The Mathematical Monthly, Volum 2John Daniel Runkle John Bartlett, 1860 "A complete catalogue of the writings of Sir John Herschel": v. 3, p. 220-227. |
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Resultat 1-5 av 56
Side 7
... parallel to 0 0 " P ' , and O'B ' to Q'Q " , produced , parallel to the same , we have OB : 0 Q ' = OP : 0P - R : R. . OB = ; O Q ' XR ' also O B ' 0 " Q R O'Q ' × R ' = Ø ′ Q " : 0 ′′ Q " : - R R ' : R " . . O'B ' = But OQ O ' Q ' = R ...
... parallel to 0 0 " P ' , and O'B ' to Q'Q " , produced , parallel to the same , we have OB : 0 Q ' = OP : 0P - R : R. . OB = ; O Q ' XR ' also O B ' 0 " Q R O'Q ' × R ' = Ø ′ Q " : 0 ′′ Q " : - R R ' : R " . . O'B ' = But OQ O ' Q ' = R ...
Side 8
... parallel , Ο since they are both perpendicular to AA ' . Through E draw EL parallel to O A , and sup- pose AB produced to meet EL produced in C , and A'B ' to meet in C ' . The similar B ' C'E give EC O ' A ' = EB ' O'B ' EC AO = EB OB ...
... parallel , Ο since they are both perpendicular to AA ' . Through E draw EL parallel to O A , and sup- pose AB produced to meet EL produced in C , and A'B ' to meet in C ' . The similar B ' C'E give EC O ' A ' = EB ' O'B ' EC AO = EB OB ...
Side 35
... parallel to DE . By reason of these parallels , the triangle BED is similar to BPA , EBC to ELF , FLD to DSA , and C'HF to CPA ; whence the proportions : ( 1 ) ... BE : ED = BP : PA ( 3 ) .... LF : DL = SD : SA in which R D F B H E B ...
... parallel to DE . By reason of these parallels , the triangle BED is similar to BPA , EBC to ELF , FLD to DSA , and C'HF to CPA ; whence the proportions : ( 1 ) ... BE : ED = BP : PA ( 3 ) .... LF : DL = SD : SA in which R D F B H E B ...
Side 37
... parallel AP ' . For R. F H E the triangle DEF is half of the paral- lelogram DH ; having the same base and altitude ... parallel to DB ' , intersecting C'D in 0 , we must have RO AB . For the triangles CDB and D RO are similar , having ...
... parallel AP ' . For R. F H E the triangle DEF is half of the paral- lelogram DH ; having the same base and altitude ... parallel to DB ' , intersecting C'D in 0 , we must have RO AB . For the triangles CDB and D RO are similar , having ...
Side 40
... parallel to p F ; then or TL : Tp TX : TF ; r : P = TX : r ' . .. TX = TE = ę . This construction is by FREDERIC T. HAMPTON . CHAUNCEY Wright . 1 TRUMAN HENRY SAFFORD . SOLUTION OF PROBLEMS IN MAXIMA AND MINIMA BY ALGEBRA . 40.
... parallel to p F ; then or TL : Tp TX : TF ; r : P = TX : r ' . .. TX = TE = ę . This construction is by FREDERIC T. HAMPTON . CHAUNCEY Wright . 1 TRUMAN HENRY SAFFORD . SOLUTION OF PROBLEMS IN MAXIMA AND MINIMA BY ALGEBRA . 40.
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A₁ Algebra astronomers atmosphere axis b₁ body c₁ centre CHAUNCEY WRIGHT circle coefficients College computation Conic Sections constant cos² cot B cot curve December 18 denote distance divided ellipse equal equation force formula fraction geometry given gives Hamilton College hence hyperbola inscribed integral logarithms Marietta College Mass Mathematical Monthly maximum Mercury motion multiplied oxen parabola perihelion perpendicular Perry City plane polygon Prize is awarded PRIZE PROBLEMS PRIZE SOLUTION probability Probs proposition quantities quaternions R₁ radii radius ratio rectangle result rhombs right angles roots sides SIMON NEWCOMB sin² SOLUTION OF PROBLEM sphere spherical square suppose surface tan² tangent Theorem tion triangle TRUMAN HENRY SAFFORD vector velocity whole number
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