The Common School Algebra

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Taggard & Thompson, 1865
 

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Page 198 - Divide the first term of the dividend by the first term of the divisor, and write the result as the first term of the quotient. Multiply the whole divisor by the first term of the quotient, and subtract the product from the dividend.
Page 96 - Multiply all the numerators together for a new numerator, and all the denominators together for a new denominator.
Page 73 - ANOTHER. 1. Divide the coefficient of the dividend by the coefficient of the divisor. 2.
Page 168 - There will be as many figures in the root as there are periods in the given number.
Page 219 - That is, in any proportion either extreme ' is equal to the product of the means divided by the other extreme ; and either mean is equal to the product of the extremes divided by the other mean.
Page 197 - Multiply the divisor, with the term last annexed, by the last term of the root, and subtract the product from the last dividend.
Page 144 - Which proves that the square of a number composed of tens and units contains, the square of the tens plus twice the product of the tens by the units, plus the square of the units.
Page 223 - In any proportion the terms are in proportion by Composition and Division; that is, the sum of the first two terms is to their difference, as the sum of the last two terms is to their difference.
Page 235 - ... multiply the last term by the ratio, subtract the first term from this product, and divide the remainder by the ratio diminished by unity.
Page 72 - Now, since the quotient must be such a quantity as multiplied by the divisor will produce the dividend, the co-efficient of the quotient multiplied by 7 must give 35 ; hence, it is 5. Again, the exponent of each letter in the quotient must be such that when added to the exponent of the same letter in the divisor, the sum will be the exponent of that letter in the dividend. Hence, the exponent of a in the quotient is 3, the exponent of b is 1, that of c is 1, and the required quotient is 5u3bc.

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