A Complete Arithmetic ...Werner School Book Company, 1899 - 447 σελίδες |
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Σελίδα 7
... perimeter of the triangle is feet . and 20. 15 feet divided by 2 feet , means , find how many times 2 feet are contained in 15 feet . I can change 15 ft . and 2 ft . to half - feet ; 15 ft . = half - feet ; 2 feet = half - feet ; half ...
... perimeter of the triangle is feet . and 20. 15 feet divided by 2 feet , means , find how many times 2 feet are contained in 15 feet . I can change 15 ft . and 2 ft . to half - feet ; 15 ft . = half - feet ; 2 feet = half - feet ; half ...
Σελίδα 35
... perimeters . Find the area and the perimeter of each of the following rectan- gular surfaces : ( a ) 23 feet by 6 feet . ( c ) 15 yards by 8 yards . ( e ) 30 inches by 20 inches . ( b ) 25 inches by 7 inches . ( d ) 32 feet by 23 feet ...
... perimeters . Find the area and the perimeter of each of the following rectan- gular surfaces : ( a ) 23 feet by 6 feet . ( c ) 15 yards by 8 yards . ( e ) 30 inches by 20 inches . ( b ) 25 inches by 7 inches . ( d ) 32 feet by 23 feet ...
Σελίδα 39
... perimeter of each of the fol- lowing rectangular surfaces : 7 feet by 5 feet . ( p ) 27 feet by 12 feet . ( q ) 15 inches by 15 inches . 8. If 2 gallons of molasses are worth 604 , 3 gallons are worth — cents . ( r ) If 2 loads of brick ...
... perimeter of each of the fol- lowing rectangular surfaces : 7 feet by 5 feet . ( p ) 27 feet by 12 feet . ( q ) 15 inches by 15 inches . 8. If 2 gallons of molasses are worth 604 , 3 gallons are worth — cents . ( r ) If 2 loads of brick ...
Σελίδα 59
... lb. of sugar are worth 454 , at the same rate are worth 8 lb. cents . ( n ) If 12 boxes of soap are worth $ 22.50 , how much are 8 boxes worth at the same rate ? MISCELLANEOUS PROBLEMS . 1. The perimeter of a rectangular surface PART I. 59.
... lb. of sugar are worth 454 , at the same rate are worth 8 lb. cents . ( n ) If 12 boxes of soap are worth $ 22.50 , how much are 8 boxes worth at the same rate ? MISCELLANEOUS PROBLEMS . 1. The perimeter of a rectangular surface PART I. 59.
Σελίδα 60
Frank H. Hall. MISCELLANEOUS PROBLEMS . 1. The perimeter of a rectangular surface 7 feet by 9 feet is feet . ( a ) How many rods of fence are required to enclose a rectangular field 38 rods by 54 rods ? 2. If the sun rises at 7 o'clock ...
Frank H. Hall. MISCELLANEOUS PROBLEMS . 1. The perimeter of a rectangular surface 7 feet by 9 feet is feet . ( a ) How many rods of fence are required to enclose a rectangular field 38 rods by 54 rods ? 2. If the sun rises at 7 o'clock ...
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40 rods acres acres of land Algebra altitude bought bushels Change coal COMMON FRACTIONS common multiple contained cords cube cubic centimeters cubic feet cubic foot DECIMAL FRACTIONS decimal order decimeter DENOMINATE NUMBERS diameter discount Divide dividend dollars earn equal exact divisor exactly divisible Express the ratio feet long figure Find the amount Find the area Find the cost Find the interest Find the number Find the product Find the quotient Find the sum foot gain hundred hundredths improper fractions integral number least common multiple less lowest terms means measured meters miles MISCELLANEOUS PROBLEMS multiplicand NOTE number story Observe PERCENTAGE perimeter piece of land pounds primary units prime factors pupil rectangular rhombus right angles side sold solid content specific gravity square feet square inches square rods square root subtract tenths third thousandths tons triangle weigh