Question Number 45422 by Tawa1 last updated on 12/Oct/18 $$\mathrm{Three}\:\mathrm{consecutive}\:\mathrm{terms}\:\mathrm{of}\:\mathrm{a}\:\mathrm{G}.\mathrm{P}\:\mathrm{are}\:\mathrm{the}\:\mathrm{3rd},\:\mathrm{5th}\:\mathrm{and}\:\mathrm{8th}\:\mathrm{term}\:\mathrm{of}\:\mathrm{an}\:\mathrm{A}.\mathrm{P}. \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{common}\:\mathrm{ratio}. \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 12/Oct/18 $${A}+\left(\mathrm{3}−\mathrm{1}\right){D}={a} \\ $$$${A}+\left(\mathrm{5}−\mathrm{1}\right){D}={ar} \\…
Question Number 176485 by Shrinava last updated on 20/Sep/22 Commented by mr W last updated on 20/Sep/22 $${x}=\mathrm{4}\:{is}\:{the}\:{solution}. \\ $$ Answered by a.lgnaoui last updated…
Question Number 176472 by mr W last updated on 19/Sep/22 $${solve}\:{for}\:{x},{y},{z}\:{with} \\ $$$${x}+{y}+{z}={x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} ={x}^{\mathrm{3}} +{y}^{\mathrm{3}} +{z}^{\mathrm{3}} =\mathrm{5} \\ $$ Answered by behi834171 last…
Question Number 176482 by Vynho last updated on 19/Sep/22 $${L}\left({E}\right)\:{est}\:{l}'{algebre}\:{des}\:{endomorphisme} \\ $$$${continus}\:{d}'{un}\:{espace}\:{de}\:{Banach}\:{E}, \\ $$$${muni}\:{de}\:{la}\:{norme}\:{d}'{application}\:{lineaire} \\ $$$$;\:{GL}\left({E}\right)\:{est}\:{le}\:{sous}\:{ensemble}\:{des} \\ $$$${elements}\:{inversibles}\:{de}\:{L}\left({E}\right) \\ $$$$\left.{a}\right){Montrer}\:{que}\:{GL}\left({E}\right)\:{est}\:{ouvert}\:{dans} \\ $$$${L}\left({E}\right) \\ $$$$\left.{b}\right){montrer}\:{que}\:{l}'{application}\: \\…
Question Number 45399 by Rio Michael last updated on 12/Oct/18 $${using}\:{your}\:{knowlege}\:{on}\:{Arithmetic}\:{progressions}, \\ $$$${show}\:{that}\:\:{A}=\:{p}\left(\mathrm{1}+\frac{{r}}{\mathrm{100}}\right)^{{n}} \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 12/Oct/18 $${principle}={p} \\ $$$${rate}\:{of}\:{interest}={r\%}…
Question Number 176453 by mnjuly1970 last updated on 19/Sep/22 $$ \\ $$$$\:\:\:{If}\:,\:\:\alpha\:,\:\beta\:,\:\gamma\:\in\:\left(\:\mathrm{0}\:\:,\:\:\mathrm{1}\:\right)\:\:,\:\:{then}\: \\ $$$$\:\:\:\: \\ $$$$\:\:\:\:{prove}\:\:{that}\::\:\: \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\sqrt{\left(\mathrm{1}\overset{} {−}\alpha\:\right).\left(\mathrm{1}\overset{} {−}\beta\:\right).\:\left(\mathrm{1}\overset{} {−}\gamma\:\right)}\:+\sqrt{\overset{} {\alpha}.\overset{} {\beta}.\overset{}…
Question Number 176458 by mathlove last updated on 19/Sep/22 $${x}^{\mathrm{2}} +{x}=\mathrm{1} \\ $$$$\frac{{x}^{\mathrm{5}} +\mathrm{8}}{{x}+\mathrm{1}}=? \\ $$ Answered by Rasheed.Sindhi last updated on 19/Sep/22 $${x}^{\mathrm{2}} +{x}=\mathrm{1};\:\:\:\frac{{x}^{\mathrm{5}}…
Question Number 176448 by peter frank last updated on 19/Sep/22 $$\mathrm{Suppose}\:\mathrm{a}^{\mathrm{3}} +\mathrm{b}^{\mathrm{3}} +\mathrm{c}^{\mathrm{3}} =\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} \\ $$$$=\mathrm{a}+\mathrm{b}+\mathrm{c}\:\:\mathrm{Prove}\:\mathrm{that}\:\mathrm{abc}=\mathrm{0} \\ $$$$ \\ $$ Commented by mr…
Question Number 110897 by bemath last updated on 31/Aug/20 $$\left(\mathrm{1}\right)\mathrm{4x}−\mathrm{4}\:\leqslant\:\mid\mathrm{x}^{\mathrm{2}} −\mathrm{3x}+\mathrm{2}\:\mid\: \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{solution}\:\mathrm{set}\: \\ $$$$\left(\mathrm{2}\right)\:\frac{\mathrm{1}+\mathrm{cos}\:\left(\frac{\alpha}{\mathrm{2}}\right)−\mathrm{sin}\:\left(\frac{\alpha}{\mathrm{2}}\right)}{\mathrm{1}−\mathrm{cos}\:\left(\frac{\alpha}{\mathrm{2}}\right)−\mathrm{sin}\:\left(\frac{\alpha}{\mathrm{2}}\right)}=? \\ $$ Answered by john santu last updated on 31/Aug/20…
Question Number 45364 by Tawa1 last updated on 12/Oct/18 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{n}\:\mathrm{terms}:\:\:\frac{\mathrm{1}}{\mathrm{1}.\mathrm{3}}\:+\:\frac{\mathrm{1}}{\mathrm{3}.\mathrm{5}}\:+\:…\:+\:\frac{\mathrm{1}}{\left(\mathrm{2n}\:−\:\mathrm{1}\right)\left(\mathrm{2n}\:+\:\mathrm{1}\right)}\:\:=\:? \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 12/Oct/18 $${T}_{{n}} =\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{2}{n}−\mathrm{1}\right)} \\ $$$${T}_{{n}} =\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\left(\mathrm{2}{n}+\mathrm{1}\right)−\left(\mathrm{2}{n}−\mathrm{1}\right)}{\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{2}{n}−\mathrm{1}\right)}\right] \\…