should not be the area of △ABC, but △ABP. If it's △ABC, I'll quit. No matter where point P goes, we only take the bottom AB=8 as the bottom of the triangle. It can be seen that when X=4-9, y is unchanged, but the bottom is unchanged, so we can deduce that the height is unchanged, so when X=4-9, point P moves on the edge CD, when X=4, point P is at point C, and when X=9, point P is at.
Now you can find the equation of x and y.
When X=0-4, y =1/2 * ab * BP =1/2 * 8 * x = 4x.
When X=4-9, y =1/2 * ab * BC =1/2 * 8 * 4 =16.
When X=9- 14, y =1/2 * ab * (AP * cos37) = 4cos37x.
8. If the meaning of the question is 3/(X- 1) = (1-k)/(1-x), if both sides are multiplied by x-1,then K=4 can be obtained through simplification. If k is not equal to 4, the equation has no solution and has roots. Because there is a restriction in front, x is not equal to 1 and K=4.