A Key to Chase's Common School Arithmetic: With Explanations and Remarks Upon the Peculiar Features of the Work, and Operations of the More Difficult ExamplesA. Hutchinson, 1853 |
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Page 3
... applications , unencumbered with diffuse explanations and un- necessary details , which rather serve to prevent the learner from obtaining a clear and ready perception of the thing to be acquired . In the study of arithmetic , it is ...
... applications , unencumbered with diffuse explanations and un- necessary details , which rather serve to prevent the learner from obtaining a clear and ready perception of the thing to be acquired . In the study of arithmetic , it is ...
Page 5
... the science and of their application , and to prepare him , by an easy progress , for the more difficult portions . * A 5 6 SUGGESTIONS TO TEACHERS AND OTHERS . Some cases have SUGGESTIONS TO TEACHERS AND OTHERS. ...
... the science and of their application , and to prepare him , by an easy progress , for the more difficult portions . * A 5 6 SUGGESTIONS TO TEACHERS AND OTHERS . Some cases have SUGGESTIONS TO TEACHERS AND OTHERS. ...
Page 6
... applications may be extended to almost all oper- ations in business , and also because its use affords one of the most thorough exercises for the discipline of the pupil's mind . It is the application of this rule that constitutes the ...
... applications may be extended to almost all oper- ations in business , and also because its use affords one of the most thorough exercises for the discipline of the pupil's mind . It is the application of this rule that constitutes the ...
Page 54
... application of ¶ 133 . Write in the numerator of a fraction the term which is of the same kind as the required answer . Take any two terms that are alike , ( of the same denomi- nation , ) and consider whether , from their relation to ...
... application of ¶ 133 . Write in the numerator of a fraction the term which is of the same kind as the required answer . Take any two terms that are alike , ( of the same denomi- nation , ) and consider whether , from their relation to ...
Page 92
... application of the rule in its most common uses . 1. Required the contents of a load of wood 9 ft . 6 in . long , 4 ft . wide , and 4 ft . high . Ans . 152 cu . ft . , = 1 % cords . - 2. What are the contents , in cubic feet , cord feet ...
... application of the rule in its most common uses . 1. Required the contents of a load of wood 9 ft . 6 in . long , 4 ft . wide , and 4 ft . high . Ans . 152 cu . ft . , = 1 % cords . - 2. What are the contents , in cubic feet , cord feet ...
Other editions - View all
A Key to Chase's Common School Arithmetic: With Explanations and Remarks ... Admiral Paschel Stone No preview available - 2017 |
A Key to Chase's Common School Arithmetic: With Explanations and Remarks ... Admiral Paschel Stone No preview available - 2017 |
A Key to Chase's Common School Arithmetic: With Explanations and Remarks ... Admiral Paschel Stone No preview available - 2016 |
Common terms and phrases
11 gall 11 spaces 20 gall 9 mo A.'s distance A.'s share Add 11 gall Amount due Amount of $1 Assume 1 lb assume 1 oz Assume 10 gall bank discount carats cent circ COMPLEX ANALYSIS complex fraction compound interest COMPOUND NUMBERS cords cost cube root Decillions Decimals Deficiency denominator diameter divide dividend Duodecillions DUODECIMALS Excess 10 ct feet frac gain Hence hour inches last root figure last term least common multiple length Mensuration miles mill MISCELLANEOUS EXAMPLES mixture Nonillions obtain Octillions Octodecillions operation of Example perform Powers Present worth principal pupil Quindecillions quotient ratio remainder Required the contents rods Septillions square SUBTRACTION Sum of products Table Take 1 less teacher text book tons Tredecillions trial divisor Vigintillions Whole Numbers worth of $1
Popular passages
Page 92 - RULE. — Multiply the length (in feet) by the width (in inches) and divide the product by 12 — the result will be the contents in square feet.
Page 24 - ... thirds, and we wish to divide it into 6ths : We have, therefore, simply to reduce thirds to sixths. 2 sixths make a third, for the unit is divided into twice as many parts, and therefore the parts are one-half as large. Hence the RULE. Divide the required denominator by the denominator of the given fraction, and multiply the quotient by the numerator. The product will be the required numerator. Art. 58. — To reduce a whole number to an equivalent fraction, having a given denominator. 1. Reduce...
Page 92 - But. in measuring timber, you may multiply the breadth in inches, ami the depth in inches, and that product by the length in feet, and divide the last product by 144, which will give the solid content in ftet, &c.
Page 91 - С in. long, 14ft. wide, and 10ft. high. The room contains 4 windows, each 3 ft. 6 in. by 5 ft. 8 in. ; 2 doors, each 6 ft. 4 in. by 2 ft.
Page 69 - The roots of fractions are obtained by extracting the root of the numerator, and of the denominator, separately.
Page 44 - The remainder will form a new principal, upon which interest is to be cast to the time of the next payment.