An Introduction to Algebra: Upon the Inductive Method of InstructionHilliard, Gray & Company, 1837 - 276 σελίδες |
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Αποτελέσματα 1 - 5 από τα 23.
Σελίδα 3
... solve several by himself . The most simple combinations are given first , then those which are more difficult . The learner is expected to derive most of his knowledge by solving the examples himself ; therefore care has been taken to ...
... solve several by himself . The most simple combinations are given first , then those which are more difficult . The learner is expected to derive most of his knowledge by solving the examples himself ; therefore care has been taken to ...
Σελίδα 5
... solve every question in course , and do it algebraically . If he finds a ques- tion which he can solve as easily without the aid of algebra as with it , he may be assured , this is what the author expected . If he first solves a ...
... solve every question in course , and do it algebraically . If he finds a ques- tion which he can solve as easily without the aid of algebra as with it , he may be assured , this is what the author expected . If he first solves a ...
Σελίδα 9
... solving abstruse and complicated questions consists in discovering how the operations are to be applied . It is often difficult , and sometimes absolutely impossible to discover , by the ordinary modes of reasoning , how the funda ...
... solving abstruse and complicated questions consists in discovering how the operations are to be applied . It is often difficult , and sometimes absolutely impossible to discover , by the ordinary modes of reasoning , how the funda ...
Σελίδα 13
... solving or reducing the equation . No rules can be given for putting questions into equations ; this must be learned by practice ; but rules may be found for solving most of the equations that ever occur . After the preceding questions ...
... solving or reducing the equation . No rules can be given for putting questions into equations ; this must be learned by practice ; but rules may be found for solving most of the equations that ever occur . After the preceding questions ...
Σελίδα 26
... solve than any of the preceding . In the first place I subtract 76 from both members , so as to remove it from the first member . Then to get 3x out of the second member , which is there subtracted , I add 3 x to both members ; then the ...
... solve than any of the preceding . In the first place I subtract 76 from both members , so as to remove it from the first member . Then to get 3x out of the second member , which is there subtracted , I add 3 x to both members ; then the ...
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Άλλες εκδόσεις - Προβολή όλων
An Introduction to Algebra Upon the Inductive Method of Instruction Warren Colburn Πλήρης προβολή - 1826 |
An Introduction to Algebra Upon the Inductive Method of Instruction Warren Colburn Πλήρης προβολή - 1831 |
An Introduction to Algebra upon the Inductive Method of Instruction Warren Colburn Πλήρης προβολή - 1844 |
Συχνά εμφανιζόμενοι όροι και φράσεις
12 rods 3d power 3d root 5th power a b c A's share a² b² a² b³ ac² added algebra algebraic quantities apples approximate root Arith arithmetic becomes binomial Binomial Theorem bought breadth bushels coefficient compound interest compound quantities consisting contained decimal difference divide the number dividend division divisor equal equation example exponent expression factor figure formula fourth fraction gallons gives greater Hence length less Let the learner letter logarithm merator miles multiplicand negative quantity number of terms observe pears question quotient remainder required to find rule second power second root second term shillings sold subtracted Suppose third power third root twice unknown quantity whole number yards zero
Δημοφιλή αποσπάσματα
Σελίδα 186 - The 3d power of (2 a — rf)4 is (2a — rf)^«+« = (2a — d)4x3=(2a — d)". That is, any quantity, which is already a power of a compound quantity, may be raised to any power by multiplying its exponent by the exponent of the power to which it is to be raised. 7. Express the 2d power of (3 b — c)4. 8. Express the 3d power of (a — c -J- 2 d)*. 9. Express the 7th power of (2 a* — 4 c3)3.
Σελίδα 2 - DISTRICT OF MASSACHUSETTS, TO WIT: District Clerk's Office. BE IT REMEMBERED, that on the...
Σελίδα 101 - 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.
Σελίδα 92 - It will be seen by the above section that if both the numerator and denominator be multiplied by the same number, the value of the fraction will not be altered...
Σελίδα 2 - District Clerk's Office. BE IT REMEMBERED, That on the seventh day of May, AD 1828, in the fifty-second year of the Independence of the UNITED STATES OF AMERICA, SG Goodrich, of the said District, has deposited in this office the...
Σελίδα 21 - A cask, which held 146 gallons, was filled with a mixture of brandy, wine, and water. In it there were 15 gallons of wine more than there were of brandy, and as much water as both wine and brandy. What quantity was there of each...
Σελίδα 232 - I, n, d, and. S; any three of which being given, the other two may be found, by combining the two equations. I shall leave the learner to trace these ' himself as occasion may require. Examples in Progression by Difference.
Σελίδα 35 - How many days did he work, and how many days was he idle ? Let x = the number of days he worked.
Σελίδα 229 - Hence, any term may be found by adding the product of the common difference by the number of terms less one, to the first term.
Σελίδα 273 - A gentleman bought a rectangular lot of valuable land, giving 10 dollars for every foot in the perimeter. If the same quantity had been in a square, and he had bought it in the same way, it would have cost him $33 less ; and if he had bought a square piece of the same perimeter he would have had 12^ rods more.