Kneel and beg math masters! The algorithm of a math problem, and the result! 5

It is discussed in four situations.

1: There is only one of the 4 books drawn, so there are 6 ways to take it.

2: There are only two kinds of 4 books in total, so there are 2 situations again.

1) Two books of each kind, there are 6*5=30 ways to take them.

2) One book, one book, one book, 6 * 5 = 30 ways to take it.

3: There are only three ways to draw out 4 books, so it can only be 112 ways to take it, and there are 6 * 5 * 4 = 120 ways to take it.

4: There are four kinds of 4 books to draw out, so there are 6*5*4*3=360 ways to take them.

So there are a total of 6 + 30 + 30 + 120 + 360 = 546 ways to take it.

Solution: If each inner angle of the polygon increases by the same degree, then the corresponding number of outer angles decreases by the same degree. Let the number of sides of this polygon be x

The minimum number of external angles is: 180-140 = 40 (degrees);

The maximum number of external angles is: 180-100=80 (degrees).

According to the definition of the sum of the outer angles of the polygon, it can be seen that:

40+80)*x/2=360

Solution: x=6

A: The number of sides of this polygon is 6

It can be assumed that the damage value per time obeys an even distribution of 50-150.

The total value of 45 independent injuries is an even distribution of 2250-6750.

Then p(n>5000) = (6750-5000) (6750-2250)=

Just mathematical probability, (150-5000 45) 100 is approximately equal to ?

Depends on how you operate it, haha.

1/(1*3)=(1/2)*(1/1)-(1/3))1/(3*5)=(1/2)*(1/3)-(1/5))1/(5*7)=(1/2)*(1/5)-(1/7))1/(7*9)=(1/2)*(1/7)-(1/9))1/(9*11)=(1/2)*(1/9)-(1/11))1/(11*13)=(1/2)*(1/11)-(1/13))1/(13*15)=(1/2)*(1/13)-(1/15))1/(15*17)=(1/2)*(1/15)-(1/17)).

All the way up to the last addition, equally.

What are these things, can you use all numbers and symbols!

The purchase price is regarded as 1, and the selling price is 1 + 40% = 140%.

Pricing: 140% 20% = 700%.

Profit margin: 700%-1=600%.

If x cows can eat all the grass in 126 hectares of pasture, the original amount of grass on each hectare of pasture is A, the daily grass growth on each hectare of pasture is B, and the amount of grass eaten by each cow per day is C. It can be obtained according to the title.

12*28c=

21*63c=

Solution: b=30c a=2520c

x*126c=

Solution x=36

A: 36 cows can eat all the grass on hectares of pasture in 126 days.

21*63-36*28) (63-36)=9 This is the newly grown grass in hectares of meadow.

1008-9*28=756 This is the original grass of the hectare of meadow.

756* (head.)

First, solve the coordinates of the p(x,y) point: using Equation|po|=focal length=(a 2+b 2) 1 2, and x 2 a 2-y 2 b 2=1 to calculate the coordinates of p points represented by a and b (a total of four points meet the requirements, due to hyperbolic symmetry, you may wish to let p be the solution of the first quadrant);

Utilization|pf1|=2|pf2|List equations containing a and b as equations (1);

The straight line ab satisfying the problem should be the intersection point of the first and fourth quadrants of the quasiline through the point p, set to a(m,b a*m),b(n,-b a*n), and use a,b,p three points to list the equation (2);

Use oa*ob=-27 4 to list equation (3);

Use 2pa=-pb to list equation (4);

Four equations, a, b, m, n four parameters, can be solved.

18 BC B goes to work for A 18 BC A rests.

19bc 19bc a rest.

20bc c for a job 20bc a rest.

21ac normal 21ac b rest.

22bc a for b to work 22ac b rest.

23ab normal 23ab c rest.

24ac 24ac b rest.

25bc a for c to work 25ab c rest.

26ab 26ab c rest.

27ac 27ac b rest.

28 BC A can take a break 28 BC A rest.

On the 18th, the original AC B was on holiday, and on the 19th, the original BC A was on vacation.

On the 20th, the original AB C took a shift A and C shifts, that is, in a cycle of three people in shifts, A had to take three consecutive days off, so he had to change shifts with B and C for one day each, and then return two days of leave.

Only one day is taken in a three-day cycle, and one person's leave can be returned, and the next six days (in two cycles, that is, the 21st-26th) can be paid off for the two days of shift.

On the seventh day, I normally go to work according to the schedule, which is the AC shift.

On the eighth day (the 28th, the second day of the third cycle), you can rest in the normal order.

"A can change shifts with B or C" Then let BC work for 3 days in a row, and then A will help them each take 1 shift.

Let A and B go to work on the first day Let B and A go to work on the second day On the third day, let A and C go to work On the fourth day, let A and C go to work, On the fifth day, let A and B go to work A works for six consecutive days, B and C have one day off, After six days, A can rest for three days.

It feels like it's not very well written