Common formulas Lots, lots of drops!.. (screams) The formula of the equation, to the original drop

Common formulas Lots, lots of drops!.. (screams) The formula of the equation, to the original drop

Mathematics concept of grade one in junior high school
Real number:
Rational numbers and irrational numbers are called real numbers
Rational number:
Integers and fractions are called rational numbers
Irrational number:
Irrational number refers to the infinite non cyclic decimal
Natural number:
The number of objects 0, 1, 2, 3, 4 ～ (0 included) are called natural numbers
Number axis:
A straight line that specifies a dot, a positive direction, and a unit length is called a number axis
Opposite number:
Two numbers with different symbols are opposite to each other
reciprocal:
Two numbers whose product is 1 are reciprocal to each other
Absolute value:
The absolute value of a is the distance between a point on the number axis and a circle point. The absolute value of a positive number is itself, the absolute value of a negative number is its opposite, and the absolute value of 0 is 0
Mathematical theorem formula
The algorithm of rational number
(1) addition rule: add two numbers with the same sign, take the same sign, and add the absolute value; add two numbers with different signs, take the sign of the addend with the larger absolute value, and subtract the smaller absolute value from the larger absolute value, and add two numbers that are opposite to each other to get 0
Subtraction rule: subtracting a number is equal to adding the opposite number of the number
(3) multiplication rule: if two numbers are multiplied, the same sign will get positive, the different sign will get negative, and the absolute value will be multiplied; if any number is multiplied by 0, it will get 0
(4) division rule: divide a number by the reciprocal of the number; divide two numbers by the same sign to get positive, different sign to get negative, and divide the absolute value by each other; divide 0 by any number that is not equal to 0 to get 0

Mathematics application formula, not too much, not too little, not more than grade 6 level

Total quantity △ total copies = average
[formula of general travel problem]
Average speed × time = distance;
Distance △ time = average speed;
Distance / average speed = time
[formula of reverse travel problem]
The problem of reverse travel can be divided into two kinds: meeting problem (two people start from two places and walk in opposite direction) and separation problem (two people walk in opposite direction)
(speed and) × encounter (departure) time = encounter (departure) distance;
Distance of encounter (departure) / (speed sum) = encounter (departure) time;
Distance of meeting (leaving) and time of meeting (leaving) = speed and distance
[formula of travel in the same direction]
Catch up (pull out) distance (speed difference) = catch up (pull out) time;
Catch up (pull away) distance △ catch up (pull away) time = speed difference;
(speed difference) × overtaking time = overtaking distance
[formula of train crossing bridge problem]
(bridge leader + train leader) △ speed = bridge crossing time;
(bridge leader + train leader) △ bridge crossing time = speed;
Speed × crossing time = the sum of bridge and vehicle length
[formula of sailing problem]
(1) general formula:
Static water speed (ship speed) + current speed (water speed) = downstream speed;
Ship speed water speed = upstream speed;
(downstream speed + upstream speed) △ 2 = ship speed;
(downstream velocity - upstream velocity) 2 = water velocity
(2) the formula of two ships sailing in opposite directions:
Ship a's downstream speed + ship B's upstream speed = ship a's still water speed + ship B's still water speed
(3) the formula of two ships sailing in the same direction:
The static water velocity of fore (AFT) ship - the static water velocity of fore (AFT) ship = the speed of narrowing (widening) the distance between two ships
After calculating the speed of the distance between the two ships, answer the question according to the above formula
[engineering problem formula]
(1) general formula:
Work efficiency × working hours = total amount of work;
Total amount of work △ working hours = work efficiency;
Total amount of work △ work efficiency = working hours
(2) the formula for solving engineering problems by assuming that the total amount of work is "1":
1 △ working time = fraction of the total amount of work completed in unit time;
1 △ what percentage of work can be completed per unit time = working time
(Note: to solve engineering problems with hypothesis method, you can arbitrarily assume that the total amount of work is 2, 3, 4, 5 Especially when the total amount of work is assumed to be the least common multiple of several working hours, the fractional engineering problem can be transformed into a relatively simple integer engineering problem, and the calculation will become relatively simple.)
[profit and loss formula]
(1) if there is surplus (surplus) at one time and insufficient (deficit) at one time, the following formula can be used:
(profit + loss) / (the difference between the two distributions) = the number of people
For example, "each child has 10 peaches less than 9, and each child has 8 peaches more than 7. Question: how many children and how many peaches are there?"
Solution (7 + 9) / (10-8) = 16 / 2
= 8 Number of people
10 × 8-9 = 80-9 = 71 Peach
Or 8 × 8 + 7 = 64 + 7 = 71
(2) there is surplus (surplus) in two times
(big profit - small profit) / (the difference between the number of people allocated twice) = the number of people
For example, "when soldiers carry bullets for marching training, each person carries 45 rounds, an additional 680 rounds; if each person carries 50 rounds, an additional 200 rounds will be provided. Question: how many soldiers are there? How many bullets are there?"
Solution (680-200) △ 50-45) = 480 △ 5
= 96 (person)
45 × 96 + 680 = 5000
Or 50 × 96 + 200 = 5000 (hair)
(3) two times are not enough
(big loss - small loss) / (the difference between the number of people allocated twice) = the number of people
For example, "if you distribute a batch of books to students, each of them will receive 10 copies, with a difference of 90 copies; if each of them receives 8 copies, there will still be a difference of 8 copies. How many students and how many copies are there?"

