Get two math problems with a high score of 55 and two math problems with a high score

1 The odds of getting 5 yellow balls are: 5 10*4 10*3 10*2 10*1 10... Count yourself...

The odds of getting 3 yellow balls are 5 10*4 10*3 10*5 10*4 10... Count yourself... 2The odds of getting 6 yellow balls are 6 12*5 12*4 12*3 12*2 12*1 12...

Count yourself... The odds of getting 5 yellow balls are 6 12*5 12*4 12*3 12*2 12*6 12... Count yourself...

The odds of getting 4 yellow balls are 6 12*5 12*4 12*3 12*6 12*5 12... Count yourself... The odds of getting 3 yellow balls are 6 12*5 12*4 12*6 12*5 12*4 12

The odds of five yellow balls are 1 30240, the odds of four yellow balls are 1 6048, and the odds of 3 yellow balls are 1 1512

The odds of 6 yellow balls are 1 665280, the odds of 5 yellow balls are 1 110880, 4 yellow balls are 1 22176, and 3 yellow balls are 1 5544.

That's too annoying, isn't it? Let's draw a tree by yourself**.

Solution: If there are 6x ducks, 9x chickens and 8x geese.

Columnable equation: 9x-8x=500

Solution: x=500

So: 6 500 = 3000 ducks.

If the number of chickens is 3x and the number of ducks is 2x, the number of geese is 8x 3, and the equation 3x-8x 3=500 is listed

Solution x=1500

So raise 3,000 ducks.

Duck x chicken x 3 2 goose x 4 3

Three-thirds x minus four-thirds x equals 500

Solution x=3000

3000 pcs.

Answer: Let the duck be x and the chicken be y, then the goose will be (y-500)y x=3 2

x/(y-500)=3/4

Using the perfect square formula: (a-b) 2=a 2-2ab+b 2

So, the original = (

This is not a formula, (Formula a 2-2ab+b 2=(a-b) 2

The figure is omitted, and it is easy to prove that EOB and Coa are congruent (corner edges), i.e., OE=OC

c(2,1) easily obtains e(-1,2).

The equation is y=3x+5

(1) Because (ob -3)+|oa-1|=0, so there is ob=3, oa=1, because a, b are on the x-axis y-axis positive semi-axis, so there are a(1,0),b(0, 3).

2) It can be found that BC=2 3, AB=2, and AC=1+3=4, we can get that ΔABC is a right triangle, ABC=90 degrees.

Point P proceeds from point C and moves along the ray Cb at a velocity of 1 unit per second, from which it can be obtained: cp=t, and t [0,2 3].

s = sδabp = pb*ab 2 = (bc-pc)*2 2 = 2 3-t, where t [0,2 3].

3) If there is a point P that makes ΔABP similar to ΔAAB, then from PB=90 degrees, it can be concluded that PB and AB are the two right-angled edges of ΔABP, and their proportions should meet the ratio of the two right-angled edges in ΔAOB, and since OA, OB are the two right-angled edges of ΔAOB, they are not equal to each other, OB 0A= 3 1 = 3, so the ratio of the two right-angled edges Pb and AB in ΔPAB should also be equal to 3, but it is impossible to determine which of them is longer and which is shorter, and it needs to be classified and discussed.

If PB is longer than AB, then there is PB AB= 3, then PB= 3*2=2 3, T=PC=BC-PB=2 3-2 3=0, it can be seen that in this case, point P coincides with point C, and the coordinates of P are (-3,0).

If AB is longer than PB, then AB PB= 3, PB= 3*2 3=2 3 3, T = 2 3-2 3 3=4 3 3, which satisfies the value range of T, so this point also exists.

The equation of the straight line through the two points b(0, 3) and c(-3,0) can be found as y= 3x 3+ 3, and p is located on this, and yp=t 2=2 3 3 can be obtained from the geometric relation, and xp=-1 can be obtained by substituting the linear equation

So the p coordinates are (-1, 2, 3, 3).

Question 1a,m=[0,1],n=[0,1).

The second question D, just substitute the calculation.

Question B, it should all be vectors. BC*(AC+BA+AB+CB) = BC*BA=0 BC Vertical BA

Question 1a,m=[0,1],n=[0,1).

The second question d,z 2=cos2+2icos*sin-sin2

Question 3 B, BC*(AC+BA+AB+CB) = BC*BA=0 BC Vertical BA