Number and Its Algebra: Syllabus of Lectures on the Theory of Number and Its Algebra Introductory to a Collegiate Course in AlgebraD.C. Heath, 1903 - 219 σελίδες |
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Σελίδα 105
... coefficients . ( Vide § 73. ) If possible , first reduce by the principle of Sections 159 , 160 , e.g. , 1/3 √32 - √18 + 3 √64 = 1/3 √ ( 16 ) ( 2 ) — √ ( 9 ) ( 2 ) + 3 √ ( 4 ) ( 2 ) = 4/3 √2-3 √2 +6 √2 = ( 4/3 − 3 + 6 ) √2 ...
... coefficients . ( Vide § 73. ) If possible , first reduce by the principle of Sections 159 , 160 , e.g. , 1/3 √32 - √18 + 3 √64 = 1/3 √ ( 16 ) ( 2 ) — √ ( 9 ) ( 2 ) + 3 √ ( 4 ) ( 2 ) = 4/3 √2-3 √2 +6 √2 = ( 4/3 − 3 + 6 ) √2 ...
Σελίδα 132
... coefficients of expansion need to be accurately known for all substances employed . The origi- nal ( 1799 ) French standard metre is a platinum bar end - standard about 1 inch wide and 4 inch thick . End- standards are objectionable ...
... coefficients of expansion need to be accurately known for all substances employed . The origi- nal ( 1799 ) French standard metre is a platinum bar end - standard about 1 inch wide and 4 inch thick . End- standards are objectionable ...
Σελίδα 157
... coefficient . Care must be taken to supply by zeros any lacking terms in a particular case . EXAMPLE . Divide 3 x 5 x2 + 3 + 0 + 59 +11 +6 + 12 + 34 + 50 + 3 + 6 +17 + 25 + 61 • • 9x11 by x 2 . ( Coefficients of dividend ) . ( Each ...
... coefficient . Care must be taken to supply by zeros any lacking terms in a particular case . EXAMPLE . Divide 3 x 5 x2 + 3 + 0 + 59 +11 +6 + 12 + 34 + 50 + 3 + 6 +17 + 25 + 61 • • 9x11 by x 2 . ( Coefficients of dividend ) . ( Each ...
Σελίδα 160
... coefficients are independent of the varia- bles , if two integral functions are equal as an identity ( vide § 40 ) , and the coefficients of one are determined by any means , then these coefficients are determined once for all . This ...
... coefficients are independent of the varia- bles , if two integral functions are equal as an identity ( vide § 40 ) , and the coefficients of one are determined by any means , then these coefficients are determined once for all . This ...
Σελίδα 161
... more untrue , the more numerous the terms ; but if the remainder be added at any stage , we we have 1 — 9 have a true equation : - −1 = 1 + 10 + 100 + 1000_9 + 90 + 900 = 9 1000 9 = • or 1 ( 1 + 1 ) ( 1 + UNDETERMINED COEFFICIENTS . 161.
... more untrue , the more numerous the terms ; but if the remainder be added at any stage , we we have 1 — 9 have a true equation : - −1 = 1 + 10 + 100 + 1000_9 + 90 + 900 = 9 1000 9 = • or 1 ( 1 + 1 ) ( 1 + UNDETERMINED COEFFICIENTS . 161.
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