The Principles of the Solution of the Senate-house 'riders,' Exemplified by the Solution of Those Proposed in the Earlier Parts of the Examinations of the Years 1848-1851Macmillan & Company, 1851 - 116 sider |
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Resultat 1-5 av 23
Side 13
... axis - major is equal to the same line . ( B ) . Shew that the axis - major is greater than any other diameter . Let PCP ' ( fig . 16 ) be any diameter . Join P and P ' with the foci . and Then , since two sides of a triangle are ...
... axis - major is equal to the same line . ( B ) . Shew that the axis - major is greater than any other diameter . Let PCP ' ( fig . 16 ) be any diameter . Join P and P ' with the foci . and Then , since two sides of a triangle are ...
Side 14
... axis produced in U , and PN be drawn perpen- dicular to the minor axis , then CN : CB :: CB : CU . ( B ) . If a series of ellipses be described having the same major axis , the tangents at the extremities of their latera- recta will all ...
... axis produced in U , and PN be drawn perpen- dicular to the minor axis , then CN : CB :: CB : CU . ( B ) . If a series of ellipses be described having the same major axis , the tangents at the extremities of their latera- recta will all ...
Side 15
... axis - major as diameter . ( B. ) Employ this proposition to find the locus of the intersection of a pair of tangents at right angles to each other . ( C ) . Deduce from the proposition an analogous one for the parabola . ( B ) . Let SY ...
... axis - major as diameter . ( B. ) Employ this proposition to find the locus of the intersection of a pair of tangents at right angles to each other . ( C ) . Deduce from the proposition an analogous one for the parabola . ( B ) . Let SY ...
Side 17
... axis with the ordinate and tangent respectively , is equal to the square of the semi - axis major . ( CN.CT = AC2 ) . ( B ) . Through N draw NQ , parallel to AP , to meet CP in Q ; prove that AQ is parallel to the tangent at P. Since CN ...
... axis with the ordinate and tangent respectively , is equal to the square of the semi - axis major . ( CN.CT = AC2 ) . ( B ) . Through N draw NQ , parallel to AP , to meet CP in Q ; prove that AQ is parallel to the tangent at P. Since CN ...
Side 18
... axis , is a parabola . ( B ) . The foci of all parabolic sections which can be cut from a given right cone lie upon the surface of another cone . In the investigation of ( 4 ) we obtain AL2 PN2 .AN . ( fig . 24 ) . BL See Goodwin's ...
... axis , is a parabola . ( B ) . The foci of all parabolic sections which can be cut from a given right cone lie upon the surface of another cone . In the investigation of ( 4 ) we obtain AL2 PN2 .AN . ( fig . 24 ) . BL See Goodwin's ...
Andre utgaver - Vis alle
The Principles of the Solution of Senate-house 'riders': Exemplified by the ... Francis James Jameson Uten tilgangsbegrensning - 1851 |
The Principles of the Solution of the Senate-House 'Riders: Exemplified by ... Francis J. Jameson Ingen forhåndsvisning tilgjengelig - 2018 |
The Principles of the Solution of the Senate-House 'Riders: Exemplified by ... Francis J. Jameson Ingen forhåndsvisning tilgjengelig - 2015 |
Vanlige uttrykk og setninger
AC² AN.NM Arithmetic arithmetical progression axis bisects body C₁ Cambridge centre of gravity chord CHURCHILL BABINGTON circle cloth cone Conic Sections conjugate hyperbola constant curvature curve cycloid describe diameter direction directrix distance drawn Edition ellipse equations equilibrium Fellow of St fluid focus geometrical given point Hence horizontal hyperbola inches inclined inscribed John's College joining latus-rectum least common multiple Lemma length locus meet mirror move number of seconds oscillation parabola parallel parallelogram particle perpendicular plane polygon pressure prop proportional proposition prove pullies quadrilateral quantity radius ratio rays rectangle refraction right angles sewed shew sides specific gravity spherical square straight line string surface tan² tangent triangle ABC Trinity College tube V₁ vary vertex vertical W₁ weight
Populære avsnitt
Side 4 - To divide a given straight line into two parts, so that the rectangle contained by the whole and one of the parts, shall be equal to the square of the other part.
Side 6 - The angle at the centre of a circle is double of the angle at the circumference upon the same base, that is, upon the same part of the circumference.
Side 11 - AB is a diameter, and P any point in the circumference of a circle; AP and BP are joined and produced if necessary ; if from any point C of AB, a perpendicular be drawn to AB meeting AP and .BP in points D and E respectively, and the circumference of the circle in a point F, shew that CD is a third proportional of CE and CF.
Side 9 - IF the angle of a triangle be divided into two equal angles, by a straight line which also cuts the base; the segments of the base shall have the same ratio which the other sides of the triangle have to one another...
Side 4 - In any right-angled triangle, the square which is described upon the side subtending the right angle, is equal to the squares described upon the sides which contain the right angle.