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12 rods A's share added algebraic ALGEBRAIC QUANTITIES amount answer arithmetical arithmetical progression arithms bought bushels called cents coefficient compound interest denominator denote Divide the number dividend divisor dollars equa equal equation equidifference evident exponent Extract the square extracting the root factors feet find a number find the square Find the values following rule formula fraction gain gallons gentleman given number gives greater last term learner least common multiple less letter logarithm man's share manner metic miles monomials multiplied negative quantity number of terms obtain operations perfect square perform persons piece polynomials positive pound progression by difference progression by quotient proportion proposed ratio received reducing remainder result rods second degree sheep shillings solution square root subtracted third tion traveled twice unknown quantity value of x whence WILLIAM SMYTH
Page 188 - A labourer dug two trenches, one of which was 6 yards longer than the other, for 17/. 16s., and the digging of each of them cost as many shillings per yard as there were yards in its length. What was the length of each ? Ans. 10, and 16 yards.
Page 119 - In a series of equal ratios, the sum of the antecedents is to the sum of the consequents as any antecedent is to its consequent.
Page 93 - What fraction is that, to the numerator of which if 1 be added, the value...
Page 139 - Multiply £ the sum of the extremes by the number of terms, and the product will be the answer 10.
Page 61 - 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 57 - Divide the coefficient of the dividend by the coefficient of the divisor.
Page 179 - Divide the number 14 into two such parts, that the quotient of the greater divided by the less, shall be to the quotient of the less divided by the greater as 16 to 9.
Page 13 - It is required to divide the number 99 into five such parts, that the first may exceed the second by 3, be less than the third by 10, greater than the fourth by 9, and less than the fifth by 16. Let x= the first part.
Page 192 - This result indicates, that there is some absurdity in the conditions of the question proposed, since in order to obtain the value of x, we must extract the root of a negative quantity, which is impossible. In order to see in what this absurdity consists, let us examine into what two parts a given number should be divided, in order that the product of these parts may be the greatest possible. Let us represent the given number by p, the product of the two parts by q, and the difference of the two...