1 The problem of planting trees on unclosed lines can be divided into the following three situations:
(1) If trees are planted at both ends of the non-closed line, then:
Number of plants = number of nodes+1 = total length-1.
Total length = plant spacing × (number of plants-1)
Plant spacing = total length ÷ (number of plants-1)
2 If you want to plant trees at one end of the unclosed line and not at the other end, then:
Number of plants = number of segments = total length ÷ plant spacing
Total length = plant spacing × number of plants
Plant spacing = total length/number of plants
(3) If no trees are planted at both ends of the non-closed line, then:
Number of plants = number of nodes-1 = total length-1.
Total length = plant spacing × (number of plants+1)
Plant spacing = total length ÷ (number of plants+1)
The quantitative relationship of planting trees on the closed line is as follows
Number of plants = number of segments = total length ÷ plant spacing
Total length = plant spacing × number of plants
Plant spacing = total length/number of plants
The question of profit and loss
(Profit+Loss) ÷ Difference between two distributions = number of shares participating in distribution.
(Big profit-small profit) ÷ Difference between two distributions = number of shares participating in distribution.
(big loss-small loss) ÷ The difference between two distributions = the number of shares participating in the distribution.
encounter a problem
Meeting distance = speed × meeting time
Meeting time = meeting distance/speed and
Speed Sum = Meeting Distance/Meeting Time
Catch up with the problem
Catch-up distance = speed difference× catch-up time
Catch-up time = catch-up distance ÷ speed difference
Speed difference = catching distance ÷ catching time
Tap water problem
Downstream velocity = still water velocity+current velocity
Countercurrent velocity = still water velocity-current velocity
Still water velocity = (downstream velocity+countercurrent velocity) ÷2
Water velocity = (downstream velocity-countercurrent velocity) ÷2
Concentration problem
Solute weight+solvent weight = solution weight.
The weight of solute/solution × 100% = concentration.
Solution weight × concentration = solute weight
Solute weight-concentration = solution weight.
Profit and discount problem
Profit = selling price-cost
Profit rate = profit/cost × 100% = (selling price/cost-1) × 100%.
Up and down amount = principal × up and down percentage
Discount = actual selling price ÷ original selling price× 1 00% (discount <1)
Interest = principal × interest rate× time
After-tax interest = principal × interest rate × time × (1-20%)
Time unit conversion
1 century = 100 1 year =65438+ February.
The big month (3 1 day) includes:1\ 3 \ 5 \ 7 \ 8 \10 \ 65438+February.
Abortion (30 days) includes: April \ June \ September \165438+1October.
February 28th in a normal year and February 29th in a leap year.
There are 365 days in a normal year and 366 days in a leap year.
1 day =24 hours 1 hour =60 minutes.
1 min = 60s 1 hr = 3600s product = bottom area × height V=Sh Primary school mathematical formula:
1, the perimeter of the rectangle = (length+width) ×2 C=(a+b)×2.
2. The circumference of a square = side length ×4 C=4a.
3. Area of rectangle = length× width S=ab
4. Square area = side length x side length s = a.a = a.
5. Area of triangle = base × height ÷2 S=ah÷2.
6. parallelogram area = bottom x height S=ah
7. trapezoidal area = (upper bottom+lower bottom) × height ÷ 2s = (a+b) h ÷ 2.
8. Diameter = Radius× 2D = 2r Radius = Diameter ÷2 r= d÷2
9. The circumference of a circle = π× diameter = π× radius× 2c = π d = 2π r.
10, circular area = pi × radius× radius? =πr
1 1, the surface area of a cuboid = (length× width+length× height+width× height) × 2.
12, cuboid volume = length× width× height V =abh.
13, the surface area of the cube = side length × side length× ×6 S =6a.
14, volume of cube = side length x side length x side length v = a.a.a = a.
15, lateral area of cylinder = circumference of bottom circle × height S=ch.
16, surface area of cylinder = upper and lower bottom area+side area.
s = 2πr+2πRH = 2π(d÷2)+2π(d÷2)h = 2π(c÷2÷π)+Ch
17, cylinder volume = bottom area × height V=Sh
V=πr h=π(d÷2) h=π(C÷2÷π) h
18, volume of cone = bottom area × height ÷3.
v = sh÷3 =πr h÷3 =π(d÷2)h÷3 =π(c÷2÷π)h÷3
19, cuboid (cube, cylinder)
1, number of copies × number of copies = total number of copies/number of copies = total number of copies/number of copies = number of copies.
2. 1 multiple × multiple = multiple ÷ 1 multiple = multiple