A Treatise of Algebra: Wherein the Principles are Demonstrated ... To which is Added, the Geometrical Construction of a Great Number of Linear and Plane Problems ...M. Carey & sons, 1821 - 408 σελίδες |
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Σελίδα 247
... parallel to CF , meeting AO in n , and BH and mv parallel to AO , meeting DG in H and v : then , because the arches BC and CD are equal to each D B H AE Fn GO other , OC is not only perpendicular to BD , but also bi- sects it ( Euc . 3 ...
... parallel to CF , meeting AO in n , and BH and mv parallel to AO , meeting DG in H and v : then , because the arches BC and CD are equal to each D B H AE Fn GO other , OC is not only perpendicular to BD , but also bi- sects it ( Euc . 3 ...
Σελίδα 251
... parallel to FB ) it is plain , that , if BD be made the ra- dius , BF will be the tangent of BDF , and DE the tangent of DBE but , because of the similar triangles CFB and CDE , CF : CD :: BF : DE ; that is , as the sum of the sides AC ...
... parallel to FB ) it is plain , that , if BD be made the ra- dius , BF will be the tangent of BDF , and DE the tangent of DBE but , because of the similar triangles CFB and CDE , CF : CD :: BF : DE ; that is , as the sum of the sides AC ...
Σελίδα 254
... parallel or perpendicular to other lines in the figure , or so as to form similar triangles ; and if an angle be given , let the perpendicular be opposite to that angle , and also fall from the end of a given line , if possible . 2o ...
... parallel or perpendicular to other lines in the figure , or so as to form similar triangles ; and if an angle be given , let the perpendicular be opposite to that angle , and also fall from the end of a given line , if possible . 2o ...
Σελίδα 269
... parallel to CQ ; moreover , from the centre F , let FA , FB , and FC be drawn ; also let CE be drawn ( parallel to AB ) . Put the sine of the given angle ACB , to the radius 1 , = m , its co - sinen , the semi- base BD = a , the bisect ...
... parallel to CQ ; moreover , from the centre F , let FA , FB , and FC be drawn ; also let CE be drawn ( parallel to AB ) . Put the sine of the given angle ACB , to the radius 1 , = m , its co - sinen , the semi- base BD = a , the bisect ...
Σελίδα 275
... parallel to BA ; then CH being = and CK = mc n n our equation will be changed to x2 + xx CH = AC × CK , or to CDxCD + CH = AC × CK . Upon CH as a diameter let the circle CTHQ be described , in which in- scribe CG AK ; and in CG produced ...
... parallel to BA ; then CH being = and CK = mc n n our equation will be changed to x2 + xx CH = AC × CK , or to CDxCD + CH = AC × CK . Upon CH as a diameter let the circle CTHQ be described , in which in- scribe CG AK ; and in CG produced ...
Συχνά εμφανιζόμενοι όροι και φράσεις
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Δημοφιλή αποσπάσματα
Σελίδα 241 - The circumference of every circle is supposed to be divided into 360 equal parts, called degrees ; and each degree into 60 equal parts, called minutes ; and each minute into 60 equal parts, called seconds ; and these into thirds, &c.
Σελίδα 53 - Multiply the numerators together for a new numerator, and the denominators together for a new denominator.
Σελίδα 64 - ... then, by adding, or subtracting, the two equations thus obtained, as the case may require, there will arise a new equation, with only one unknown quantity in it, which may be resolved as before.
Σελίδα 251 - ... the sum of the segments of the base is to the sum of the sides as the difference of the sides to the difference of the segments of the base.
Σελίδα 87 - A composition of copper and tin containing 100 cubic inches weighed 505 ounces. How many ounces of each metal did it contain, supposing a cubic inch of copper to weigh of ounces, and a cubic inch of tin to weigh 4т ounces ? Ans. 420 of copper, and
Σελίδα 88 - ... half of what he had left, and half a sheep over ; and, soon after this, a third party met him, and used him in the same manner, and then he had only five sheep left. It is required to find what number of sheep he had at first, Ans, 47 sheep.
Σελίδα 254 - The following particular directions, however, may be of some use. 1st, In preparing the figure, by drawing lines, let them be either parallel or perpendicular to other lines in the figure, or so as to form similar triangles. And if an angle be given, it will be proper to let the perpendicular be opposite to that angle, and to fall from one end of a given line, if possible.