Daboll's Complete Schoolmaster's Assistant: Being a Plain Comprehensive System of Practical Arithmetic, Adapted to the Use of Schools in the United States... |
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Page 3
... received a greater share of public patronage than Daboll's Schoolmaster's Assistant : But it must be obvious to teachers of Arithmetic , and also to men of business , that Arith- metics written some thirty or forty years ago , however ...
... received a greater share of public patronage than Daboll's Schoolmaster's Assistant : But it must be obvious to teachers of Arithmetic , and also to men of business , that Arith- metics written some thirty or forty years ago , however ...
Page 4
... received that attention which their importance demands ; being simplified and illustrated in such a manner as to render the study of them pleasing and interesting to the pupil . In Simple Interest several short rules are given . Also ...
... received that attention which their importance demands ; being simplified and illustrated in such a manner as to render the study of them pleasing and interesting to the pupil . In Simple Interest several short rules are given . Also ...
Page 26
... received 411 dollars , 88 cents and B received just three times as much money as A received ; how much mo- ney did B receive ? Ans . 1235,64 . 12. B paid me $ 98,56 , C paid me just four times as much as B , and D paid me just as much ...
... received 411 dollars , 88 cents and B received just three times as much money as A received ; how much mo- ney did B receive ? Ans . 1235,64 . 12. B paid me $ 98,56 , C paid me just four times as much as B , and D paid me just as much ...
Page 32
... received $ 135,41 and Henry received just three times as much lacking 94 cents ; how much money did Henry receive ? Ans . $ 405,29 , 23. A man received $ 1500 , and out of which he paid to A $ 511,45 , to B $ 251,15 , to C $ 180,39 ...
... received $ 135,41 and Henry received just three times as much lacking 94 cents ; how much money did Henry receive ? Ans . $ 405,29 , 23. A man received $ 1500 , and out of which he paid to A $ 511,45 , to B $ 251,15 , to C $ 180,39 ...
Page 42
... received 2943 cents for melons that he sold at 9 cents apiece ; how many did he sell ? Ans . 327 . 16. How many times is 7 contained in 6680 , and how many över ? Ans . 954 times , and 2 over . 17. A merchant has $ 5122 to purchase ...
... received 2943 cents for melons that he sold at 9 cents apiece ; how many did he sell ? Ans . 327 . 16. How many times is 7 contained in 6680 , and how many över ? Ans . 954 times , and 2 over . 17. A merchant has $ 5122 to purchase ...
Other editions - View all
Daboll's Complete Schoolmaster's Assistant: Being a Plain Comprehensive ... Nathan Daboll No preview available - 2016 |
Daboll's Complete Schoolmaster's Assistant: Being a Plain Comprehensive ... Nathan Daboll No preview available - 2017 |
Common terms and phrases
2qrs 3qrs acres annex annuity answer Arithmetic barrels broadcloth bushels called cent per annum ciphers common denominator common difference common multiple compound interest contained cords cost cube root cubic diameter divi dividend divisor dollars dols equal EXAMPLES farthings Federal money find the amount Find the value frustrum gain gallons given number given sum greatest common divisor hogshead hundred improper fraction inches last term least common multiple leave length lowest terms merchant bought miles mills mixed number months multiplicand Multiply Note number of terms payment pence pint pound present worth principal PROB proportion quantity quarts quotient figure rate per cent ratio Reduce remainder right hand roods Rule of Three separatrix shillings simple sold solid contents square rods square root subtract subtrahend sugar tare tens thousand units VULGAR FRACTIONS weight whole number wine yards of cloth
Popular passages
Page 207 - Divide the difference of the extremes by the common difference, and the quotient increased by 1 is the number of terms.
Page 191 - Multiply the divisor, thus augmented, by the last figure of the root, and subtract the product from the dividend, and to the remainder bring down the next period for a new dividend.
Page 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.
Page 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.
Page 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...
Page 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'.
Page 141 - Compute the interest to the time of the first payment ; if that be one year or more from the time the interest commenced , add it to the principal, and deduct the payment from the sum total. If there be after payments made, compute the interest on the balance due, to the next payment, and then deduct the payment as above ; and in like manner from one payment to another till all the payments are absorbed ; provided the time between one payment and another be one year or more.
Page 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.
Page 228 - What difference is there between the interest of 500Z. at 5 per cent, for 12 j'ears, and the discount of the same sum at the same rate, and for the same time ? Ans.
Page 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.