Elements of AlgebraA. S. Barnes, 1838 - 355 sider |
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Resultat 1-5 av 29
Side 60
... satisfy the equation ; that is , render the first member equal to the second . This latter part is called the solution of the equation . 82. An equation is said to be verified , when such a value is sub- stituted for the unknown ...
... satisfy the equation ; that is , render the first member equal to the second . This latter part is called the solution of the equation . 82. An equation is said to be verified , when such a value is sub- stituted for the unknown ...
Side 82
... satisfy these new equations . Now , if the first be multiplied by 9 , the second by 2 , and the re- sults be added together , we find 419x = 1257 , whence x = 3 . We might , by means of the two equations involving 82 ALGEBRA .
... satisfy these new equations . Now , if the first be multiplied by 9 , the second by 2 , and the re- sults be added together , we find 419x = 1257 , whence x = 3 . We might , by means of the two equations involving 82 ALGEBRA .
Side 86
... satisfy it equally well . If we had two equations involving three unknown quantities , we could in the first place eliminate one of the unknown quantities by means of the proposed equations , and thus obtain an equation , which ...
... satisfy it equally well . If we had two equations involving three unknown quantities , we could in the first place eliminate one of the unknown quantities by means of the proposed equations , and thus obtain an equation , which ...
Side 96
... satisfies both the equation and enunciation . 2. A father has lived a number a of years , his son a number of years expressed by b . Find in how many years the age of the son will be one fourth the age of the father . Let the required ...
... satisfies both the equation and enunciation . 2. A father has lived a number a of years , his son a number of years expressed by b . Find in how many years the age of the son will be one fourth the age of the father . Let the required ...
Side 97
... satisfy the enunciation of the problem , understood in its algebraic sense . 3d . If the enunciation is to be understood in its arithmetical sense , in which the quantities referred to are always supposed to be positive , then this ...
... satisfy the enunciation of the problem , understood in its algebraic sense . 3d . If the enunciation is to be understood in its arithmetical sense , in which the quantities referred to are always supposed to be positive , then this ...
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affected algebraic quantities arithmetical arranged binomial cents co-efficient common factor consequently contain continued fraction contrary signs cube root decimal deduce denominator denote divide dividend division entire number enunciation equa equal equation becomes equation involving example exponent expression extract the square figure Find the greatest find the values formula fourth fraction given number gives greater greatest common divisor greyhound Hence inequality irreducible fraction last term leaps least common multiple less letters logarithm manner monomial multiplicand multiplied negative nth root number of terms obtain operations perfect square positive roots preceding problem proposed equation proposed polynomials quotient radical reduced remainder result satisfy second degree second member second term square root substituted subtract suppose take the equation taken tens third tion transformation transposing unity unknown quantity verified whence whole number
Populære avsnitt
Side 115 - Multiply the divisor, thus augmented, by the last figure of the root, and subtract the product from the dividend, and to the remainder bring down the next period for a new dividend.
Side 148 - B, departed from different places at the same time, and travelled towards each other. On meeting, it appeared that A had travelled 18 miles more than B ; and that A could have gone B's journey in 15| days, but B would have been 28 days in performing A's journey. How far did each travel ? Ans.
Side 174 - It is required to divide the number 24 into two such parts, that their product may be equal to 35 times their difference.
Side 28 - Multiply each term of the multiplicand by each term of the multiplier, and add the partial products.
Side 183 - To express that the ratio of A to B is equal to the ratio of C to D, we write the quantities thus : A : B : : C : D ; and read, A is to B as C to D.
Side 112 - 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.
Side 190 - That is, the last term of a geometrical progression is equal to the first term multiplied by the ratio raised to a power whose exponent is one less than the number of terms.
Side 228 - Divide the first term of the remainder by three times the square of the root already found, and write the quotient for the next term of the root.
Side 92 - If A and B together can perform a piece of work in 8 days, A and c together in 9 days, and B and c in 10 days, how many days will it take each person to perform the same work alone.
Side 116 - ... brought down, there is no remainder, the proposed number is a perfect square. But if there is a remainder, you have only found the root of the greatest perfect square contained in the given number, or the entire part of the root sought. For example, if it were required to extract the square root of...