2. Calculation process of 2n- 1 of n square 1 square 2 * 1- 1 = 1 square 2 * 2- 1 = 3/4 square 2 * 3-1
The result is 2n- 1 of the square of n).
3. There are 5 cows, 1 pig and 94 sheep.
Suppose I bought x cows, y pigs and (100-x-y) sheep. 1000x+300y+50 (100-x-y) =10000 is simplified to 19x+5y= 100, 5y and1. Therefore, x=5, y= 1, 100-x-y=94.
4 "three angles" and "several Ho", * * * counts as nine angles. "Three exchanges" and "Jihe"?
Answer: The geometry is: 9-3=6 angles.
5.A=4, B=2, C=8, D=5, E=7 3E mantissa is 1, E=7, 3e = 3 * 7 = 2 1: 3D+2 mantissa is 7, D = 5, 3D+2 = 3. 3C+ 1 mantissa is 5, C=8, 3c+1= 3 * 8+1= 25; The mantissa of 3B+2 is 8, B=2, 3b+3 B+2+3 * 2+2 = 8；; The mantissa of 3A is 2, a = 4, 3 * 4 = 12.
So, A=4, B=2, C=8, D=5 and E=7.
6. Put two pots in the middle of each of the four sides and use eight pots. Put 1 basin and *** 10 basin at both ends of any diagonal.
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8. 13 The area of a big square is equal to the sum of the areas of four right triangles and the small square in the middle. The area of each right triangle is 3. The area of four right-angled triangles is 12, the side length of the middle small square is 3-2 = 1, and the area is 1. So the area of the big square is 3× 4+ 1 = 60.
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10.48km/h step: let the distance be x, and the total time be t x/40+x/60=x/24=t t t is the time to go back and forth twice, so the average speed is 2x/t=2x/24 x=48.
1 1.-
12.450km method 1. Let the distance between the two places be x kilometers. 5 hours, 2 cars are x kilometers away. In 4 hours, the distance traveled by two cars is X-90km, and the speed of the two cars is unchanged, so the distance traveled is proportional to time 5/4=X/(X-90)X = 450.
Method two. Let the speed of a car be x km/h, the speed of b car be y km/h, and the distance between the two places be s km. According to the proposal, (X+Y)*5 = S(X+Y)*9 = 2S-90, and if X+Y is eliminated, S=450.
Method three. Suppose the distance between the two places is S kilometers, then the speed of car A is S/9, and the speed of car B is (S-90)/95 hours. When two cars meet, (S/9+(S-90)/9) * 5 = S is solvable, and S=450.
On 13.28 days, frogs can be afraid of 1 meter every day, and can climb1= 27*3 of 9 meters in 27 days. At this time, the frog has stayed at a height of 9 meters and can climb 1 meter during the day on the 28th.
14. 15:22/2+2+2= 15 1 1:22/2*2/2= 1 1 28:22+2+2+2=28 1232 1=(222/2)^2= 1232 1
15.(6+(-3))*2*4=24 ((-3)-(-5))*2*6=24