The Youth's Assistant in Theoretic and Practical Arithmetic: Designed for the Use of Schools in the United States |
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Page 6
A man bought a pair of horses for 216 dollars , a sleigh for 84 dollars , and a harness for 63 dollars ; what did they all cost him ? Here we write down the numbers as before , and be . 216 dolls . gin with the right hand colunm - 3 and ...
A man bought a pair of horses for 216 dollars , a sleigh for 84 dollars , and a harness for 63 dollars ; what did they all cost him ? Here we write down the numbers as before , and be . 216 dolls . gin with the right hand colunm - 3 and ...
Page 8
Now it frequently happens that the numbers to be added are all equal , in wbich case the operation may be abridged by a process called Multiplication . 3. If a book cost 5 cents , what rv : ll 4 such books cost ? are 20 . B4 , 8 .
Now it frequently happens that the numbers to be added are all equal , in wbich case the operation may be abridged by a process called Multiplication . 3. If a book cost 5 cents , what rv : ll 4 such books cost ? are 20 . B4 , 8 .
Page 9
Four booke will evidently cost four times Addition . as much as one book ; and to answer the Multipucation 5 question by Addition , we should write 5 down 4 fives , and add them , as at the left 5 hand . By Multiplication we should pro5 ...
Four booke will evidently cost four times Addition . as much as one book ; and to answer the Multipucation 5 question by Addition , we should write 5 down 4 fives , and add them , as at the left 5 hand . By Multiplication we should pro5 ...
Page 11
If a man drink - a glass there are 27 rows of trees , of spirits 3 times a day , and and 15 trees in each row , I will be the cost for a year ? each glass cost 6 cents , what how many trees are there ? Ans . 405 . Ans . 6570 cts .
If a man drink - a glass there are 27 rows of trees , of spirits 3 times a day , and and 15 trees in each row , I will be the cost for a year ? each glass cost 6 cents , what how many trees are there ? Ans . 405 . Ans . 6570 cts .
Page 12
A man . bought 17 cows for 15. dollars apiece ; what did they all cost ? If we muliiply 17 by 5 , we find the cost at 5 dollars apiere , Operation , and since 15 is 3 times 5 , the cost , at 15 dollars apicce , will 17 manifestly be 3 ...
A man . bought 17 cows for 15. dollars apiece ; what did they all cost ? If we muliiply 17 by 5 , we find the cost at 5 dollars apiere , Operation , and since 15 is 3 times 5 , the cost , at 15 dollars apicce , will 17 manifestly be 3 ...
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Common terms and phrases
acres added Addition amount ANALYSIS answer bush bushels called cash cents Change ciphers column common compound contains cost cube cubic decimal denominator denoted diameter difference distance divide dividend division divisor dollars dolls equal evidently example expressed factors feet figures foot four fraction gain gallon give given greater half Hence hundred hundredths inches interest least left hand length less mean measure method miles months multiplicand multiply names operation payment period person pound principal proceed proportion quantity QUESTIONS FOR PRACTICE quotient ratio receive Reduce remainder right hand rods root rule share shillings side simple solid square square root subtract supposed tens tenths third tion units vulgar weight whole worth write written yard
Popular passages
Page 82 - Multiply each payment by its term of credit, and divide the sum of the products by the sum of the payments ; the quotient will be the average term of credit.
Page 89 - The greatest common divisor of two or more numbers, is the greatest number which will divide them without a remainder. Thus 6 is the greatest common divisor of 12, 18, 24, and 30.
Page 118 - PROBLEM II. The first term, the last term, and the number of terms given, to find the common difference. RULE. — Divide the difference of the extremes by the number of terms less 1 , and the quotient will be the common diffcrenct.
Page 111 - Subtract the square number from the left hand period, and to the remainder bring down the next period for a dividend. III. Double the root already found for a divisor ; seek how many times the divisor is contained in the dividend...
Page 94 - It will be seen that we multiply the denominator of the dividend by the numerator of the divisor for the denominator of the quotient, and the numerator of the dividend by the denominator of the divisor for the numerator of the quotient.
Page 120 - Add together the most convenient indices to make an index less by 1 than the number expressing the place of the term sought. 3. Multiply the terms of the geometrical series together belonging to those indices, and make the product a dividend. 4. Raise...
Page 115 - 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 31 - RULE. Divide as in whole numbers, and from the right hand of the quotient point off as many places for decimals as the decimal places in the dividend exceed those in the divisor.
Page 2 - Los números cardinales 0: zero 1: one 2: two 3: three 4: four 5: five 6: six 7: seven 8: eight 9: nine 10: ten 11: eleven 12: twelve 13: thirteen 14: fourteen 15: fifteen 16: sixteen 17: seventeen 18: eighteen 19: nineteen 20: twenty...
Page 93 - Multiply the numerators together for a new numerator, and the denominators together for a new denominator.
Bibliographic information
| Title | The Youth's Assistant in Theoretic and Practical Arithmetic: Designed for the Use of Schools in the United States |
| Author | Zadock Thompson |
| Publisher | H. Johnson, 1838 |
| Original from | Harvard University |
| Digitized | 29 Jan 2007 |
| Length | 164 pages |
| Export Citation | BiBTeX EndNote RefMan |