A System of Practical Arithmetic: Applicable to the Present State of Trade, and Money Transactions: Illustrated by Numerous Examples Under Each Rule; for the Use of Schools |
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Page 26
... course the true answers to these sums are 58566 , and 272,400 . NOTE . * When nothing remains after the division , the figures that were cut off will be the true remainder ; thus 76543246 1200 = 637867100 . EXAMPLES . Ex . 1. -347652160 ...
... course the true answers to these sums are 58566 , and 272,400 . NOTE . * When nothing remains after the division , the figures that were cut off will be the true remainder ; thus 76543246 1200 = 637867100 . EXAMPLES . Ex . 1. -347652160 ...
Page 29
... the divisor ; under the last of these let him place a dot , and likewise under every subsequent figure , as he makes use of it , to the end of the dividend of course . EXAMPLES . Ex . 2 . 5694327897 4. 8436548769 6. DIVISION . 29.
... the divisor ; under the last of these let him place a dot , and likewise under every subsequent figure , as he makes use of it , to the end of the dividend of course . EXAMPLES . Ex . 2 . 5694327897 4. 8436548769 6. DIVISION . 29.
Page 32
... course the balloon would ascend , with several persons in its boat ; because it will ascend , when the balloon and persons are together , lighter than an equal bulk . of common air , REDUCTION . REDUCTION is the method of converting ...
... course the balloon would ascend , with several persons in its boat ; because it will ascend , when the balloon and persons are together , lighter than an equal bulk . of common air , REDUCTION . REDUCTION is the method of converting ...
Page 36
... course the true remainder is 16 grains . The result would have been the same , though differently expressed , had I divided by 6 and by 4 , instead of 4 and 6 : in the former case there would have been two remainders , viz . 4 and 2 ...
... course the true remainder is 16 grains . The result would have been the same , though differently expressed , had I divided by 6 and by 4 , instead of 4 and 6 : in the former case there would have been two remainders , viz . 4 and 2 ...
Page 48
... course 32 cubic feet weigh 2000 lb. which was formerly a ton . In the year 1689 , a statute of excise was passed , which made a firkin of ale or beer , without distinction , to consist of 8 gallons : this has , however , been long in ...
... course 32 cubic feet weigh 2000 lb. which was formerly a ton . In the year 1689 , a statute of excise was passed , which made a firkin of ale or beer , without distinction , to consist of 8 gallons : this has , however , been long in ...
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A System of Practical Arithmetic, Applicable to the Present State of Trade ... Jeremiah Joyce No preview available - 2018 |
A System of Practical Arithmetic, Applicable to the Present State of Trade ... Jeremiah Joyce No preview available - 2015 |
Common terms and phrases
9 Ex acres aliquot amount annual annuity annum answer arithmetical progression Avoirdupois bill bushels common denominator compound interest containing cost course of exchange cube root cubic cyphers decimal difference ditto divide dividend divisor equal EXAMPLES farthings feet figures find the value fraction gallons geometrical progression geometrical series given number given sum gives guineas per cent hogsheads hundred improper fractions inches insure joint lives last term lease logarithm London measure miles millions mixed numbers months multiplicand Multiply the number neat weight NOTE number of terms ounces paid payment pence person aged piastre pound sterling pounds present value purchase quantity quotient Reduce remainder Rule of Three shews shillings square root sterling subtract supposing tare thousand tons tret Troy TROY WEIGHT whole number wine worth yards
Popular passages
Page 177 - Multiply each payment by the time at which it is due; then divide the sum of the products by the sum of the payments, and the quotient will be the equated time, nearly.
Page 112 - To reduce a mixed number to an improper fraction, — RULE : Multiply the whole number by the denominator of the fraction, to the product add the numerator, and write the result over the denominator.
Page 243 - Multiply each term into the multiplicand, beginning at the lowest, by the highest denomination in the multiplier, and write the result of each under its respective term ; observing to carry an unit for every 12, from each lower denomination to its next superior.
Page 92 - III. finally, multiply the second and third terms together, divide the product by the first, and the quotient will be the answer in the same denomination as the third term.
Page 150 - The first term, the last term (or the extremes) and the ratio given, to find the sum of the series. RULE. Multiply the last term by the ratio, and from the product subtract the first term ; then divide the remainder by the ratio, less by 1, and the quotient will be the sum of all the terms.
Page 113 - Multiply each numerator into all the denominators except its own for a new numerator, and all the denominators together for a common denominator.
Page 243 - In like manner, multiply all the multiplicand by the inches and parts of the multiplier, and set the result of each term one place removed to the right hand of those in the multiplicand...
Page 55 - Place the numbers so that those of the same denomination may stand directly under each other.
Page 149 - Given the first term, last term, and common difference, to find the number of terms. RULE. — Divide the difference of the extremes by the common difference, and the quotient increased by 1 is the number of terms.
Page 28 - ... the number in the quotient. Multiply the divisor by the quotient figure, and set the product under that part of the dividend used. Subtract the product, last found, from that part of the dividend under which...