Math problems, masters, high scores!

Two points at the same time at a velocity of 1cm s, ae=cf, ab=cd and ab parallel cd, eab= fcd, so abe cdfdf=be, aeb= cfd, df parallel be, bedf is a parallelogram row.

2), (16-10) (1+1)=3(s)2, (1)ad bc,de ab,abed is a parallelogram row, ab=de=dc

2), CE=BC-BE=BC-AD=2, CDE is isosceles, CF=2 2=1

The square of df = the square of (root number 5) - the square of 1 = 4df = 2

First, the marketer fell asleep before the train traveled 1 to 2

Then when she woke up, she found that the remaining distance was (1 2) * (1 3) = 1 6 while she was asleep, and the train had run 1 2 - 1 6 = 1 3, that is, the remaining 1 2 before falling asleep removed the remaining 1 6

Overall, the train went 1 2 + 1 3 = 5 6

The remaining distance is 1 2 1 3 1 6

Traveled the whole course of 1 1 6 5 6

"Halfway through the train, the salesman falls asleep" means that the train travels half of the way before going to bed.

The remaining distance is one-third of the distance the train traveled before she fell asleep", which means that the remaining distance is one-sixth of the whole journey.

Then it can be estimated that the train traveled five sixths of the way.

Five-sixths of the way the train traveled half of the way before the tower fell asleep, then she woke up and found that the rest of the journey should be one-third of the half, that is, one-sixth, then the train traveled five-sixths of the whole journey.

Knowing how many more cars were produced, and knowing that there were more cars in August than in July, there should be a unit one, using 1+1/9, and then dividing by the excess.

Equation: 500 (1+1 9)=450 (vehicles).

Remember to think about it for yourself after you finish asking the question! Don't just copy the inscription.

It is set to produce x cars in July and Y cars in August.

y-x=500 ①

y=10/9x ②

Consecutive solutions.

x=4500

by 2 thousand, 5 pcs.

The number of 10 and 6 hundredths is (, and if you are precise to you, the approximate value is (2050). is composed of (5) (1), (8) (tenth) and (4) (thousandths), if two decimal places are kept about (. The difference between the largest four-digit number and the smallest five-digit number (1).

The largest four-digit number is (9710) and the smallest four-digit number is (1079). It is made up of 8 (tenths) and 5 (one hundredths). The unit of count is (one hundredth), and it has (85) such units of counting.

The decimal point of a decimal place is first moved two places to the right and then three places to the left, so that the original decimal is (. Keeping two decimal places is (, and keeping one decimal place is (. It is expressed in simple notation as (, note:

dots on the head of three decimal places). 15847053 read as (15,847,000 and 53) and rounded to 10,000 (1585). 976820830 is measured in tens of thousands, is (, rounded to 100 million places, is about (10).

Three thousand four hundred and forty thousand three hundred and twenty writing (30400320), rewritten as a number in ten thousand (. After a decimal point is removed from a decimal place, it is larger than the original number, and the original decimal is (. A two-digit decimal place is approximated by "rounding", and the decimal place is the largest ( and the smallest is (.

I wrote it all myself! Word by word!

After setting t hours, C is exactly in A. midpoint between B;

285-24t-36t=（36t-30t）/2t=5

Answer: After X hours, C is exactly at the midpoint between A and B.

Let this: A at point C, B at point D, and C at point E, then: ac=36x, ad=30x, dc=36x 30x=6x, be=24x, ae=285 24x, de=ae ad=285 24x 30x, from dc=2de:

6x=2 285 24x 30x, the solution is: x=5, and C is exactly at the midpoint between A and B after 5 hours.

Let C be in A exactly after x hours. The midpoint between B, where C should have met A, but not yet with B.

The distance between A and C is (24+36)x-285, and the distance between B and C is 285-(24+30)x, and these two distances are equal, and the equation is established.

24+36)x-285=285-(24+30)x, x=5

That is, after 5 hours, C is exactly in A. midpoint between B;

C is exactly in A. The midpoint between B, what this sentence to understand means.

It means that C and A have met and walked n meters more, but if they have not met with B, they will meet when they are still n meters apart.

Let's set t hours later...

The time t they go is the same.

36t+24t-285=285-(24t+30t) solution: t=5

So it's 5 hours later, C is exactly in A. midpoint between B;

Clever algorithm:

The velocity of the midpoint d in A and B is: (30+36) 2=33 km, which is converted into the encounter between point D and C.

Time = 283 (33 + 24) = 5 hours.

A is faster than B, so A and C meet first, 285 (36 + 24) = 285 60 hours, A and C and B are 6 * 285 60 = km, and the final km is (30 + 24 + 24 + 36) km per hour The speed is completed before C is in the middle of A and B, and then it will be farther and farther away, the time is 285 60+ hours

Let C be in A exactly after x hours. There is a midpoint between B.

30x+1 2*(36x-30x)=285-24x, resulting in x=5

If you don't understand, you can ask, I hope it can help you.

Order of operations! The text is not accurate, so the meaning of the title is unclear.