Two high bounty math problems, math problems with a high bounty

One: I read 1 6, 300*1 6 = 50 pages on the first day, which means I read 50 pages on the first day.

The next day, I read 2 15,300*2 15=40 pages, which means that I read 40 pages the next day.

I read a total of 40 + 50 = 90 pages in two days, and on the third day, of course, I read it from page 91.

Remember that the first and second days are both about the fraction of the book read, so be sure to multiply it with the total number of pages of the book

A quarter of a 20 meter is 20*1 4=5, then its length is 20+5=25 and then subtract 1 4, then its quarter is 25*1 4=! Then its length is.

1. I read 300*1 6+300*2 15=90 two days ago, so the third day starts from page 91;

1 On the first day, I read 300 1 6 50 pages.

The next day I read: 300 2 15 40 pages.

Pages remaining: 300 50 40 210.

So, the third day starts with page 211.

2 20 (1 1 4) (1 1 4) = 20 5 4 3 4 = 75 4 meters.

First arrival time t=s b - s a +1

When s=120, a=15, b=12, t=120 12-120 15+1=3 (hours).

The practical meaning is that AB and B are 120 kilometers apart, and A and B are 15 kilometers per hour and 12 kilometers per hour respectively, if A departs from place A 1 hour before B, then A will arrive at place B 3 hours earlier than B.

s/b-s/a+1

A arrives at B 3 hours earlier than B.

Known set a=,b=,If a falls to the right below a horizontal line b and b ≠ empty set, find the value range of the real number m, your question makes me very depressed. 1. The set a=, m+1 2m-1 obtains that m is greater than or equal to 2

m m-1 is greater than or equal to 5The answer is that m is greater than or equal to 3

The second question, set x 10,000 pieces, the profit is 20x-1 2 x 2 + 2x+20, simplify, -2/1 (x-18) 2 + 142When x=18, the maximum value is taken. The answer is 180,000 pieces.

1. From (x-5)(x+2)=<0, -2==-2 and 2m-1=<5 and m+1=<2m-1 can be obtained, and 2=2 can be obtained, that is, the maximum value of 20x-(1 2)x 2-2x-20 is obtained, which is simplified.

1 2) x 2 + 18x-20, and then by the formula -b (2a) = 18, then the production of 180,000 pieces to obtain the maximum profit.

It looks like it's in the exam room, hahaha.

3an 2 Do you mean three times the square of an?

9) (8) and (7) (6) - (1) (2) and (5) (3) values are the largest.

1) (3) and (9) (6) + (2) (4) and (8) (5) have the smallest value.

50* (1-3/10)* (1-2/7).

50*7/10 * 5/7.

25 kg.

1. The value is the largest, the subtracted integer part of the algebra fraction is the largest 9, the subtracted part is the smallest 1, the fractional part is subtracted by 7 8 (the largest of the remaining numbers), and the smallest is 2 6.

9 and 7 8-1 and 2 6

The value is minimal, ditto thinking.

9 6 7-8 4 5

Each vertex has three selections.

So there are a total of 3 7 = 2187 ways to move the worm.

If you want to go through all the vertices, there are still 3 ways to move the first point.

In order to ensure that there is no turning back.

The second point is that there are only 2 ways to move.

The third point is that there are also two ways to move.

In order to ensure that all points can be walked, there will only be one way to move each point in the future, so there are a total of 3x2x2=12 ways to walk.

Therefore, the probability of walking through all the vertices is 12 2187 = 4 729

1. Use the indirect method.

Let's start with the number of four-digit positive integers: 9 kinds of thousands, 100 places, 10 places, and 10 kinds of single digits, a total of 9000.

Four-digit positive integers do not contain 2 and 3: 7 kinds of thousands, 7 kinds of hundreds, 10 places, and single digits are all 8 methods, a total of 7*8 3=3584

Therefore, the number of four positive integers with at least one digit above 2 or 3 has 9000-3584 = 5416.

2 There are 3 choices each time, 7 times for a total of 3 7 types.

After 7 times to walk through all the vertices, there are only 12 ways.

Let the cube abcd-a1b1c1d1

For example, there are only 4 routes from A to D and finally to D.

A-A1 can choose the next step at this time, there are 2 kinds of B1, or D1 does not choose A-B at this time, you can choose the next step with 2 kinds of B1, or C does not choose the same starting from A and finally there are only 4 paths to B; Starting from A and finally going to A1 with only 4 paths, the probability of walking through all the vertices after seven walks is 12 3 7 = 4 729

[Question 1].

Exclude all numbers of four-digit integers - the number without 2 or 3 cannot be 0 in the first digit

Question 2] There are 8 vertices.

7 times to go all must.

There are 3*2*[1*1*(1+1)+1*(1*0+1)]=18, go 7 times, a total of 3 7

So the probability is 2 3 5

The reason] is this:

Remember the following principle, you can't go back and you must be able to go all) You first draw a sketch cube marked with ABCD

Below is marked A1, B1, C1, D1

1.Set up from A.

Obviously, there are 3 situations where I go the same way.

*10*10*10-7*8*8*8=5416 This question can also be calculated if you use the method of LS classmates, but it is a little troublesome, and the four positive integers are a total of 9*10*10*10 Because the first digit cannot be 0, we have to calculate it below, so there is no 2 or 3 digits in the four digits.

This requires each bit to be free of 2 or 3

So each multiplier subtracts 2 separately

9*10*10*10-7*8*8*8=54162 First of all, the total number is 3 7 because each 3 choices require 7 walks through all the vertices, and there are only 6 ways.

a-a1-d1-c1-b1-b-c-d; a-b-c-c1-a-b1-a1-d

The same goes for b and c.

So 6 3 7 = 2 729

1, assuming that the first bit is 2, the last three are a total of 10 * 10 * 10 possibilities, 10 is 0-9 fold 10 numbers, the second place is 2, a total of 9 * 10 * 10 possibilities, 9 is because 0 can not be the first place, that is, the first place can only be 1-9 these 9 numbers, the same is true, and finally sum 2,

1.Since it is a four-digit integer, the first cannot be 0, then the total possible chance is 9*10*10*10=9000

There is a probability that there is no 2 or 3 is 7*8*8*8=3584, and the probability of at least one 2 or 3 is 9000-3584=5416

2.The probability is (3*2 6) 3 7=

1. The numbers that are 2 and 3 in the four positive integers are: 7 8 8 8 = 3584 The total number of four positive integers is: 9000

A number with at least one of the four positive integers above 2 or 3 is 9000 3584 54162

The first counts the number of numbers that do not have 2 or 3; The second one is a bit troublesome, and the mobile phone code word is so tired.