Cambridge Senate-house Problems and Riders for the Year 1860: With SolutionsMacmillan, 1860 - 198 sider |
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Resultat 1-5 av 65
Side 2
... perpendicular to DE . Since AB . AD = AC . AE , a circle may be described about BCED . Therefore the angle BDE = BCA . Hence if A and B be fixed while C moves round the circle , the angle ADE will be constant and the locus of E will be ...
... perpendicular to DE . Since AB . AD = AC . AE , a circle may be described about BCED . Therefore the angle BDE = BCA . Hence if A and B be fixed while C moves round the circle , the angle ADE will be constant and the locus of E will be ...
Side 4
... radii CP , PO are perpendicular to the sections AB , DE , therefore VP bisects the angle between these sections . See fig . 2 . TUESDAY , Jan. 3. 1 to 4 . SENIOR MODERATOR 4 [ Jan. 3 . SENATE - HOUSE PROBLEMS AND RIDERS .
... radii CP , PO are perpendicular to the sections AB , DE , therefore VP bisects the angle between these sections . See fig . 2 . TUESDAY , Jan. 3. 1 to 4 . SENIOR MODERATOR 4 [ Jan. 3 . SENATE - HOUSE PROBLEMS AND RIDERS .
Side 12
... perpendicular to the plane r of ring , and W sin a along this plane ; .. R = W cosa and μR = W sin a sin 0 , resolving perpendicular to rod ; therefore by division sin 0 = μ cot a . vii . A point , moving with a uniform acceleration ...
... perpendicular to the plane r of ring , and W sin a along this plane ; .. R = W cosa and μR = W sin a sin 0 , resolving perpendicular to rod ; therefore by division sin 0 = μ cot a . vii . A point , moving with a uniform acceleration ...
Side 14
... perpendicular to MN . Fig . 8 . With P , Qas centres and PM , QN as radii , describe two circles ; these will in general intersect in two points S , S ' , which will be the foci of the two parabolic paths . If , however , the two ...
... perpendicular to MN . Fig . 8 . With P , Qas centres and PM , QN as radii , describe two circles ; these will in general intersect in two points S , S ' , which will be the foci of the two parabolic paths . If , however , the two ...
Side 25
... perpendicular which falls upon the opposite side . Let ABC be the triangle , O the centre of the inscribed circle , and P be the point of contact of the tangent in ques- tion so that OP makes the angle 0 with AO , ( 0 being the centre ...
... perpendicular which falls upon the opposite side . Let ABC be the triangle , O the centre of the inscribed circle , and P be the point of contact of the tangent in ques- tion so that OP makes the angle 0 with AO , ( 0 being the centre ...
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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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