Daboll's Complete Schoolmaster's Assistant Being a Plain Comprehensive System of Practical Arithmetic

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Bolles and Williams, 1843 - 240 Seiten
 

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Seite 195 - Find the first figure of the root by trial, and subtract its power from the left hand period of the given number. 3. To the remainder bring down the first figure in the next period, and call it the dividend. 4. Involve the root to the next inferior power to that which is given, and multiply it by the number denoting the given power, for a divisor.
Seite 167 - Multiply all the numerators together for a new numerator, and all the denominators for a new denominator; and they will form the fraction required.
Seite 183 - ... subtract it therefrom, and to the remainder bring down the next period for a dividend. 3. Place the double of the root already found, on the left hand of the dividend for a divisor. 4. Seek how often the divisor is contained...
Seite 106 - Let the farthings in the given pence and farthings possess the second and third places ; observing to increase the second place or place of hundredths, by 6 if the shillings be odd ; and the third place by 1 "when the farthings exceed 12, and by 2 when they exceed 36. EXAMPLES. 1. Find the decimal of 7s. 9fd. by inspection. ,3 =4 6s. 5 for the odd shillings. 39=the farthings in 9|d. 2 for the excess of 36. £. ,391=dechnal required'.
Seite 90 - To reduce an improper fraction to a whole or mixed number. RULE. Divide the numerator by the denominator, and the quotient will be the whole or mixed number sought.
Seite 233 - To measure a Parallelogram, or long square. RULE. Multiply the length by the breadth, and the product will be the area or superficial content.
Seite 44 - If any partial dividend will not contain the divisor, place a cipher in the quotient, and bring down the next figure of the dividend, and divide as before.
Seite 126 - ... multiply the second and third terms together, and divide the product by the first for the answer, which will always be of the same denomination as the third term.
Seite 119 - Then multiply the second and third terms together, and divide the product by the first term: the quotient will be the fourth term, or answer.
Seite 205 - ... the terms, RULE. Multiply the sum of the extremes by the number of terms, and half the product will be the sum of the terms.

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