The National Arithmetic on the Inductive System Combining the Analytic and Synthetic Methods Forming a Complete Course of Higher ArithmeticLeach, Shewell and Sanborn, 1863 - 444 σελίδες |
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Σελίδα 37
... cost of 4 books at 9 dollars each , we cannot multiply books and dollars together , which would be absurd , but we can , by regarding the 4 as an abstract number , take the 9 dollars , or cost of 1 book 4 times , and the product , 36 ...
... cost of 4 books at 9 dollars each , we cannot multiply books and dollars together , which would be absurd , but we can , by regarding the 4 as an abstract number , take the 9 dollars , or cost of 1 book 4 times , and the product , 36 ...
Σελίδα 40
... cost at 73 dollars per acre ? Ans . $ 26645 . 18. What will 97 tons of iron cost at 57 dollars a ton ? Ans . $ 5529 . 19. What will 397 yards of cloth cost at 7 dollars per yard ? Ans . $ 2779 . 20. What will 569 hogsheads of molasses cost ...
... cost at 73 dollars per acre ? Ans . $ 26645 . 18. What will 97 tons of iron cost at 57 dollars a ton ? Ans . $ 5529 . 19. What will 397 yards of cloth cost at 7 dollars per yard ? Ans . $ 2779 . 20. What will 569 hogsheads of molasses cost ...
Σελίδα 42
... cost of 1 acre . 7 2 2 12 dollars , cost of 7 acres . 5 Ans . 11060 dollars . The factors of 35 are 7 and 5. Now , if we multiply the price of one acre by 7 , we get the cost of 7 acres ; and , then , by multiplying the cost of 7 acres ...
... cost of 1 acre . 7 2 2 12 dollars , cost of 7 acres . 5 Ans . 11060 dollars . The factors of 35 are 7 and 5. Now , if we multiply the price of one acre by 7 , we get the cost of 7 acres ; and , then , by multiplying the cost of 7 acres ...
Σελίδα 43
... cost at 40 dollars per bale ? Ans . 28000 dollars . OPERATIΟΝ . Multiplicand 700 Multiplier Product 40 28000 The multiplicand we resolve into the factors 7 and 100 , and the multiplier into the factors 4 and 10. Now , it is evident ...
... cost at 40 dollars per bale ? Ans . 28000 dollars . OPERATIΟΝ . Multiplicand 700 Multiplier Product 40 28000 The multiplicand we resolve into the factors 7 and 100 , and the multiplier into the factors 4 and 10. Now , it is evident ...
Σελίδα 45
... cost of 1 pen- cil , when 4 pencils cost 24 cents , we separate or divide the 24 cents into 4 equal parts , of which one part is 6 cents . Hence , 1 pencil costs 6 cents , when 4 pencils cost 24 cents . 71. The remainder will always be ...
... cost of 1 pen- cil , when 4 pencils cost 24 cents , we separate or divide the 24 cents into 4 equal parts , of which one part is 6 cents . Hence , 1 pencil costs 6 cents , when 4 pencils cost 24 cents . 71. The remainder will always be ...
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25 per cent 30 days acres annexed annuity Arithmetic balance bank discount barrels becoming due bill bought bushels ciphers circumference column common denominator common fraction compound interest compound numbers contain continued fraction cost cube root decimal places decimal point diameter difference dividend division dollars equal equivalent EXAMPLES expressed decimally feet long figures gallons given number given rate grains greatest common divisor Hence hogshead hundred hundredths improper fraction inches least common multiple less longitude lowest terms miles minuend mixed number months multiplicand Multiply NOTE number denoting number of days number of terms obtain order or place paid par value payable payment pounds premium present worth prime factors principal proceeds proportion purchased quantity quotient rate per cent ratio Reduce remainder repetend rods RULE SECOND OPERATION share shillings sold square root subtract tens third thousand thousandths tons units weight whole number write yards