Question Number 206230 by BaliramKumar last updated on 09/Apr/24 $$\mathrm{Expand}\:\:\:\:{x}^{\mathrm{2}} \:+\:\mathrm{2}{x}\:+\:\mathrm{3}\:\:\:{respect}\:{to}\:{x}\:=\:−\mathrm{2}. \\ $$$$\left(\mathrm{a}\right)\:\left({x}\:−\:\mathrm{2}\right)^{\mathrm{2}} −\mathrm{2}\left({x}\:+\:\mathrm{2}\right)\:+\:\mathrm{3} \\ $$$$\left(\mathrm{b}\right)\:\left({x}\:+\:\mathrm{2}\right)^{\mathrm{2}} −\mathrm{2}\left({x}\:+\:\mathrm{2}\right)\:+\:\mathrm{3} \\ $$$$\left(\mathrm{c}\right)\:\left({x}\:+\:\mathrm{2}\right)^{\mathrm{2}} +\:\mathrm{2}\left({x}\:+\:\mathrm{2}\right)\:+\:\mathrm{3} \\ $$$$\left(\mathrm{d}\right)\:\left({x}\:−\:\mathrm{2}\right)^{\mathrm{2}} −\mathrm{2}\left({x}\:−\:\mathrm{2}\right)\:−\:\mathrm{3} \\ $$$$…
Question Number 206038 by BaliramKumar last updated on 05/Apr/24 $$\sqrt{\mathrm{1}\:+\:\mathrm{2023}\sqrt{\mathrm{1}\:+\:\mathrm{2024}\sqrt{\mathrm{1}+\:\mathrm{2025}\sqrt{\mathrm{1}\:+\:\mathrm{2026}\sqrt{\mathrm{1}\:+\:…………..\infty}}}}}\:=\:?\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$ Answered by mr W last updated on 05/Apr/24 $${x}+\mathrm{1}=\sqrt{\mathrm{1}+{x}\sqrt{\mathrm{1}+\left({x}+\mathrm{1}\right)\sqrt{\mathrm{1}+\left({x}+\mathrm{2}\right)\sqrt{\mathrm{1}+…}}}} \\ $$$$\Rightarrow\mathrm{2024}=\sqrt{\mathrm{1}\:+\:\mathrm{2023}\sqrt{\mathrm{1}\:+\:\mathrm{2024}\sqrt{\mathrm{1}+\:\mathrm{2025}\sqrt{\mathrm{1}\:+\:\mathrm{2026}\sqrt{\mathrm{1}\:+\:…………..\infty}}}}} \\ $$…
Question Number 205893 by BaliramKumar last updated on 02/Apr/24 Commented by BaliramKumar last updated on 02/Apr/24 $$−\mathrm{33}\:\mathrm{or}\:−\mathrm{1}\:\:?? \\ $$ Commented by A5T last updated on…
Question Number 205842 by BaliramKumar last updated on 31/Mar/24 $$\frac{\mathrm{32}^{\mathrm{32}^{\mathrm{32}} } }{\mathrm{9}}\:\overset{\mathrm{R}} {\equiv}\:? \\ $$ Answered by A5T last updated on 01/Apr/24 $$\mathrm{32}^{\mathrm{32}^{\mathrm{32}} } \overset{\mathrm{9}}…
Question Number 205794 by mnjuly1970 last updated on 30/Mar/24 $$ \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\mathrm{A}\:=\:\left\{\:\frac{{k}}{\mathrm{2}^{{n}} }\:\mid\:\mathrm{1}\leqslant\:{k}\:\leqslant\:\mathrm{2}^{{n}} \:,\:{n}\in\mathbb{N}\:\right\} \\ $$$$\:\:\:\:\:{find}\:.\:\:\overset{\:−} {\mathrm{A}}\:=\:? \\ $$$$ \\ $$ Commented by…
Question Number 205577 by BaliramKumar last updated on 25/Mar/24 $$\mathrm{is}\:\infty\:\mathrm{a}\:\mathrm{real}\:\mathrm{number}? \\ $$ Commented by mr W last updated on 25/Mar/24 $${it}\:{is}\:{only}\:{a}\:{symbol}.\:{it}\:{may}\:{have}\:\: \\ $$$${different}\:{meanings}\:{depending}\:{on}\: \\ $$$${in}\:{which}\:{context}\:{it}\:{is}\:{used}.\:…
Question Number 205574 by Red1ight last updated on 24/Mar/24 $${S}=\frac{\mathrm{1}}{\mathrm{2}!}−\frac{\mathrm{1}}{\mathrm{3}!}+\frac{\mathrm{1}}{\mathrm{4}!}−\frac{\mathrm{1}}{\mathrm{5}!}… \\ $$$${S}=? \\ $$ Commented by Frix last updated on 25/Mar/24 $${S}=\frac{\mathrm{1}}{\mathrm{e}} \\ $$ Answered…
Question Number 205502 by MATHEMATICSAM last updated on 22/Mar/24 $$\mathrm{If}\:\mathrm{two}\:\mathrm{roots}\:\mathrm{of}\:{ax}^{\mathrm{2}} \:+\:{bx}\:+\:{c}\:=\:\mathrm{0}\:\mathrm{are}\:\alpha\:\mathrm{and}\: \\ $$$$\beta\:\mathrm{then}\:\frac{\mathrm{1}}{\left({a}\alpha^{\mathrm{2}} \:+\:{c}\right)^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{\left({a}\beta^{\mathrm{2}} \:+\:{c}\right)^{\mathrm{2}} }\:=\:? \\ $$ Answered by Rasheed.Sindhi last updated on…
Question Number 205457 by BaliramKumar last updated on 21/Mar/24 Answered by Rasheed.Sindhi last updated on 21/Mar/24 $${x}+\mathrm{log}_{\mathrm{15}} \left(\mathrm{1}+\mathrm{3}^{{x}} \right)={x}\mathrm{log}_{\mathrm{15}} \mathrm{5}+\mathrm{log}_{\mathrm{15}} \mathrm{12} \\ $$$$\mathrm{log}_{\mathrm{15}} \mathrm{15}^{{x}} +\mathrm{log}_{\mathrm{15}}…
Question Number 205456 by BaliramKumar last updated on 21/Mar/24 Answered by Rasheed.Sindhi last updated on 21/Mar/24 $${First}\:{digit}\:{can}\:{be}\:{occuied}\:{in}\:\mathrm{5}\:{ways} \\ $$$${Second}\:{digit}\:{can}\:{be}\:{occuied}\:{in}\:\mathrm{4}\:{ways} \\ $$$${Third}\:{digit}\:{can}\:{be}\:{occuied}\:{in}\:\mathrm{3}\:{ways}. \\ $$$${Total}\:{ways}:\:\mathrm{5}×\mathrm{4}×\mathrm{3}=\mathrm{60} \\ $$$$\left({d}\right)\:{is}\:{correct}.…