Mathematics problem: to match the formula, urgent
1. The length of a cylindrical steel bar is 5m, the diameter of its cross section is 0.6dm, and each cubic decimeter of steel weighs 7.8kg. How many kg does this steel bar weigh
2. A cylindrical bucket without cover, measured from the inside, has a bottom diameter of 40cm and a height of 50cm. Fill the bucket with water to water the flowers, with an average of 0.4L for each flower. Who can water the most flowers in this bucket?

1、5m=50dm
V=h*π（d/2）^2=50*(0.6/2)^2π=4.5π
4.5*7.8π≈110.3kg
2、V=h*π（d/2）^2=50*π*（40/2）^2=20000πcm^3=20πdm^3=20πL
20 π / 0.4 = 50 π≈ 157 trees

The solution of the equation {2x + y = m + 6 {x + 2Y = 2m satisfies that x + y is greater than 0, and the range of M is obtained

Equations 2x + y = m + 6 and X + 2Y = 2m are added on both sides
3x+3y=3m+6
So x + y = m + 2 > 0
The solution is: m ＞ - 2

What is the sum of 3 / 5 divided by 3 plus 20% of 20?

Sum = 3 / 5 △ 3 + 20 × 20% = 4.2

A cuboid wooden box is decimeter long, 7 / 10 wide and 5 / 7 high. How many square decimeter boards do you need to make this cuboid wooden box?
A rectangular wooden box is seven tenths of its width and five tenths of its height. How many square decimeters of wood is needed to make this rectangular wooden box?

Width = 10x7 / 10 = 7 decimeters
Height = 7 △ 5 / 7 = 9.8 decimeters
Surface area = 10x7x2 + 10x9.8x2 + 7x9.8x2 = 473.2 square decimeters
So the cuboid wooden box needs at least 473.2 square decimeters of wood

N is a four digit number less than 3000. It is divided by 11-5, 13-6, 17-8 to find out what n is
Write as many answers as you have

A simple way to solve this problem is:
N is divided into 11 and 5, so 2n + 1 is divided into 11 and 0. Similarly, 2n + 1 is divided into 13 and 0, 2n + 1 is divided into 17 and 0
So 2n + 1 = 11 * 13 * 17 * k = 2431 * k,
When k = 1, n = (2431-1) / 2 = 1215
When k = 2, n is not an integer, and when k = 3, n > 3000
So 1215 is the only solution satisfying the condition

Sum: 1 / 1 * 4 + 1 / 4 * 7 + 1 / 7 * 10 +... 1 / (3n-2) (3N + 1)

This, feeling, the title is 1 / (1 * 4) + 1 / (4 * 7) +. + 1 / (3n-2) (3N + 1). First of all, let's see, 1 / 1-1 / 4 = 3 / 4, 1 / 4 - 1 / 7 = 3 / 28,. 1 / (3n-2) - 1 / (3N + 1) = 3 / (3n-2) (3N + 1); so, let's change the original formula to 1 / 3 [1 / 1-1 / 4 + 1 / 4-1 / 7 +. + 1 / (3n-2) - 1 /

The formula is used to solve the problem that 2x minus x plus one sixteenth equals 0

2X & # 178; + X + 1 / 16 = 0
A = 2, B = 1, C = 1 / 16
Δ = 1 & # 178; - 4 × 2 × 1 / 16 = 1 / 2
X = quarter (- 1 ± √Δ)
X = 4 parts (- 1 ± 2 parts √ 2)
X = - 1 / 4 ± 2 / 8

When m + n = 3, find the square of the algebraic formula 2m + 4mm + 2n-6

2m²+4mn+2n²-6
=2(m²+2mn+n²)-6
=2(m+n)²-6
=2*3²-6
=18-6
=12