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" Hence we derive the following RULE. Divide the coefficient of the dividend by the coefficient of the divisor, and strike out the letters of the divisor from the dividend. "
An Introduction to Algebra Upon the Inductive Method of Instruction - Σελίδα 81
των Warren Colburn - 1831 - 276 σελίδες
Πλήρης προβολή - Σχετικά με αυτό το βιβλίο

A First Course in Elementary Algebra

Jacob William Albert Young - 1908
...from that of the dividend. For example: o»_ tf"_ 149. To find the quotient of two positive monomials divide the coefficient of the dividend by the coefficient of the divisor, and to this quotient annex each letter with an exponent equal to the exponent of that letter in the dividend,...

Elementary Algebra

Jacob William Albert Young, Lambert Lincoln Jackson - 1908 - 438 σελίδες
...subtract the exponent of the divisor from that of the dividend. 149. To find the quotient of two monomials divide the coefficient of the dividend by the coefficient of the divisor, and to this quotient annex each letter with an exponent equal to ike exponent of that letter in the dividend,...

First Course in Algebra: With Eight Thousand Examples Including Three ...

Albert Harry Wheeler - 1908 - 664 σελίδες
...the radicands of the dividend and divisor. for the coefficient of the radical part of the quotient divide the coefficient of the dividend by the coefficient of the divisor, and for the radicand of the quotient divide the radicand of the dividend by the radicand of the divisor....

Elementary Algebra

Frederick Howland Somerville - 1908 - 407 σελίδες
...а3a?у2. . = _ 5 a5-3¡,:3-32^-2 _ _ 5 aгyí Result. 7, Hence, to divide a monomial by a monomial : 81. Divide the coefficient of the dividend by the coefficient of the divisor, annexing to the result the literal factors, each 1vith an exponent equal to its exponent in the dividend...

School Algebra

James William Nicholson - 1909 - 316 σελίδες
...-s. ó = 1, because 1 xb = b c3 -H e = с2, because с2 x с = e3. Hence, the quotient is - 3 а"с2. Divide the coefficient of the dividend by the coefficient of the divisor, observing the law of signs. To this quotient annex each literal factor of the dicidend with an exponent...

A Grammar School Algebra

Fletcher Durell, Edward Rutledge Robbins - 1909 - 258 σελίδες
...Arts. 62, 63, 64, we have the following general process for the division of one monomial by another : Divide the coefficient of the dividend by the coefficient of the divisor ; trading the exponent of each letter in the divisor from the exponent of the same letter in the dividend...

Durell's Introductory Algebra

Fletcher Durell - 1912 - 286 σελίδες
...in Arts. 60, 61, and 62, we have the following method for the division of one monomial by another: Divide the coefficient of the dividend by the coefficient of the divisor; Obtain the exponent of each literal factor in the quotient by subtracting the exponent of each letter...

Elementary Algebra Revised

Frederick Howland Somerville - 1913 - 447 σελίδες
...monomial. Illustration : 1. Divide - 35 aVy4 by 7 aWy2. Hence, to divide a monomial by a monomial: 81. Divide the coefficient of the dividend by the coefficient of the divisor, annexing to the result the literal factors, each with an exponent equal to its exponent in the dividend...

Durell's School Algebra

Fletcher Durell - 1914 - 523 σελίδες
...in Arts. 60, 61, and 62, we have the following method for the division of one monomial by another: Divide the coefficient of the dividend by the coefficient of the divisor; Obtain the exponent of each literal factor in the quotient by subtracting the exponent of each letter...

Durell's Algebra: Two Book Course. Book One

Fletcher Durell - 1914 - 393 σελίδες
...in Arts. 60, 61, and 62, we have the following method for the division of one monomial by another: Divide the coefficient of the dividend by the coefficient of the divisor; Obtain the exponent of each literal factor in the quotient by subtracting the exponent of each letter...




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