1 hard math problem, 1 hard math problem

3 people spent a total of 9*3 27 yuan; The 2 yuan taken by the waiter is actually 3*9=27 yuan; The boss received 25 yuan, and the waiter took 2 yuan and earned 25 yuan + 2 yuan 27 yuan, so for the customer, they paid 9 yuan for the accommodation, so 3 x 9 yuan = 27 yuan + returned) 3 yuan = 30 yuan; For the boss (paid by the customer) 30 yuan - the waiter hid it) 2 yuan - (returned by the waiter) 3 yuan = (actual income) 25 yuan; For the waiter (paid by the customer) 30 yuan - (returned to the customer) 3 yuan - (handed over to the boss) 25 yuan = (the waiter hid) 2 yuan expenditure There is no problem with income.

27+2 is meaningless, 27 is 25 of pay = boss + 2 of waiter; Add 2 more repeats.

A total of 30 yuan: 25 for the boss, 2 for the waiter, and 3 for three people

Three people out of a total of 3 * 9 = 27: the boss 25, the waiter 2 does not need to add 2, meaningless.

It's just a mistake.

The boss charged 25 yuan, and the waiter stole 2 yuan, so the three customers paid a total of 27 yuan.

It has nothing to do with that dollar, it's in the wrong direction.

This question

Almost messed up by LZ

Let's start with it: After the waiter steals 2 yuan, he can't use 10-1 9 Each person does not spend 9 yuan, but 30-2 3 yuan (approximately.

So each person pays yuan, multiplies 3 28, plus 2 yuan stolen, and finally adds 30 yuan.

Let the radius of the bottom of the cone be r and the height be h

According to the volume formula of the cone, there is 1/3 * *r 2*h, while the volume of the upper cone is: 1/3 * *r) 2* h, then the volume of water containing is: 1/3* r 2*h - 1/3 * *r) 2* h = 7/24 * 2*h = 1

There is 2*h = 24/7

Therefore, 1/3 * * r) 2 * h = 1 7, so 1 7 kg of water needs to be injected.

x+y+z=3.Then the square of (x+y+z) = 9And (x+y+z) is squared after x 2+y 2+z 2+2xy+2xz+2yz, and because (x square + y square + z square = 3, so 2xy+2xz+2yz=6, so xy+xz+yz=3, then there is.

x square + y square + z flat pin square = xy + xz + yz, multiply this 2 side cherry blossom brother by 2, turn it into 2x 2 + 2y 2 + 2z 2 = 2xy + 2xz + 2yz, and then turn it into 2x 2 + 2y 2 + 2z 2-2xy-2xz-2yz=o, continue to simplify 2x 2 + 2y 2 + 2z 2-2xy-2xz-2yz=(x-y) 2+(x-z) 2+(x-z) 2=o, so you can spine out (x-y) 2=(x-z) 2=(y-z) 2=o, so x=y=z=1

1.This is a math problem for finding the least common multiple, combined with the calculation of the date, the difficulty is average. From the inscription, it can be known that the four daughters, from the oldest to the youngest, have one day off every two days, every three days, every four days, every five days, and every six days.

Find the least common multiple of , that is, the period of rest at the same time, this least common multiple is 4x5x6 = 120 days; Then, from 1 January 1998, it was postponed by 120 days, which can be seen as May 1, 1998. 2.This is a matter of finding the least common multiple purely:

Find their least common multiple, and the common multiple is 120, and since there is always one more, you can tell that there are at least 121 eggs in the basket.

I want to start by saying that the question you gave is wrong, the first one: it should be the third daughter, each three days off and one day off. First of all, (1) the five daughters have only a certain periodicity, their cycle is 2, 3, 4, 5, 6, then the days of the five daughters resting together are the least common multiple of 2, 3, 4, 5, 6.

That is, 60, then they rest together after 60 days. That is, the 61st day of 1998, note that February of this year is 28 days, and January is 31 days, so the date of 61 days is March 3. (2) This is also a problem of the least common multiple, that is, the least common multiple of 4, 5, 6, plus 1.

The result was 61.