For painting training, Xiao Qing goes every six days, and Xiao Hong goes every eight days. After they train at the same time on July 10, at least how many months will they meet again? Party A and Party B go to borrow books. Party A goes every six days and Party B goes every eight days. If they meet in the library on March 5, what month and day will they go to the library next time?

For painting training, Xiao Qing goes every six days, and Xiao Hong goes every eight days. After they train at the same time on July 10, at least how many months will they meet again? Party A and Party B go to borrow books. Party A goes every six days and Party B goes every eight days. If they meet in the library on March 5, what month and day will they go to the library next time?

I calculated it myself
Xiaoqing and Xiaohong meet again on September 11
Xiaoqing is every six days, that is, once a week. Xiaohong is every eight days, that is, once every nine days. The common multiple of 7 and 9 is 63. 63 days after July 10 is September 11
Party A and Party B meet again on March 29
Once every 6 days in a, once every 8 days in B. the common divisor of 6 and 8 is 24. The 24 days after March 5 is March 29

Xiaohua goes to the library once every four days, Xiaohong once every five days, and Xiaolan once every six days. The three of them met in the library on March 10
When did they meet again in the library?

Because their least common multiple is 60, it takes 60 days for them to meet again
So the correct answer should be June 9

1 × 2 of 2005 + 2 × 3 of 2005 + 3 × 4 of 2005 + 4 × 5 of 2005 +. + 2004 × 2005 of 2005 = ()
Fill it in

1 × 2:2005 + 2 × 3:2005 + 3 × 4:2005 + 4 × 5:2005 +. + 2004 × 2005:2005
= 2005 × (1 × 2 / 1 + 2 × 3 / 1 + 3 × 4 / 1 + 4 × 5 / 1 + + 2004 × 2005 (1 / 2)
= 2005 × [(1-2 / 1) + (1-3 / 2) + (1-4 / 3) + (1-5 / 4) + + (1 / 2004-1 / 2005)]
= 2005 × [1-2 / 1 + 2 / 1-3 / 1 + 3 / 1-4 / 1 + 4 / 1-5 / 1 + + 1 / 2004-1 / 2005]
= 2005 × [1-1 / 2005]
= 2005 × 2004 / 2005
＝2004

The distance between the two vehicles is 260 km. A vehicle travels 80 km per hour, B vehicle 70 km per hour, B vehicle starts 0.5 hours later
Car a meets car B a few hours after departure. (car B starts 0.5 hours first)

260-70 * 0.5 = 225 km 225 divided by (80 + 70) = 1.5 hours
Then Party A and Party B can meet in another 1.5 hours

X / a = Y / b = Z / C = 2 for (2x-3y + Z) / (2a-3b + C)

Let X / a = Y / b = Z / C = 2m, so x = 2mA, y = 2MB, z = 2mc, substituting, the result is 2

When car a is 54 kilometers above the midpoint, car B is 36 kilometers away from the midpoint. It is known that the speed ratio of car a and car B is 3:2
How many kilometers did the two of them travel?

The distance ratio between a and B is (3:2)
Whole journey: (54 + 36) △ 3-2) × 2 × 2 + 36 × 2 = 432km
Or,
Whole journey (54 + 36) △ 3-2 × (3 + 2) - (54-36) = 432km

There is a decimal. If the decimal point is moved one place to the right, and then it is added to the original decimal, the sum is 21.12; if it is subtracted from the original decimal, the difference is 17.28
What was the original decimal?

Set the original decimal place X
X+10X=21.12
The solution is x = 1.92
Listed as
Because, the decimal point moves 1 place to the right, the original decimal expands 10 times
So 21.12 △ 11 = 1.92
So the original decimal place is 1.92

A and B cars drive from two cities at the same time. After 3 hours, the two cars meet at the midpoint of 18 km. The distance between a and B is 2:3?
Do not use the ordinary algorithm to solve the equation

Distance difference between two cars: 18 + 18 = 36 km
2+3＝5
Distance between the two cities: 36 (3 / 5-2 / 5) = 180 km
Speed of car a: 180 × 2 / 5 △ 3 = 24 km
Speed of vehicle B: 180 × 3 / 5 △ 3 = 36 km

If the root of the equation is three? Sin (2x + one-third) + sinx-t = 0, there is always a real number solution to find the value range of the real number t

Wrong. It's - sin2x
√3sin2xcosπ/3+√3cos2xsinπ/3-sin2x=t
√3/2*sin2x+1/2*cos2x=t
sin2xcosπ/6+cos2xsinπ/6=t
sin(2x+π/6)=t
-1

When a car goes 360 kilometers, it takes 160 kilometers in the first two hours?

It will take (360-160) / (160 △ 2) = 2.5 hours to arrive