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Passenger cars and freight cars (mathematical problems) The passenger cars and freight cars leave from a and B at the same time. The passenger cars arrive in three hours and the freight cars arrive in four hours. One hour later, the distance between the two cars is 30km. What's the distance between a and B?
30÷（1-1/3-1/4)=72
The distance between the two places is 72 kilometers
1-1/3-1/4=5/12
30/（5/12）=72
a. B the distance between the two places is 72km
One hour later, the bus drove 1 / 3, the truck drove 1 / 4, their difference was 1-3-1 / 4 = 5 / 12
So 5 / 12 of the total length is 30km, and the total length is 72km
Seventy-two
The canteen used 6 tons of coal for two days. 2.4 tons of coal was used in the first day, which is equivalent to 23 tons of coal used in the second day. Has the coal in the canteen been used up?
2.4 △ 23 + 2.4 = 3.6 + 2.4, = 6 (tons). A: six tons of coal were used up in two days
There is a column of numbers, 1, 3, 4, 7, 11, 18, 29
What is the remainder of the 2009 number in this column divided by 6?
The answer is as follows
The remainder of this column is divided by 6: 1,3,4,1,5,0,5,4,3,1,4,5,3,2,5,1,0,1,1,2,3,5,2; 1,3,4,1,5
2009 / 24 = 83 + 17
The 17th number in the above remainder sequence is 1
The answer is 1
N≥3
Yn=(Yn-1)+(Yn-2)
=2(Yn-2)+(Yn-3)
=3(Yn-3)+(Yn-4)
=4(Yn-4)+(Yn-5)
=2005(Yn-2005)+(Yn-2006)
When n = 2009, the value is 2005 * 7 + 4 = 14039
In this paper, N, n-1, n-2, n-3, ~ ~ ~ ~ n-2005, n-2006 are subscripts, a numerical value and a bracket. Is that ok?
The passenger car and the freight car leave from station a and B at the same time, and meet 5 hours later. After meeting, the two cars still move at the original speed. When they are 196 km apart, the freight car runs 80% of the whole journey. The ratio of the distance that the passenger car has traveled to the distance that it has not traveled is 3:2
196 (80% + 33 + 2-1), = 196 (45 + 35-1), = 196 (25), = 196 × 52, = 98 × 5, = 490 (km); a: the distance between a and B is 490 km
The bottom areas of the two cylinders shown in Figure 1 and Figure 2 are equal. Can you find the volume of Figure 2?
Figure 1: side area 16dm and 179; height 4.5dm
Figure 2: 6dm
The side area of Figure 1 is 162 DM and 179;
16/4.5=32/9dm,
2pir=32/9
r=0.566dm
s=pir^2=1dm^2
v=sh=3.6dm^3
3*3-3.14*3*3/4=1.935(4-3+4)*(4+3)/2-(4-3)*3=14.53*3+4*4=2525-1.935-14.5=8.565
3 * 3 / 2-3 * (3 / 2) / 2 = 2.25
There is a column of numbers 6,12,18 ~ (1). If the sum of three adjacent numbers is 306, which three numbers are they
This sequence is: 6,12,18,24,30,36,42,48,54. Let 6x be the smallest, followed by 6 (x + 1), and 6 (x + 2) be the largest. 6x + 6 (x + 1) + 6 (x + 2) = 3066x + 6x + 6 + 6x + 12-306 = 018x + 18-306 = 018x = 288x = 16. Therefore, the first number: 6 × 16 = 96, the second number: 6 × (16 + 1) = 102
Mathematical problems of a bus and a truck
A passenger car and a freight car leave from ab at the same time. After 6 hours of meeting, the two cars advance at the same speed. After 4 hours, the passenger car arrives at B. at this time, the freight car is 188 kilometers away from A. how many kilometers is the distance between AB and a?
According to the topic, it took 10 hours for the bus to arrive, so it took 1 / 10 of the whole journey every hour
It takes a total of X hours to set up a truck, 1 / x per hour
（1/10+1/x）*6=1
3/5+6/x=1
6/x=2/5
x=15
188÷[(1/10-1/15)*10]
=188÷1/3
=564 km
Suppose the speed of freight car is a, then the speed of passenger car is (6 / 4) a = 1.5A, and the distance between two places is X
There are equations
（a+1.5a）*6=x
(4+6)a+188=x
A = 188 / 5, x = 564
There are some books. If you give them to girls, you can get 15 books. If you give them to boys, you can get 10 books. If you give them to the whole class, how many books can you get?
There are several tons of grain in the two warehouses of Party A and Party B. from Party A to Party B, Party A moves another quarter to Party A. what is the percentage that Party A has?
1. The total number of books is Z, the number of boys is x, and the number of girls is y
So z = 10 * x, z = 15 * y
So x = Z / 10, y = Z / 15, and Z ^ (x + y) = 6
Each student in the class will get 6 copies
2. The second question seems wrong. Check again
1÷（1÷15＋1÷10）=6
Sixty
1 1÷（1÷15＋1÷10）=6
17 / 215
1÷（1÷15＋1÷10）=6
The second question is incomplete.
The two sides of the equal sign are not equal. Change some two positions on the left side of the equal sign to make the equation equal. 2 * 0.07 * 5 + 0.5 * 0.09 + 0.18 + 0.3 = 1
The original title: 2 * 0.07 * 5 + 0.5 * 0.09 + 0.18 + 0.3 = 12. Now it is changed to: 2 * 0.07 * 0.5 + 5 * 0.09 + 0.18 + 0.3 = 1
The bus and the truck leave from the two places at the same time. They meet in 9 hours. When they meet, the bus just takes 60% of the whole journey. The bus is 15 kilometers faster than the truck every hour. How many kilometers is the distance between the two places?
15 × 9 ^ [60% - (1-60%)] = 135 ^ 20% = 675 (km) a: the distance between the two places is 675 km
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