Question Number 133451 by mohammad17 last updated on 22/Feb/21 Commented by mohammad17 last updated on 22/Feb/21 $${question}\:\mathrm{6}\:{please} \\ $$ Commented by mohammad17 last updated on…
Question Number 133446 by aurpeyz last updated on 22/Feb/21 $$ \\ $$$${an}\:{object}\:\mathrm{3}{cm}\:{high}\:{is}\:{placed}\:\mathrm{5}{cm}\:{away} \\ $$$${from}\:{the}\:{pole}\:{of}\:{a}\:{concave}\:{spherical} \\ $$$${mirror}\:{of}\:{radius}\:{of}\:{curvature}\:\mathrm{25}{cm}. \\ $$$${The}\:{position}\:{and}\:{orientaion}\:{and}\:{size}\: \\ $$$${of}\:{the}\:{image}\:{are} \\ $$$$\left({a}\right)−\mathrm{8}.\mathrm{33}{cm}\:{upright}\:{and}\:\mathrm{5}{cm} \\ $$$$\left({b}\right)\mathrm{5}{cm}\:{upright}\:{and}\:\mathrm{8}.\mathrm{33}{cm} \\…
Question Number 2367 by Filup last updated on 18/Nov/15 $$\mathrm{What}\:\mathrm{exactly}\:\mathrm{does}\:{f}:\mathbb{C}\rightarrow\mathbb{C}\:\mathrm{mean}? \\ $$ Answered by RasheedAhmad last updated on 18/Nov/15 $${A}\:{function}\:\:{f}\:\:{whose}\:{domain}\:{and} \\ $$$${range}\:{both}\:{are}\:\mathbb{C}\:\left({set}\:{of}\:{complex}\right. \\ $$$$\left.{numbers}\right) \\…
Question Number 2334 by Syaka last updated on 16/Nov/15 $$\int\frac{{dx}}{\left({x}^{\mathrm{2}} \:+\:\mathrm{3}{x}\:+\:\mathrm{9}\right)\sqrt{{x}^{\mathrm{2}} \:+\:\mathrm{5}{x}\:+\:\mathrm{7}}\:}\:\:=\:? \\ $$ Commented by prakash jain last updated on 16/Nov/15 $${x}^{\mathrm{2}} +\mathrm{5}{x}+\mathrm{7}=\left({x}+\frac{\mathrm{5}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{4}}…
Question Number 2326 by Filup last updated on 16/Nov/15 $$\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}×\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}+\sqrt{\mathrm{2}}}}×\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}}}}}×…=? \\ $$ Commented by Rasheed Soomro last updated on 16/Nov/15 $$\mathcal{N}{ice}\:\mathcal{A}{pproach}! \\ $$ Commented by…
Question Number 67844 by aliesam last updated on 01/Sep/19 $${x}^{\mathrm{2}} +\mid{x}\mid−\mathrm{6}=\mathrm{0} \\ $$ Commented by gunawan last updated on 01/Sep/19 $$\mathrm{case}\:\mathrm{1}\:{x}<\mathrm{0} \\ $$$${x}^{\mathrm{2}} −{x}−\mathrm{6}=\mathrm{0} \\…
Question Number 2300 by Syaka last updated on 14/Nov/15 $${if}\:{f}\left(\mathrm{4}{xy}\right)\:=\:\mathrm{2}{y}\left({f}\left({x}\:−\:{y}\right)\:+\:{f}\left({x}\:+\:{y}\right)\right)\:{and}\:{f}\left(\mathrm{5}\right)\:=\:\mathrm{3} \\ $$$${what}\:{is}\:{the}\:{value}\:{of}\:{f}\left(\mathrm{2015}\right)\:=\:? \\ $$ Commented by Rasheed Soomro last updated on 17/Nov/15 $$\mathcal{EXCELLENT}\:! \\ $$…
Question Number 133368 by otchereabdullai@gmail.com last updated on 21/Feb/21 $$\mathrm{Assuming}\:\mathrm{that}\:\mathrm{the}\:\mathrm{world}\:\mathrm{population} \\ $$$$\mathrm{is}\:\mathrm{6billion}\:\mathrm{and}\:\mathrm{total}\:\mathrm{world}\:\mathrm{trip}\:\mathrm{is} \\ $$$$\mathrm{4}.\mathrm{8billion}. \\ $$$$\left(\mathrm{a}\right)\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{country}\:\mathrm{potential}\: \\ $$$$\mathrm{generation}\:\mathrm{index}\:\left(\mathrm{CPGI}\right)\:\mathrm{of}\:\mathrm{the}\: \\ $$$$\mathrm{country} \\ $$$$\left(\mathrm{b}\right)\:\mathrm{interprete}\:\mathrm{your}\:\mathrm{results}\:\mathrm{in}\:\left(\mathrm{a}\right)\:\mathrm{above}. \\ $$$$ \\…
Question Number 2292 by Filup last updated on 14/Nov/15 $$\mathrm{This}\:\mathrm{isn}'\mathrm{t}\:\mathrm{a}\:\mathrm{question}. \\ $$$$\mathrm{Just}\:\mathrm{wanted}\:\mathrm{to}\:\mathrm{say}\:\mathrm{that}\:\mathrm{since}\:\mathrm{I}\:\mathrm{joined} \\ $$$$\mathrm{here}\:\mathrm{I}\:\mathrm{have}\:\mathrm{learnt}\:\mathrm{so}\:\mathrm{much}.\:\mathrm{You}\:\mathrm{guys} \\ $$$$\mathrm{are}\:\mathrm{awesome}! \\ $$ Answered by Rasheed Soomro last updated on…
Question Number 2286 by Filup last updated on 14/Nov/15 $$\mathrm{Can}\:\mathrm{you}\:\mathrm{evaluate}: \\ $$$$\frac{\mathrm{1}}{\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}}}+\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}}}}+…+\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{…}} \\ $$ Commented by Yozzi last updated on 14/Nov/15 $${u}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{1}},\:{u}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}}}=\frac{\mathrm{1}}{\mathrm{1}+{u}_{\mathrm{1}} }…