Two monkeys found a push of peaches. The big monkey ate half and half of the peaches. The little monkey ate the other half and half, and there were two left. How many were there

[(2 + 1 / 2) * 2 + 1 / 2)] * 2 = 11

A pile of peaches. The old monkey ate half of them and took another one. The big monkey ate the remaining half and took another one. The little monkey also ate them
The old monkey ate half of the peaches and took another one. The big monkey ate the remaining half and took another one,
The little monkey also ate half of it and took another one. As a result, there was only one left. How many peaches were there in this pile?
Thank you for explaining every step carefully, OK? I didn't understand the online reference answer. Is this the score type of applied problem? How many ways to solve this problem

There are three kinds. First, the little monkey takes one and leaves one, which means that the big monkey has four left

In the four mixed operations with brackets, we should first calculate the one in (), then calculate the one in (), and finally calculate ()

In the four mixed operations with brackets, we should first calculate the one inside (small brackets), then the one inside (large brackets), and finally the one outside the brackets

As shown in the figure, the length of homogeneous long rod AB is l, the mass is m, and the angular velocity is ω

According to the theorem of kinetic energy of rigid body rotation e = 1 / 2 * J * W ^ 2, consider Jao = 3 / 4 * m * (3L / 8) ^ 2 = 27ml ^ 2 / 256jbo = 1 / 4 * m * (L / 8) ^ 2 = = ml ^ 2 / 256. Therefore, e = 1 / 2 * (Jao + jbo) * W ^ 2 = 7ml ^ 2 * W ^ 2 / 128

The sum of the first n terms of the arithmetic sequence an and Sn = (2 ^ n + 1) - 2.1 are known. Find the general term formula of an? 2. Let BN = an + an + 1 (1 is small 1), find the general term formula of BN

Sn=（2^n+1）-2
S（n-1)=2^n-2
Subtraction
an=2^n
bn=an+an+1=3*2^n

A trolley car is accelerating along a straight line with an acceleration of 2.0 M / S ^ 2. Q: what is the displacement of the car after 8 s

a=2.0m/s^2
v0=0m/s
t=8s
Using the formula x = v0t + 2 / 1At & # 178, the displacement is x = 64M
If you have any questions

Xiao Li walks a quarter more than Xiao Ming, but Xiao Ming walks a sixth more than Xiao Li
The speed ratio of Xiao Li and Xiao Ming is () {finding the whole process}

This is how it is calculated. First, according to the conditions, we can get the distance ratio of two people: Xiao Li - Xiao Ming = (1 + 1 / 4): 1 = 5:4
The time ratio of the two is: Xiao Li - Xiao Ming = 1: (1 + 1 / 6) = 6:7
So we can find out the speed ratio: Xiao Li - Xiao Ming = 5 / 6:4 / 7 = 35:24

A problem on Polynomials
It is known that a is a cubic polynomial and B is also a cubic polynomial. How many polynomials are a + B and A-B?
(there should be a process)
I mean to make a case

A + B must be a cubic polynomial, because if we combine the similar terms, the coefficient will not be 0, and the degree will not be affected
A-B will be a polynomial equal to or less than degree 3, because the coefficient of merging similar terms may be 0, which affects the degree

As shown in Figure Su, there are three points a, B and C on the number axis. Now we need to move two of them so that the numbers represented by the three points are the same. There are several kinds of transfers. How
Move? Figure: A on negative 6, B on negative 3, C on 4

There are three points in total. To move two of them so that the number of the three points is the same, so the other point should not be moved. Therefore, there are three methods
A does not move, BC moves to a, that is - 6
B does not move, AC moves to B, that is - 3
C does not move, AB moves to C, that is 4
This topic is not difficult to understand. Give me a point

A tailor makes a children's dress, a pair of trousers and a coat. The ratio of all time is 1:2:3. He can make 2 children's clothes, 3 pairs of trousers and 4 coats a day. How many days does it take for him to make 2 coats, 10 pairs of trousers and 14 pieces of children's clothes?

The ratio of time is 1:2:3;
Two children's clothes, three pairs of trousers and four coats are what he can make every day, 2 * 1 + 3 * 2 + 4 * 3 = 20
There are 2 * 3 + 10 * 2 + 14 * 1 = 40 pieces of 2 coats, 10 pants and 14 children's wear
40 / 20 = 2 days

Two monkeys found a push of peaches. The big monkey ate half and half of the peaches. The little monkey ate the other half and half, and there were two left. How many were there