An Elementary Treatise on Algebra: To which are Added Exponential Equations and Logarithms |
From inside the book
Results 1-5 of 88
Page 6
... Find the sum of the similar positive terms by add- ing their coefficients , and in the same way the sum of the ... 4 - Ans . 13 a4 . 5 a to its simplest form . 2. Reduce the polynomial 5 a4 b + 3ab2c - 7ab + 17 ab + 2ab2c - 6a2b ...
... Find the sum of the similar positive terms by add- ing their coefficients , and in the same way the sum of the ... 4 - Ans . 13 a4 . 5 a to its simplest form . 2. Reduce the polynomial 5 a4 b + 3ab2c - 7ab + 17 ab + 2ab2c - 6a2b ...
Page 7
... Find the sum of a and a . Ans . 2 a . 2. Find the sum of 11 x and 9 x . Ans . 20 x . 3. Find the sum of 11 z and 9 π . Ans . 2 x . 4. Find the sum of - 11 x and 9x . Ans . 2 x . 6. Find the sum of a and b . 5. Find the sum of - 11x and - 9 ...
... Find the sum of a and a . Ans . 2 a . 2. Find the sum of 11 x and 9 x . Ans . 20 x . 3. Find the sum of 11 z and 9 π . Ans . 2 x . 4. Find the sum of - 11 x and 9x . Ans . 2 x . 6. Find the sum of a and b . 5. Find the sum of - 11x and - 9 ...
Page 8
To which are Added Exponential Equations and Logarithms Benjamin Peirce. Subtraction . 10. Find the sum of 7 x2 - 6√2 + 5xz + 3 - g -x23I -8 - g -x2 + √ - 3x3z - 1 + 7g -2x2 + 3 + 3xz - 1 - g 22 + 81-5x3z + 9 - g Ans . 4 x2 + 3√ + 2 + ...
To which are Added Exponential Equations and Logarithms Benjamin Peirce. Subtraction . 10. Find the sum of 7 x2 - 6√2 + 5xz + 3 - g -x23I -8 - g -x2 + √ - 3x3z - 1 + 7g -2x2 + 3 + 3xz - 1 - g 22 + 81-5x3z + 9 - g Ans . 4 x2 + 3√ + 2 + ...
Page 9
... 4. From 7 a take 12 a . Ans . - 5 a . 5. From 19 a take - 20 a . Ans . a . 6. From 12 take - 7 . Ans . 19 . 7 ... find the continued product of several monomials . Solution . The required product is indicated by writing the given ...
... 4. From 7 a take 12 a . Ans . - 5 a . 5. From 19 a take - 20 a . Ans . a . 6. From 12 take - 7 . Ans . 19 . 7 ... find the continued product of several monomials . Solution . The required product is indicated by writing the given ...
Page 10
... 4 a b3 c . 6. Find the continued product of 5 a3 64 , a2 68 , and Ans . 20 a6 615 с . 7. Find the continued product of am br cq , an brc , and am + n b . Ans . a2m + 2 n bp + r + 1cq + 30. Problem . To find the product of two poly ...
... 4 a b3 c . 6. Find the continued product of 5 a3 64 , a2 68 , and Ans . 20 a6 615 с . 7. Find the continued product of am br cq , an brc , and am + n b . Ans . a2m + 2 n bp + r + 1cq + 30. Problem . To find the product of two poly ...
Other editions - View all
Common terms and phrases
126 become zero 3d root arithmetical mean arithmetical progression Binomial Theorem coefficient commensurable roots common difference contained continued fraction continued product Corollary deficient terms denote derivative Divide dividend division equal roots equal to zero equation x² factor Find the 3d Find the 4th Find the continued Find the greatest Find the number Find the square Find the sum Free the equation Geometrical Progression given equation gives greatest common divisor Hence imaginary roots last term least common multiple letter logarithm monomials multiplied number of real number of terms polynomial positive roots preceding article Problem quantities in example quotient radical quantities ratio real roots reduced remainder required equation required number row of signs Scholium Second Degree Solution Solve the equation square root Sturm's Theorem subtracted Theorem unity unknown quan unknown quantity variable whence
Popular passages
Page 48 - 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 55 - There is a number consisting of two digits, the second of which is greater than the first, and if the number be divided by the sum of its digits, the quotient is 4...
Page 130 - The rule of art. 28, applied to this case, in which the factors are all equal, gives for. the coefficient of the required power the same power of the given coefficient, and for the exponent of each letter the given exponent added to itself as many times as there are units in the exponent of the required power. Hence...
Page 127 - Multiply the divisor, thus increased, by the last figure of the root; subtract the product from the dividend, and to the remainder bring down the next period for a new dividend.
Page 159 - A certain capital is let at 4 per cent. ; if we multiply the number of dollars in the capital, by the number of dollars in the interest for 5 months, we obtain 11?041§.
Page 172 - Ans. 15 and 26. 31. What two numbers are they, whose sum is a, and the sum of whose squares is b 1 Ans.
Page 232 - An equation of any degree whatever cannot have a greater number of positive roots than there are variations in the signs of Us terms, nor a greater number of negative roots than there are permanences of these signs.
Page 63 - A term may be transposed from one member of an equation to the other by changing its sign.
Page 45 - Given three terms of a proportion, to find the fourth. Solution. The following solution is immediately obtained from the test. When the required term is an extreme, divide the product of the means by the given extreme, and the quotient is the required extreme. When the required term is a mean, divide the product of the extremes by the given mean, and the quotient is the required mean.
Page 196 - Hence, to find the sum, multiply the first term by the difference between unity and that power of the ratio whose exponent is equal to the number of terms, and divide the product by the difference between unity and the ratio. Examples in Geometrical Progression.