Higher Arithmetic, Or, the Science and Application of Numbers: Combining the Analytic and Synthetic Modes of Instruction : Designed for Advanced Classes in Schools and Academies |
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Page 31
... bushels of wheat of one man , 4952 bushels of another , and 3273 bushels of another : how many bushels did he buy of all ? Operation . 7864 4952 3273 Ans . 16089 bu . Write the numbers under each other , so that units may stand under ...
... bushels of wheat of one man , 4952 bushels of another , and 3273 bushels of another : how many bushels did he buy of all ? Operation . 7864 4952 3273 Ans . 16089 bu . Write the numbers under each other , so that units may stand under ...
Page 33
... bushels , which is the same as in the solution above . Thus , it is evident , when the sum of a column exceeds 9 , the right hand figure denotes units of the same order as the column added , and the tens or left hand figure denotes ...
... bushels , which is the same as in the solution above . Thus , it is evident , when the sum of a column exceeds 9 , the right hand figure denotes units of the same order as the column added , and the tens or left hand figure denotes ...
Page 34
... QUEST . - Note . Why add the columns downwards , instead of upwards ? Can addition be proved by any other methods ? * Wallis ' Arithmetic , Oxford 1657 . tained 6725 bushels , the second 7208 , the third 34 [ SECT . II . ADDITION .
... QUEST . - Note . Why add the columns downwards , instead of upwards ? Can addition be proved by any other methods ? * Wallis ' Arithmetic , Oxford 1657 . tained 6725 bushels , the second 7208 , the third 34 [ SECT . II . ADDITION .
Page 35
... bushels , the second 7208 , the third 5047 , the fourth 12386 , and the fifth 10391 bushels : how many bushels did he buy ? 3. A tavern - keeper bought six loads of hay which weighed as follows : 1725 pounds , 2163 pounds , 1581 pounds ...
... bushels , the second 7208 , the third 5047 , the fourth 12386 , and the fifth 10391 bushels : how many bushels did he buy ? 3. A tavern - keeper bought six loads of hay which weighed as follows : 1725 pounds , 2163 pounds , 1581 pounds ...
Page 45
... bushels of it : how many bushels had he left ? 3. A speculator laid out 50000 dollars in wild land , and after- wards sold it at a loss of 19046 dollars : how much did he get for his land ? 4. A man owning a block of buildings worth ...
... bushels of it : how many bushels had he left ? 3. A speculator laid out 50000 dollars in wild land , and after- wards sold it at a loss of 19046 dollars : how much did he get for his land ? 4. A man owning a block of buildings worth ...
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Common terms and phrases
acres amount Analysis annexed answer required Arithmetic avoirdupois bank discount barrel bbls bought bushels called canceling ciphers CIRCULATING DECIMALS common denominator common fraction composite number compound numbers contains cost decimal figures decimal places denotes difference dividend division dollars dolls Dry Measure equal expressed farthings Federal Money gain gallons gals given fractions given number greatest common divisor Hence hhds hundred hundredths improper fraction insured interest of $1 least common multiple miles millionths mills mixed number months multiplicand Multiply number of days number of terms Operation paid partial product payable pence period pound premium present worth prime factors principal proportion pwts quantity quotient rate per cent ratio remainder rods shillings sold subtract tenths thousandths Troy Troy pound Troy weight units usury weight whole number wine measure yard
Popular passages
Page 373 - Now (4)2+(3)2=25 sq. ft. ; and the square described on BC also contains 25 sq. ft. Hence, the square described on the hypothenuse of any right-angled triangle, is equal to the sum of the squares described on the other two sides.
Page 374 - A mean proportional between two numbers is equal to the square root of their product.
Page 272 - ... the interest on this amount for the next year, or specified time, and add it to the principal as before. Proceed in this manner with each successive year...
Page 265 - 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 385 - Divide the difference of the extremes by the common difference, and the quotient increased by 1 will be the number of terms.
Page 68 - The number to be divided is called the dividend. The number by which we divide is called the divisor.
Page 338 - X 5'" = 20"'". Hence the RULE. I. Write the several terms of the multiplier under the corresponding terms of the multiplicand. II. Multiply each term of the multiplicand by each term of the multiplier...
Page 208 - 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 263 - If the payment be less than the interest, the surplus of interest must not be taken to augment the principal; but interest continues on the former principal until the period when the payments, taken together, exceed the interest due, and then the surplus is to be applied towards discharging the principal; and interest is to be Computed on the balance, as aforesaid.
Page 370 - Multiply the divisor, thus increased, by the last figure of the root; subtract the product from the dividend, and to the remainder bring down the next period for a new dividend.