On the number axis, an ant starts from the origin and crawls 1 unit length to the right for the first time, then 2 units to the left for the second time, 3 units to the right for the third time, and 4 units to the left for the fourth time, Question: (1) where is the last position of the ant on the number axis? (2) what is the sum of the numbers corresponding to the positions reached after each crawl? (3) how long has the ant crawled

On the number axis, an ant starts from the origin and crawls 1 unit length to the right for the first time, then 2 units to the left for the second time, 3 units to the right for the third time, and 4 units to the left for the fourth time, Question: (1) where is the last position of the ant on the number axis? (2) what is the sum of the numbers corresponding to the positions reached after each crawl? (3) how long has the ant crawled

It's OK to have the concept of the number axis. With mathematical induction, the first time 1, the second time, 1-2 = - 1, the third time - 1 + 3 = 2, the fourth time, 2-4 = - 2, the fifth time - 2 + 5 = 3 2009 / 2 = 1004 So at position 1005, 1 + (- 1) + 2 + (- 2) + +1005 = 1005 total distance = 1 + 2 + 3 + +2009=(1...

Paint the surface of a cube with 8cm edges, and then cut them into small cubes with edges of 2cm. What is the number of small cubes with no paint on any surface

(8 / 2) * (8 / 2) * (8 / 2) = 64 (total number of small cubes)
(8 / 2) * (8 / 2) = 16, 16 * 2 = 32 (4 + 4 + 2 + 2) * 2 = 24
32 + 24 = 56 (number of small cubes with painted surface)
64-56 = 8 (number of small cubes without paint surface)
If you don't understand the steps, ask me. The answer must be right

|-0.2 | - (- 3) - (+ 8) | - | - 8-2 + 10 | should have a complete process PS: first change the minus sign into a plus sign! 5.75 - (- 3 / 8) + (- 0.5) - (- 5 / 6) + 4 2 / 3-1 / 8 (42 / 3 is 4 and 2 / 3) If | a | a | = 3, | B | = 1, | C = 5, and | a + B | = a + B, | a + C | = - (a + C), calculate the value of A-B + C |1 / 2-1 | + | 1 / 3-1 / 2 | + | 1 / 4-1 / 3 +. + | 1 / 2010-1 / 2009 |. To get the number (formula)

|-PS: we must change the minus sign into a plus sign first! | - 0.2 | - (+ 8) | - (- 3) - (+ 8) | - 8-2 + 10 | = 0.2-11-0 = -10.85.75 - (- 3 / 8) + (- 0.5) - (- 5 / 6) + 4 4 4 (4): 4.4-11-0 = -10.85.75 - (- 3 / 8) + (- 0.5) - (- 5 / 6) + 4 4 4 4) + (4.5) - (- 5 / 6) + 4 4 4 + 4) + 4.5.5) - (- 5 / 6) + 4 + 4 + 4 + 4 + 4 + 4.5.5) - (- 5 / 6) + 4 + 4 + 4 + 4 + 4.5.5.6 2 / 3-1 / 8 (4 2 / 3 is 4 and 2 / 3)

Help to do a math problem in junior high school According to the demand, a circular paperboard needs to be cut into several sectors by several times. The operation is as follows: in the first cutting, the circular paperboard is divided into four sectors; in the second cutting, one of the fan-shaped surfaces obtained last time is equally divided into four sectors, and then it is carried out according to the method of the second cutting. (1) calculate the total number of sectors after the nth cutting (2) After the 2003 operation, how many fan-shaped cardboard can the original circular cardboard be cut into?

