Question 9 and Question 10, the solution process

The first count is 72, and the 3 times are all 6

So that's 20, and that's the one who starts with 4.

Number 5 on the left: Yes, no.

9 questions 610 5.

The approach is similar. 9 Questions:

That is, count (20 7 = 2) times.

On the third occasion, there were more people than numbers.

So the count stops at 6, which is the sixth. 10 questions.

Similar to question 9.

A total of (25 5 = 5) times was counted, and the fifth time was counted from left to right, so it stood in 5th place.

The exact number can be said to be so.

The first time is from left to right.

The second time is from right to left.

It can be concluded that the odd number of times is left.

Even times is right.

There is a remainder to indicate that the count stopped halfway through, and most of them should be at once.

The previous time is the quotient, and most of the several are remainders.

Of course, there is another algorithm.

The number of people remains the same, the way of counting is 1 7, the leftmost one is not counted, and then the rightmost one is not counted, that is, 7 + 6 + 5 + 4 + 3 + 2 + 1 can be counted at most: < 28 so this number can only be made, 7 + 6 + 5 + 4 is 22 so it is counted 4 times to 2 to the end, so it is when there are three people in the middle, that is, the person in the middle, is the fourth person. It's a hassle to explain.

25, that 5+4+3+2+1=15 counts less than 25, so it's not counted that way.

Or the third algorithm.

1 7 folds back 8 13 and then 14 19 so that it is sixth.

Here's the algorithm.

13 divided by 6 = 2....1

Fold back twice with 6.

The odd number of times there is no team tail, and the even number of times there is no team leader (or left and right) counts twice, that is, this time from the end of the team, then the count to 20 is exactly 7-1 6th place.

The one of 25.

20 divided by 4 = 5

Standing in fourth place is fine.

The three algorithms of 20 and the two algorithms of 25 are given to you, you can see if it works.

Point oblique type, first find the slope, and then bring the point to Okai.

When I was in school, I just couldn't understand this kind of question.

As Liang Xintu demonstrated.

You might get it if you go to the homework gang.

Find the oblique asymptote equation for the curve y=(x-1) [3(x+1)].

Solution: This is the oblique asymptote equation.

I'm sorry math teacher, it's been 6 years since I graduated, no.

<> see the diagram of the remaining beams of the row excavation and rolling gear.

If the rank is 2, then the determinant is zero, and the determinant is found