Question Number 50744 by Necxx last updated on 19/Dec/18 $${If}\:{the}\:{perimeter}\:{of}\:{a}\:{rectangle}\:{is} \\ $$$${a}\:\mathrm{2}−{digit}\:{number}\:{which}\:{unit}\:{digit}\mathscr{L} \\ $$$${and}\:{tens}\:{digit}\:{represents}\:{its}\:{length} \\ $$$${and}\:{breadth}\:{respectively}.{Find}\:{its} \\ $$$${area}\:{in}\:{constant}. \\ $$ Answered by Rasheed.Sindhi last updated…
Question Number 50730 by ajfour last updated on 19/Dec/18 $$\sqrt{{x}−{a}}+\sqrt{{x}−{b}}+\sqrt{{x}−{c}}+{x}\:=\:{d} \\ $$$${solve}\:{for}\:{x}. \\ $$ Answered by behi83417@gmail.com last updated on 21/Dec/18 $${after}\:{squaring}\:{and}\:{symplifing}\:{i}\:{got} \\ $$$${this}\:{equtition}: \\…
Question Number 116259 by Khalmohmmad last updated on 02/Oct/20 Answered by TANMAY PANACEA last updated on 02/Oct/20 $${a}+{a}+\mathrm{4}{d}=\mathrm{30}\:\:{considering}\:{A}.{P}\:{series} \\ $$$${a}+\mathrm{2}{d}+{a}+\mathrm{6}{d}=\mathrm{120} \\ $$$$\mathrm{8}{d}+\mathrm{30}−\mathrm{4}{d}=\mathrm{120} \\ $$$${d}=\frac{\mathrm{90}}{\mathrm{4}}=\frac{\mathrm{45}}{\mathrm{2}} \\…
Question Number 181794 by Shrinava last updated on 30/Nov/22 $$\mathrm{f}\left(\mathrm{0}\right)\:=\:\mathrm{0} \\ $$$$\mathrm{f}\left(\mathrm{1}\right)\:=\:\mathrm{e}^{\mathrm{3}} \\ $$$$\mathrm{f}\:\in\:\mathbb{C}^{\mathrm{2}} \:\left(\mathbb{R}\right) \\ $$$$\mathrm{f}\:^{''} \:\left(\mathrm{x}\right)\:−\:\mathrm{5}\:\mathrm{f}\:^{'} \left(\mathrm{x}\right)\:+\:\mathrm{6}\:\mathrm{f}\left(\mathrm{x}\right)\:=\:\mathrm{0}\:\:,\:\:\forall\:\mathrm{x}\:\in\:\mathbb{R} \\ $$$$\mathrm{Find}:\:\:\:\boldsymbol{\Omega}\:=\underset{\boldsymbol{\mathrm{x}}\rightarrow\infty} {\mathrm{lim}}\:\left(\mathrm{1}\:−\:\frac{\mathrm{1}}{\mathrm{f}\:\left(\mathrm{x}\right)}\right)^{\boldsymbol{\mathrm{x}}} \\ $$ Answered…
Question Number 50717 by Tawa1 last updated on 19/Dec/18 Answered by tanmay.chaudhury50@gmail.com last updated on 19/Dec/18 $${x}+\sqrt{{y}}\:−\sqrt{{x}}\:−{y}=\mathrm{4} \\ $$$$\left(\sqrt{{x}}+\sqrt{{y}}\:\right)\left(\sqrt{{x}}\:−\sqrt{{y}}\:\right)−\left(\sqrt{{x}}\:−\sqrt{{y}}\:\right)=\mathrm{4} \\ $$$$\left(\sqrt{{x}}\:−\sqrt{{y}}\:\right)\left(\sqrt{{x}}\:+\sqrt{{y}}\:−\mathrm{1}\right)=\mathrm{4} \\ $$$$\:{by}\:{logic}\:{trial}\:{x}=\mathrm{9}\:\:{y}=\:\mathrm{4}\:\:{so}\:\:\:{z}=\mathrm{1} \\ $$$$…
Question Number 116253 by I want to learn more last updated on 02/Oct/20 $$\left(\mathrm{1}\right)\:\:\:\:\mathrm{Show}\:\mathrm{that}\:\:\:\:\underset{\mathrm{i}\:\:=\:\:\mathrm{0}} {\overset{\mathrm{n}} {\sum}}\:\mathrm{L}_{\mathrm{i}} \left(\mathrm{x}\right)\:\:\:=\:\:\:\mathrm{1} \\ $$$$\left(\mathrm{2}\right)\:\:\:\:\mathrm{Show}\:\mathrm{that}\:\:\:\:\underset{\mathrm{i}\:\:=\:\:\mathrm{0}} {\overset{\mathrm{n}} {\sum}}\:\mathrm{L}_{\mathrm{i}} \left(\mathrm{x}\right).\:\mathrm{x}_{\mathrm{i}} ^{\mathrm{k}} \:\:\:=\:\:\:\mathrm{x}^{\mathrm{k}} ,\:\:\:\:\:\:\:\:\mathrm{k}\:\leqslant\:\mathrm{n}…
Question Number 50710 by Tawa1 last updated on 19/Dec/18 $$\mathrm{If}\:\:\mathrm{x}\:+\:\mathrm{y}\:+\:\mathrm{z}\:=\:\mathrm{15}\:\:\mathrm{and}\:\:\mathrm{xy}\:+\:\mathrm{yz}\:+\:\mathrm{zx}\:\:=\:\mathrm{85},\:\:\:\mathrm{find}\:\:\:\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{y}^{\mathrm{2}} \:+\:\mathrm{z}^{\mathrm{2}} \\ $$ Answered by mr W last updated on 19/Dec/18 $$\left({x}+{y}+{z}\right)^{\mathrm{2}} =\left({x}+{y}\right)^{\mathrm{2}} +\mathrm{2}\left({x}+{y}\right){z}+{z}^{\mathrm{2}}…
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Question Number 50698 by Cheyboy last updated on 19/Dec/18 $$\mathrm{Solve}\:\mathrm{for}\:\mathrm{x} \\ $$$$ \\ $$$$\mathrm{4}^{\mathrm{2x}+\mathrm{1}} ×\mathrm{5}^{\mathrm{x}−\mathrm{2}} =\:\mathrm{6}^{\mathrm{1}−\mathrm{x}} \\ $$ Answered by MJS last updated on 19/Dec/18…
Question Number 116226 by bobhans last updated on 02/Oct/20 $$\frac{\mathrm{1}}{\mathrm{2}+\sqrt{\mathrm{2}}}\:+\frac{\mathrm{1}}{\mathrm{3}\sqrt{\mathrm{2}}+\mathrm{2}\sqrt{\mathrm{3}}}\:+\frac{\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{3}}+\mathrm{3}\sqrt{\mathrm{4}}}+…+\frac{\mathrm{1}}{\mathrm{100}\sqrt{\mathrm{99}}+\mathrm{99}\sqrt{\mathrm{100}}} \\ $$ Commented by bemath last updated on 02/Oct/20 $$\mathrm{0}.\mathrm{9} \\ $$ Answered by john…