Cambridge Senate-house Problems and Riders for the Year 1860: With SolutionsMacmillan, 1860 - 198 sider |
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Side 2
... intersection of the two circles employed in the construction , and E the other , and AE be drawn meeting BD produced in F , prove that FAB is another isosceles triangle of the same kind . = For ADE is an isosceles triangle , and the ...
... intersection of the two circles employed in the construction , and E the other , and AE be drawn meeting BD produced in F , prove that FAB is another isosceles triangle of the same kind . = For ADE is an isosceles triangle , and the ...
Side 3
... intersections with the three straight lines are not all in one straight line . This may be easily proved by a " reductio ad absurdum . " vii . Define a parabola : and prove from the definition that it cannot be cut by a straight line in ...
... intersections with the three straight lines are not all in one straight line . This may be easily proved by a " reductio ad absurdum . " vii . Define a parabola : and prove from the definition that it cannot be cut by a straight line in ...
Side 14
... intersect in two points S , S ' , which will be the foci of the two parabolic paths . If , however , the two circles touch one another , there will be but one parabolic path . In order that this may be the case , we must have QN + PM ...
... intersect in two points S , S ' , which will be the foci of the two parabolic paths . If , however , the two circles touch one another , there will be but one parabolic path . In order that this may be the case , we must have QN + PM ...
Side 23
... intersection of these lines will ultimately coincide with the point of contact P , and we have △ APA ' = △ BPB ' ultimately ; .. AP . A'P sin APA ' = BP . B'P sin BPB ' , AP.A'P = BP.B'P ; .. AP = BP , since A'P and B'P are ultimately ...
... intersection of these lines will ultimately coincide with the point of contact P , and we have △ APA ' = △ BPB ' ultimately ; .. AP . A'P sin APA ' = BP . B'P sin BPB ' , AP.A'P = BP.B'P ; .. AP = BP , since A'P and B'P are ultimately ...
Side 26
... intersection of the string with the directions of the frictions at the points of support is an arc of a circle and a part of a straight line . Find also how the force must be applied that its inter- sections with the frictions may trace ...
... intersection of the string with the directions of the frictions at the points of support is an arc of a circle and a part of a straight line . Find also how the force must be applied that its inter- sections with the frictions may trace ...
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Cambridge Senate-house Problems and Riders for the Year 1860: With Solutions Henry William Watson,Edward John Routh Uten tilgangsbegrensning - 1860 |
Cambridge Senate-house Problems and Riders for the Year 1860: With Solutions Henry William Watson,Edward John Routh Uten tilgangsbegrensning - 1860 |
Cambridge Senate-house Problems and Riders for the Year 1860: With Solutions Henry William Watson,Edward John Routh Uten tilgangsbegrensning - 1860 |
Vanlige uttrykk og setninger
acceleration angular velocity Arabic numbers asymptotes attraction axes axis bisects Cambridge centre of force centre of gravity chord circle circular cloth co-ordinates cone conic conic section constant cos² Crown 8vo curvature cylinder described diameter differential equation direction directrix distance drawn dt dt ellipse ellipsoid equal angles equiangular spiral equilibrium extremity Fcap fixed point fluid foci focus given point Hence horizontal hyperbola inclined intersection latus rectum length locus MACMILLAN & CO.'S meridian motion normal parabola parallel particle perpendicular point of contact position pressure prove ratio respectively right angles Roman numbers SENIOR MODERATOR shew sides sin² singular solution smooth sphere spherical straight line string subtend surface tangent tension triangle Trinity College uniform vertex vertical Vis Viva weight аф
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