Math problems, trouble to give a process, math problems? Please write about the process

Actually, I think so.

If you draw 5 cards of the same suit. Then you can draw one of the same as him.

So now you can draw up to five cards of up to four suits. That's when you draw another one. It doesn't matter what suit you're drawing. There will always be a suit that becomes 6 cards. So you now need at least 21 cards to guarantee that at least 6 cards are of the same suit.

But. Don't forget. There are also two cards of big and small rings. So that adds up to 23 sheets.

Let's say you're unlucky today and draw four cards of different suits. Then when the fifth card is drawn, it must be the same suit as one of the first four cards, so at least 5 cards must be drawn to ensure that at least two cards are of the same suit.

If it is at least 6 of the same, then when 5 cards of each suit are drawn (20), then 1 more card can be drawn, plus the big and small king, it is 23 cards.

The key word for this question is: "at least", if there is no first at least, then draw all the cards. The second one, at least, limits the possibility of not considering 7 or 8 cards.

Another key is "guarantee", that is, when the luck is the worst, if there is no ***, then draw 6 cards, if you are lucky, all of the same suit.

This kind of problem involves the "drawer principle" in mathematics.

5 cards of each suit are drawn, 2 kings are drawn, and then there are 6 cards of the same suit no matter what is drawn. Luck 6 cards are the same, but you can't guarantee it, the title says to be able to guarantee ah, that is, you will be able to find 6 cards of the same suit.

a<0, b<0, c>0, then.

A Squared 》0,|a+b|becomes -a+(-b).

c-a|Change to C-A

b-c|Change to c-b

So a square|a+b|+|c-a|+|b-c|= the power of a - the square of a b + c - a + c-b

y=ln(x+a)

y'=1 Source Zen band (x+a).

y'=1, which is the same as the slope of the straight line.

1 (x+a)=1, x+a=1, x=1-a, y=ln(x+a)=ln(1-a+a)=0.

Cut hail reed point: (1-a,0).

Substituting a straight-line equation.

The tangent attack is also in a straight line.

0=1-a+1, get a=2

(1) In the triangle A'ed, because the sum of the inner angles of the triangle is 180°, there are: angle A' + angle A'ed + angle A'DE=180°; In the quadrilateral BCDE, because the sum of the internal angles of the quadrilateral is 360°, there are: angle b + angle c + angle a'ed ed + angle a'de + angle 1 + angle 2 = 360°; And because:

Angle b + angle c = 180° - angle a; Angle A = Angle A', so: 180° - Angle A' + Angle A'ed + Angle A'de + Angle 1 + Angle 2 = 360°. Compared with the first equation, the angle a=(angle 1 + angle 2) 2 (2) (180-b-c) 2+b=90-efd; c-b=2efd; The same can be proven.

When v is equal to 0, that is, when t = 3 seconds to solve the equation, the car stops.

The velocity function is then integrated between 0 and 3 of t.

I don't know if you have learned integrals, this problem first calculates that the velocity is 0 after 3 seconds, and then integrates v(t), from 0 to 3, you can get the answer.

Solution: Let B x yuan.

then A. x+5 yuan.

300-10 (x+5) = 300-10x-50 = 250-10x, these are used to buy B.

then bought. 250-10x）/x

pcs = 250 x-10

If you buy all of them.

It is possible to buy. 300/(x+5）

Buy it now. 10+250/x-10

The equation is. 250/x

300/(x+5)

250 (x + 5) - 5x (x + 5) = 300x250x + 1250-5x squared - 25x = 300x5x squared - 75x + 1250 = 0

x squared - 15x + 250 = 0

x squared - 15x + 250 = 0

x+25）（x-10）=0

x=10 another dissolution goes.

A: $10 for B and $15 for A.