Solutions of the problems and riders proposed in the Senate-house examination for 1854, by the moderators and examiners |
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Resultat 1-5 av 34
Side 23
... condition the arc drawn from perpendicular to hk = 4 ; therefore 1 = 7 tan cota + tan cotß = cota + cot3ß = p2 + q3 ... .. ( 1 ) , writing p and q for cota and cotß . Again , the equation to the cutting plane referred to OA , OB , OC as ...
... condition the arc drawn from perpendicular to hk = 4 ; therefore 1 = 7 tan cota + tan cotß = cota + cot3ß = p2 + q3 ... .. ( 1 ) , writing p and q for cota and cotß . Again , the equation to the cutting plane referred to OA , OB , OC as ...
Side 25
... condition Now ху a b when x = y = √ / 2 √ / 2 r ' y2 + a2 b2 x . = .y . 1 , ; hence has a maximum value equal to r 3 a - • 2 b a 8. Trace the curve whose equation is y2 = - x ( x - a ) ' first supposing a to be less than c , then ...
... condition Now ху a b when x = y = √ / 2 √ / 2 r ' y2 + a2 b2 x . = .y . 1 , ; hence has a maximum value equal to r 3 a - • 2 b a 8. Trace the curve whose equation is y2 = - x ( x - a ) ' first supposing a to be less than c , then ...
Side 31
... condition it will plainly represent an ellipse , its equation being of the form h2 k2 + = = n2 . a2 α b b'b : COR . Let a = a , b ' b : then the equation to the locus of P becomes ( + ) { + 2 2 = 9 . + 72 h2 k2 Solving this quadratic ...
... condition it will plainly represent an ellipse , its equation being of the form h2 k2 + = = n2 . a2 α b b'b : COR . Let a = a , b ' b : then the equation to the locus of P becomes ( + ) { + 2 2 = 9 . + 72 h2 k2 Solving this quadratic ...
Side 44
... conditions of the problem . Now x and z are to be two independent variables ; there- fore , which denotes the value of x in terms of the new variables , stands for x , and ƒ denotes the function that y is of x , z : hence do do 1 , dx ...
... conditions of the problem . Now x and z are to be two independent variables ; there- fore , which denotes the value of x in terms of the new variables , stands for x , and ƒ denotes the function that y is of x , z : hence do do 1 , dx ...
Side 47
... a ) da = π ( '17 ) nearly . 0 6. Determine the form of the function f ( 0 ) from the equation f ( 20 ) = cose ƒ ( 0 ) ; with the condition ƒ ( 0 ) = m . arc . Apply the result to find the centre of 91-121 . ] 47 PROBLEMS .
... a ) da = π ( '17 ) nearly . 0 6. Determine the form of the function f ( 0 ) from the equation f ( 20 ) = cose ƒ ( 0 ) ; with the condition ƒ ( 0 ) = m . arc . Apply the result to find the centre of 91-121 . ] 47 PROBLEMS .
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Solutions of the Problems and Riders Proposed in the Senate-House ... Exam Papers Cambridge Univ Ingen forhåndsvisning tilgjengelig - 2016 |
Vanlige uttrykk og setninger
angular velocity asymptote axes Cambridge catenary centre of force centre of gravity chord circle cloth conic section constant cos² cose cosẞ cota cotß Crown 8vo curvature curve cylinder denoting described diameter direction distance dx dy dy dx ecliptic elastic ellipse equal equation equilibrium Fellow of St fixed point fluid geometrical progression given Hence horizontal hyperbola inclined plane intersection lamina latus rectum length locus longitude M.A. Fellow major axis middle point motion orbit parabola parallel particle passing perpendicular position pressure projection prove radius refraction right angles ring shew sides Similarly sine sino sinẞ straight line string Supposing surface tana tangent tanß triangle Trinity College tube V₁ vertical
Populære avsnitt
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