An Introduction to Algebra: Upon the Inductive Method of InstructionHilliard, Gray & Company, 1841 - 276 σελίδες |
Αναζήτηση στο βιβλίο
Αποτελέσματα 1 - 5 από τα 22.
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 ...
Περιεχόμενα
13 | |
21 | |
45 | |
65 | |
74 | |
80 | |
89 | |
95 | |
193 | |
195 | |
197 | |
202 | |
208 | |
217 | |
221 | |
226 | |
102 | |
112 | |
121 | |
131 | |
142 | |
150 | |
159 | |
175 | |
182 | |
189 | |
228 | |
233 | |
239 | |
242 | |
254 | |
256 | |
260 | |
264 | |
267 | |
268 | |
Άλλες εκδόσεις - Προβολή όλων
An Introduction to Algebra: Upon the Inductive Method of Instruction Warren Colburn Πλήρης προβολή - 1837 |
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 |
Συχνά εμφανιζόμενοι όροι και φράσεις
12 rods 3d power 3d root 5th power a b c A's share a² b² a² b³ added algebra algebraic quantities apples approximate root Arith arithmetic becomes binomial Binomial Theorem bought breadth bushels cents apiece coefficient compound interest compound quantities consisting contain decimal denominator difference divide the number dividend divisor equal equation example exponent expressed factor figure formula found by multiplying fourth fraction gallons gives greater Hence horse length less Let the learner letter logarithm merator miles multiplicand number of dollars number of sheep number of terms observe pears question quotient remainder required to find rods 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 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 17 - ... of the length of the body, and the body is as long as the head and tail both. What is the whole length of the fish ? 17.
Σελίδα 82 - In each of these cases the division is to be indicated by writing . the divisor under the dividend, in the form of a fraction.
Σελίδα 83 - ... arithmetic. It was shown in arithmetic, that a fraction properly expresses a quotient. Algebraic fractions are subject to precisely the same rules as fractions in arithmetic. Many of the operations are more easily performed on algebraic fractions. In these, as in arithmetic, it must be kept in mind, that the denominator shows into how many parts a unit is divided ; and the numerator shows how many of those parts are used ; or the denominator shows into how many parts the numerator is divided....
Σελίδα 173 - B set out from two towns, which were distant 247 miles, and travelled the direct road till they met. A went 9 miles a day ; and the number of days, at the end of which they met, was greater by 3 than the number of miles which B went in a day. How many miles did each go ? • . ,.l \[ 17.
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 162 - ... 8. A man wishes to make a cistern which shall hold 500 gallons, in such a form that the length shall be to the breadth as 5 to 4, and the depth to the length as 2 to 5. Required the length, breadth, and depth.
Σελίδα 232 - The sum of all the terms. Any three of which being given, the other two may be found.
Σελίδα 270 - A detachment of soldiers from a regiment being ordered to march on a particular service, each company furnished four times as many men as there were companies in the...
Σελίδα 141 - ... 5th. Bring down the next two figures as before, to form a new dividend, and double the root already found, for a divisor, and proceed as before. The root will be doubled, if the right hand figure of the last divisor be doubled. If it happens that the divisor is not contained in the dividend when the right hand figure is rejected, a zero must be written in the root, and also at the right of the divisor ; and the next figures must be brought down, and then a new trial made. If it happens that the...