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The Formula of Math Application Problems in Grade Six
, per share × number of shares = total number of shares/share = total number of shares/number of shares = 2 per share, 1 multiple × multiple = multiple ÷ 1 multiple = multiple ÷ multiple = 1 multiple 3, speed × time = distance ÷ work efficiency × working time = total work ÷ work efficiency = working time ÷ total work. Factor × factor = product divisor ÷ divisor = quotient dividend ÷ quotient = divisor quotient × divisor = calculation formula of primary school mathematics 1, square c perimeter s area a side length perimeter = side length × 4c = 4a area = side length× side length S=a×a 2, cube v: volume a: side length. Rectangular c perimeter s area a side length perimeter = (length+width) ×2 C=2(a+b) area = length × width S=ab 4, cuboid v: volume s: area a: length b: width h: height (1) surface area = (length× width+length× height+width × Parallelogram S area A bottom H height area = bottom X height s=ah 7, trapezoid S area A top bottom B bottom H height area = (upper bottom+lower bottom) × h ∏ 2 8 circular S area C perimeter ∏ d= diameter r= radius (1) perimeter = diameter× ∏ = 2× ∏. Bottom area R: bottom radius C: bottom perimeter (1) lateral area = bottom perimeter × height (2) surface area = lateral area+bottom area× 2 (3) volume = bottom area× height (4) volume = lateral area ÷2× radius 10, and cone V: volume H. Total number ÷ Total number of copies = average (sum+difference) ÷ 2 = large number (sum-difference) ÷ 2 = decimal and multiple problem and ÷ (multiple-1) = decimal x multiple = large number (or sum-decimal = large number) difference problem ÷. The problem of planting trees on unclosed lines can be mainly divided into the following three situations: (1) If trees are to be planted at both ends of unclosed lines, Then: number of plants = number of nodes+1 = total length ÷ plant spacing-1 ÷ total length = plant spacing × (number of plants-1) plant spacing = total length ÷ (number of plants-then: number of plants = number of nodes = total length. The quantitative relationship of the tree planting problem on the closed line is as follows: the difference between two distributions = the number of shares participating in distribution (big profit-small profit) ↓ the difference between two distributions = the number of shares participating in distribution (big loss-small loss) ⊙ the difference between two distributions = the number of shares participating in distribution is satisfied: satisfaction distance = speed and x satisfaction time = satisfaction distance ÷ sum of speed and speed = satisfaction distance ÷. Hydrostatic velocity = (downstream velocity+countercurrent velocity) ÷2 Water flow velocity = (downstream velocity-countercurrent velocity) ÷2 Concentration problem: solute weight+solvent weight = solution weight ÷ solution weight × 100% = concentrated solution weight × concentration = solute weight ÷ concentration = solution weight; Profit and discount: profit = selling price-cost profit rate = profit ÷ cost × 654; . × 100% increase or decrease = principal× increase or decrease 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/-)