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Category: Geometry

given-a-ar-ar-2-ar-3-is-a-GPwith-n-r-lt-1-if-a-x-1-x-2-ar-x-3-x-4-x-5-x-6-ar-2-x-7-x-8-x-9-x-10-x-11-x-12-ar-3-where-a-x-1-x-2-ar-AP-ar-x-3-x-4-x-5-x

Question Number 79649 by john santu last updated on 27/Jan/20 $$\mathrm{given}\:\mathrm{a},\mathrm{ar},\mathrm{ar}^{\mathrm{2}} ,\mathrm{ar}^{\mathrm{3}} ,…\:\mathrm{is}\:\mathrm{a}\:\mathrm{GPwith}\: \\ $$$$\mathrm{n}\rightarrow\infty\:,\mathrm{r}\:<\:\mathrm{1} \\ $$$$\mathrm{if}\::\:\mathrm{a},\mathrm{x}_{\mathrm{1}} ,\mathrm{x}_{\mathrm{2}} ,\mathrm{ar},\mathrm{x}_{\mathrm{3}} ,\:\mathrm{x}_{\mathrm{4}} ,\mathrm{x}_{\mathrm{5}} ,\mathrm{x}_{\mathrm{6}} ,\mathrm{ar}^{\mathrm{2}} , \\…

Solve-the-Partial-fraction-3x-4-9x-3-16x-2-9x-13-x-1-2-x-2-2x-2-2-

Question Number 14071 by tawa tawa last updated on 27/May/17 $$\mathrm{Solve}\:\mathrm{the}\:\mathrm{Partial}\:\mathrm{fraction}\: \\ $$$$\frac{\mathrm{3x}^{\mathrm{4}} \:−\:\mathrm{9x}^{\mathrm{3}} \:+\:\mathrm{16x}^{\mathrm{2}} \:+\:\mathrm{9x}\:+\:\mathrm{13}}{\left(\mathrm{x}\:−\:\mathrm{1}\right)^{\mathrm{2}} \left(\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{2x}\:−\:\mathrm{2}\right)^{\mathrm{2}} } \\ $$ Commented by Tinkutara last…

Question-79572

Question Number 79572 by ajfour last updated on 26/Jan/20 Commented by ajfour last updated on 26/Jan/20 $$\bigtriangleup{ABC}\:{is}\:{equilateral}.\:{The}\:{smaller} \\ $$$${circles}\:{have}\:{radii}\:\mathrm{1}\:{and}\:{a}.\:{Find} \\ $$$${the}\:{circumcircle}\:{radius}. \\ $$ Answered by…

Prove-that-if-A-B-and-C-are-the-midpoints-of-the-sides-BC-CA-and-AB-respectively-then-AA-BB-CC-lt-AB-BC-CA-

Question Number 13888 by Tinkutara last updated on 24/May/17 $$\mathrm{Prove}\:\mathrm{that}\:\mathrm{if}\:{A}',\:{B}'\:\mathrm{and}\:{C}'\:\mathrm{are}\:\mathrm{the} \\ $$$$\mathrm{midpoints}\:\mathrm{of}\:\mathrm{the}\:\mathrm{sides}\:{BC},\:{CA}\:\mathrm{and}\:{AB}, \\ $$$$\mathrm{respectively},\:\mathrm{then} \\ $$$${AA}'\:+\:{BB}'\:+\:{CC}'\:<\:{AB}\:+\:{BC}\:+\:{CA} \\ $$ Commented by Tinkutara last updated on 24/May/17…

Question-79368

Question Number 79368 by TawaTawa last updated on 24/Jan/20 Commented by TawaTawa last updated on 24/Jan/20 $$\mathrm{31}\:+\:\mathrm{105}\:+\:\angle\mathrm{Q}\:\:\:=\:\:\mathrm{180}, \\ $$$$\Rightarrow\:\:\:\angle\:\mathrm{Q}\:\:\:=\:\:\mathrm{44} \\ $$$$\therefore\:\:\:\:\:\:\angle\:\mathrm{S}\:\:+\:\:\angle\mathrm{Q}\:\:\:\:=\:\:\mathrm{180},\:\:\:\:\:\:\:\:\:\angle\mathrm{S}\:\:=\:\:\mathrm{136} \\ $$$$ \\ $$$$\angle\mathrm{P}\:+\:\angle\mathrm{R}\:\:=\:\:\mathrm{180},\:\:\:\:\:\:\:\angle\mathrm{R}\:\:=\:\:\mathrm{180}\:−\:\mathrm{150}\:\:=\:\:\mathrm{30}\:\:\:\left(\mathrm{confuse}\:\mathrm{here}\right).…