I am a beginner. What is the principle of mathematical formula? For example, why is distance = speed x time? Tell me how to remember and why it is so determined.

general operation rule

1 ××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××× = total price ÷ total price ÷ unit price = total quantity ÷ quantity = unit price

5 working efficiency × working time = total work amount ÷ working efficiency = working time ÷ working efficiency

6 addend+addend = sum-one addend = another addend

7 minuend. Factor × factor = product ÷ one factor = another factor

9 Divider ÷ Divider = quotient Divider ÷ Quotient = divisor quotient × divisor = dividend < P > Calculation formula of primary school mathematics graphics

1 Square C perimeter S area A side length < P > perimeter = side length × 4 C. > 2 cube v: volume a: edge length

surface area = edge length× edge length× 6s table =a×a×6

volume = edge length× edge length× edge length V=a×a×a

3 rectangle c perimeter s area a side length

perimeter = (length+width )×

area = length× width S=ab

4 cuboid v: volume s: area a: length b: width h: height

surface area (length× width+length× height+width× height) ×2 S=2(ab+ah+bh).

volume = length× width× height V=abh

5 triangle s area a bottom h height

area = bottom× height ÷2 s=ah÷2

triangle height = area× 2 ÷ triangle bottom = area× 2 ÷ height

. S area a upper bottom b lower bottom h height

area = (upper bottom+lower bottom )× height ÷2 s=(a+b)× h÷2

8 circular S area c circumference ∏ d= diameter r= radius

circumference = diameter× ∏ = 2. Bottom area r: bottom radius c: bottom circumference

lateral area = bottom circumference× high surface area = lateral area+bottom area× 2

volume = bottom area× high volume = lateral area ÷2× radius

1 cone v: volume h: height s; Base area r: base radius

volume = base area × height ÷3

primary school olympiad formula

and difference problem formula

(sum+difference) ÷ 2 = large number (sum-difference) ÷ 2 = decimal

and multiple problem formula

sum. = decimal × multiple = large number (or decimal+difference = large number)

Formula of tree planting problem

1 The problem of tree planting on unclosed lines can be mainly divided into the following three situations:

(1) If trees are planted at both ends of unclosed lines, Then:

number of plants = number of segments+1 = total length ÷ plant spacing-1

total length = plant spacing × (number of plants-1)

plant spacing = total length ÷ (number of plants-1)

(2) If trees are to be planted at one end of the non-closed line and not at the other end, then: Then:

number of plants = number of segments-1 = total length ÷ plant spacing-1

total length = plant spacing × (number of plants+1)

plant spacing = total length ÷ (number of plants+1)

2 The quantitative relationship of tree planting problems on closed lines is as follows

number of plants = number of segments. ÷ the difference between the two distributions = the number of shares participating in the distribution

(big profit-small profit) ÷ the difference between the two distributions = the number of shares participating in the distribution

(big loss-small loss) ÷ the difference between the two distributions = the number of shares participating in the distribution

the formula of meeting problem

meeting distance = speed and x meeting time

meeting time = meeting distance ÷ speed and

speed sum = meeting distance ÷ meeting time

the formula of catching up with the problem

catching up with distance = speed difference x catching up. +current velocity

countercurrent velocity = still water velocity-current velocity

still water velocity = (downstream velocity+countercurrent velocity) ÷2

current velocity = (downstream velocity-countercurrent velocity) ÷2

formula of concentration problem

weight of solute+weight of solvent = weight of solution

weight of solute ÷ weight of solution × 1% = concentration

weight of solution × concentration = weight of solute

weight of solute ÷ concentration = weight of solution

profit and discount. ×1%

fluctuation amount = principal× fluctuation percentage

discount = actual selling price ÷ original selling price×1% (discount < 1)

interest = principal× interest rate× time

after-tax interest = principal× interest rate× time × (1-2%) < p