1. (3N + 1)
2. 2003 * 3 + 1 = 6010 (PCS)

(x2 + MX + 8) (x2-3x + n) after opening, what is the value of M + n? (the number next to X is the index, meaning the third power of the sum of squares of x)

The original formula = x ^ 4-3x? + NX? + MX? - 3mx? + mnx + 8x? - 24x + 8N
∵ this equation is opened without terms of X ∵ and X ᙽ 3
∴-3+m=0 8-3m+n=0
∴m=3 n=1
∴m+n=3+1=4

There are 1530t class a goods and 1150t class B goods transported by a certain company in the railway station. Now it is planned to use 50 sections of A, It is known that the freight of each type a carriage is 5000 yuan, and that of each type B carriage is 8000 yuan. 35t of class a goods and 15t of type B goods can be filled with a type a carriage, a type goods of 25t and B goods of 35t can be filled with a B type carriage. According to this requirement, a, B what kind of scheme are there for the number of cars? Please design it and explain which one has the least freight? Let's talk about the method!

If the number of A-type cargo box is x, then the number of B-type cargo box is (50-x). 35x + 25 (50-x) ≥ 153015x + 35 (50-x) ≥ 1150, the solution is: 28 ≤ x ≤ 30. ∵ x is a positive integer,

The purchase price of a product is 3000 yuan, and the price is 4500 yuan. The store requires that the profit of the product is not less than 5%, and the minimum discount is () A. 20% off B. 70% off C. 5% off D. 5% off

The discount for this product is X
From the meaning of the title: 4500 × x
10-3000=0.05×3000
X = 7
Therefore, B

It's from Beijing Normal University Press! The first volume of mathematics book, 242 pages of 12.13.14.15.16 questions (all using equations) No books, 12. There is a cube billet with edge length of 0.6m. If you want to forge it into a rectangular steel with a cross section of 0.008 m, how high is the steel forged? 13. A company sold 12200 pairs of shoes a and B last year. This year, the sales volume of type a shoes was 6% more than that of last year, and the sales volume of type B shoes decreased by 5% compared with last year. The total sales volume of two kinds of shoes increased by 50 pairs. How many pairs of shoes were sold last year? 14. There is a bamboo pole and a rope. The rope is 0.5 meters longer than the bamboo pole. After the rope is folded in half, it is 0.5 meters shorter than the bamboo pole. How many meters are the length of this bamboo pole and this rope? 15. During the Spring Festival, Xiao Ming and Xiao Ying went to see Mr. Li, who gave them sugar in a special way. Mr. Li first took Xiaoming a piece, then gave him one seventh of the sugar box, then gave Xiaoying two pieces, and then gave her the remaining one seventh of the sugar box? 16. It takes 26 seconds for the train to pass through a 256 meter long tunnel (i.e. from the front entrance to the rear exit), and the train passes through the 96 meter long tunnel in 16 seconds. Calculate the length of the train

12. 0.6×0.6×0.6÷0.008=27
13. Suppose a sold x pairs last year and B sold 12200-x pairs
X(1+6%)+(12200-X)(1-5%)=12200+50
X=6000
14. Set rope length x and bamboo pole length x-0.5
X÷2+0.5=X-0.5
X=2
15. Let x
1 + (X - 1) × 1/7 = 2 + ({X - [1 + (X - 1) × 1/7] -2 } × 1/7)
X= 36
16. Let the length be x and the speed a
26X=256+A
16X=96+A
10X=160
X=16
I hope LZ can understand how it came out

A math problem in junior high school Xiaoming and Xiaogang set out from a to B according to the same route. It is known that Xiao Ming's speed is 80 meters per minute, and Xiao Gang walks 120 meters per minute. (1) if Xiao Ming starts five minutes later, Xiao Gang starts at the same time, so they both arrive at place B. the distance between a and B remains the same. (2) the distance between the two places remains unchanged, and if they start at the same time, Xiaogang will immediately go back to pick up Xiaoming after arriving at place B, Ask them when they will meet

(1) If the time from point a to point B is t, then 80 * (T + 5) = 120 * t, then t = 10 minutes, the distance is equal to 10 * 120 = 1200 meters. (2) when Xiaogang arrives at point B, Xiaoming has already walked 800 meters, and there is still 400 meters left

Fill a 40 cm diameter and 60 cm high cylindrical bucket full of water into another cylindrical bucket with a radius of 30 cm. What is the height of the water at this time?

Let this be the height X
20*20*60=30*30*x
24000=900x
x=3/80