The Youth's Assistant in Theoretic and Practical Arithmetic: Designed for the Use of Schools in the United States |
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Page 21
Divide first by one of the component parts , and that quotient by another , and so
on , if there be more than two ; the last quotient will be the answer . 5 . Divide
31046835 by 50 = 7 | 6 . Divide 84874 by 48 = 6X8 . X8 . Quot . 55 . 1407 , Rein .
Divide first by one of the component parts , and that quotient by another , and so
on , if there be more than two ; the last quotient will be the answer . 5 . Divide
31046835 by 50 = 7 | 6 . Divide 84874 by 48 = 6X8 . X8 . Quot . 55 . 1407 , Rein .
Page 73
Now if we divide the product of the means by one of the means . the quotient is
evidently the other mean , consequently if we divide the product of the extremes
by one of the means , the quotient is the other mean . For the same reason , if we
...
Now if we divide the product of the means by one of the means . the quotient is
evidently the other mean , consequently if we divide the product of the extremes
by one of the means , the quotient is the other mean . For the same reason , if we
...
Page 81
Divide $ 160 among 41 9 . Three men hire a pasture men , so that their shares
shall for $ 100 ; A puts in 40 oxen be as 1 , 2 , 3 , and 4 . for 20 days , B 30 oxen
for 40 days , and C 50 oxen for 10 days ; how much must each man pay ? $ 32 A '
s ...
Divide $ 160 among 41 9 . Three men hire a pasture men , so that their shares
shall for $ 100 ; A puts in 40 oxen be as 1 , 2 , 3 , and 4 . for 20 days , B 30 oxen
for 40 days , and C 50 oxen for 10 days ; how much must each man pay ? $ 32 A '
s ...
Page 91
Divide the numerarator , or divide the denomina - i tor , or multiply the denomina
tor , of the fraction by the tor , of the fraction by the whole whole number ; the
result will | number ; the result will be the be the product required . required
quotient ...
Divide the numerarator , or divide the denomina - i tor , or multiply the denomina
tor , of the fraction by the tor , of the fraction by the whole whole number ; the
result will | number ; the result will be the be the product required . required
quotient ...
Page 94
To find the cost per bushel , we must divide the price by the quantity ( 154 ) , that
is , we must divide i by . But to divide a number by a fract sion , we multiply it by
the denominator , and divide the product by the numerator ( 226 ) ; hence , we ...
To find the cost per bushel , we must divide the price by the quantity ( 154 ) , that
is , we must divide i by . But to divide a number by a fract sion , we multiply it by
the denominator , and divide the product by the numerator ( 226 ) ; hence , we ...
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Common terms and phrases
acres added Addition amount ANALYSIS answer body 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 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 |