When calculating the division with remainder, 308 is wrongly written as 368, and the quotient is increased by 5 and the remainder is just right

When calculating the division with remainder, 308 is wrongly written as 368, and the quotient is increased by 5 and the remainder is just right

(368-308)/5=12
The divisor is 12
The remainder of 308 / 12 is 8

When Xiao Ming calculates the division with remainder, he wrongly writes the divisor 115 as 151, and the result quotient is 3 larger than the correct result, but the remainder is exactly the same______ ．

Because the quotient is increased by 3, the divisor (151-115) △ 3 = 36 △ 3 = 12 can be obtained 7. So the answer is: 115 △ 12 = 9 7．

When Xiao Ming calculates the division with remainder, he wrongly writes the divisor 115 as 151, and the result quotient is 3 larger than the correct result, but the remainder is exactly the same______ ．

Because the quotient is increased by 3, the divisor (151-115) △ 3 = 36 △ 3 = 12 can be obtained 7. So the answer is: 115 △ 12 = 9 7．

When Xiao Ming calculates the division with remainder, he wrongly writes the divisor 115 as 151, and the result quotient is 3 larger than the correct result, but the remainder is exactly the same______ ．

Because the quotient is increased by 3, the divisor (151-115) △ 3 = 36 △ 3 = 12 can be obtained 7. So the answer is: 115 △ 12 = 9 7．

A mathematical problem, known as B < a < 0 < C, is simplified

│a│－│b＋a│＋│c－b│－│a－c│
=-a-(-b-a)+(c-b)-(c-a)
=-a+b+a+c-b-c+a
=a

As shown in Figure 15, it is known that there are two points c and D on the line AB, and AC: CD: DB = 2:3:4, e and F are the midpoint of AC and DB respectively, if EF = 2.4cm?

Let AB = x, AC: CD: DB = 2:3:4, so AC = 2 / 9x, CD = 3 / 9x. DB = 4 / 9xef = 2.4 = EC + CD + DF=
1/2 * AC + CD + 1/2 * DB= 1/2 * 2/9 X + 3/9X + 1/2 * 4.9XX=3.6

Differences between mathematical brackets, such as (a, b) and [a, b]
Could you make it more clear?

The front one does not contain a and B
After that is the inclusion
PS: the first one is
Numbers greater than a and less than B
The second is
Numbers greater than or equal to a and less than or equal to B
I remember learning in the first day of junior high school

(a+b )-2(a-b)

Simplify evaluation questions
3x - {- 2Y + [x - (4x-3y)]}, where x = 1 / 3, y = 1 / 5

3x-{-2y+[x-(4x-3y)]}
=3x+2y+[x-(4x-3y)]
=3x+2y+x-(4x-3y)
=4x+2y-4x+3y
=5y
=5*1/5
=1

The part of "removing bracket" in mathematics of grade one in junior high school
Given | A-2 | + | B-1 | = 0, find the value of the algebraic formula a + {3B - [2B - (2a-3b)]}