Question Number 47697 by ajfour last updated on 13/Nov/18 Commented by ajfour last updated on 13/Nov/18 $${The}\:{blue}\:{sphere}\:{touches}\:{the} \\ $$$${xy},\:{yz},\:{zx},\:{and}\:{the}\:{triangular} \\ $$$${plate},\:{while}\:{the}\:{smaller}\:{pink} \\ $$$${sphere}\:{touches}\:{the}\:{yz},\:{zx},\:{larger} \\ $$$${sphere},\:{and}\:{the}\:{triangular}\:{plate}.…
Question Number 178737 by cherokeesay last updated on 21/Oct/22 Answered by Rasheed.Sindhi last updated on 21/Oct/22 Commented by Rasheed.Sindhi last updated on 21/Oct/22 $$\mathrm{Let}\:\mathrm{AC}=\mathrm{BC}=\mathrm{r} \\…
Question Number 47657 by behi83417@gmail.com last updated on 12/Nov/18 Commented by mr W last updated on 13/Nov/18 $${dear}\:{father}\:{from}\:{Behi}: \\ $$$${it}\:{took}\:{me}\:{some}\:{time},\:{but}\:{I}\:{could}\:{finally} \\ $$$${find}\:{the}\:{old}\:{question}\:#\mathrm{15969}\:{where} \\ $$$${this}\:{interesting}\:{question}\:{was}\:{once} \\…
Question Number 47659 by behi83417@gmail.com last updated on 12/Nov/18 Answered by ajfour last updated on 13/Nov/18 $${Eq}.\:{of}\:{tangent}: \\ $$$$\frac{{xx}_{\mathrm{1}} }{{a}^{\mathrm{2}} }+\frac{{yy}_{\mathrm{1}} }{{b}^{\mathrm{2}} }=\mathrm{1} \\ $$$${A}\left(\frac{{a}^{\mathrm{2}}…
Question Number 178645 by a.lgnaoui last updated on 19/Oct/22 $${determiner}\:{la}\:{surface}\:{exterieure}\:{au}\:{carre}\:{bleu}\:{dans}\:{laquelle} \\ $$$$\:{la}\:{chevre}\:{pourra}\:{circuler} \\ $$ Commented by a.lgnaoui last updated on 19/Oct/22 Commented by Frix last…
Question Number 47497 by MrW3 last updated on 10/Nov/18 Commented by MrW3 last updated on 10/Nov/18 $${Find}\:{the}\:{area}\:{of}\:{the}\:{shadow}\:{of}\:{the} \\ $$$${cylinder}\:{produced}\:{by}\:{a}\:{point}\:{source}\:{of} \\ $$$${light}\:{S}. \\ $$$$ \\ $$$${see}\:{also}\:{Q}\mathrm{47449}.…
Question Number 47449 by ajfour last updated on 10/Nov/18 Commented by ajfour last updated on 10/Nov/18 $${Find}\:{the}\:{area}\:{of}\:{the}\:{shadow}\:{of} \\ $$$${the}\:{right}\:{circular}\:{cone}\:\left({not}\right. \\ $$$$\left.{including}\:{its}\:{base}\:{area}\right). \\ $$$${S}\:{is}\:{a}\:{point}\:{source}\:{of}\:{light},\:{at} \\ $$$${a}\:{distance}\:{d}\:{from}\:{center}\:{of}…
Question Number 47407 by ajfour last updated on 09/Nov/18 Answered by ajfour last updated on 10/Nov/18 $${let}\:{BD}\:=\:{x}\:,\:\angle{A}\:=\:\phi,\:\angle{C}\:=\:\theta \\ $$$${S}_{{ABCD}} \:=\:\frac{{ab}\mathrm{sin}\:\theta}{\mathrm{2}}+\frac{{cd}\mathrm{sin}\:\phi}{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} =\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{ab}\mathrm{cos}\:\theta…
Question Number 47382 by ajfour last updated on 09/Nov/18 Commented by ajfour last updated on 09/Nov/18 $${The}\:{sides}\:{of}\:{inclined}\:{triangle} \\ $$$${are}\:{p},\:{q},\:{and}\:{r}. \\ $$$${Find}\:{P}\:{G}=\:{h}\:. \\ $$$${The}\:{sides}\:{of}\:{the}\:{prism}\:{box}\:{are} \\ $$$${squares},\:{and}\:{base}\:{equilateral}.…
Question Number 178450 by mr W last updated on 16/Oct/22 Commented by mr W last updated on 16/Oct/22 $${find}\:{the}\:{area}\:{of}\:{shaded}\:{rectangle}. \\ $$ Commented by cortano1 last…