Question Number 158054 by zainaltanjung last updated on 30/Oct/21 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{composition}\:\mathrm{function} \\ $$$$\mathrm{that}\:\mathrm{can}\:\mathrm{be}\:\mathrm{form}\:\mathrm{from}\:\mathrm{two}\:\mathrm{this} \\ $$$$\mathrm{single}\:\mathrm{functions}\:\mathrm{below}\:: \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{2x}+\mathrm{4}\: \\ $$$$\mathrm{g}\left(\mathrm{x}\right)=\:\frac{\mathrm{2x}−\mathrm{7}}{\mathrm{3x}} \\ $$$$\bullet\:\mathrm{fog}\left(\mathrm{x}\right)=……. \\ $$$$\bullet\:\mathrm{gof}\left(\mathrm{x}\right)=…… \\ $$ Answered…
Question Number 157970 by zainaltanjung last updated on 30/Oct/21 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{neext}\:\mathrm{number}\:\mathrm{for}\:\mathrm{this} \\ $$$$\mathrm{sequence}\:\mathrm{below}\: \\ $$$$\left.\mathrm{1}\right).\:\mathrm{1},\:\:\mathrm{3},\:\mathrm{6},\:\mathrm{10},\:\mathrm{15}…. \\ $$$$\left.\mathrm{2}\right).\:\mathrm{1},\:\mathrm{5}\:,\mathrm{14}\:,\mathrm{30},\:\:\mathrm{55}\:…. \\ $$$$\left.\mathrm{3}\right).\:\mathrm{1},\mathrm{7},\mathrm{17},\mathrm{31},\:\mathrm{49}\:…. \\ $$$$\left.\mathrm{4}\right).\:\:\mathrm{4},\:\mathrm{13},\mathrm{28},\:\mathrm{49},\mathrm{74}… \\ $$$$\left.\mathrm{5}\right).\:\:\mathrm{1},\mathrm{8},\mathrm{27},\mathrm{64},\mathrm{125}…. \\ $$$$ \\…
Question Number 157805 by Huy last updated on 28/Oct/21 $$\mathrm{Find}\:\mathrm{x}\in\mathbb{R}: \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{x}^{\mathrm{3}} −\mathrm{2x}+\mathrm{3}=\sqrt{\mathrm{2x}+\mathrm{5}}+\mathrm{4}\sqrt{\mathrm{x}−\mathrm{1}} \\ $$ Answered by MJS_new last updated on 28/Oct/21 $$\mathrm{rhs}\:\mathrm{defined}\:\mathrm{for}\:{x}\geqslant\mathrm{1}\:\mathrm{and}\:\mathrm{strictly}\:\mathrm{increasing} \\ $$$$\mathrm{lhs}\:\mathrm{strictly}\:\mathrm{increasing}\:\mathrm{for}\:{x}>\frac{\sqrt{\mathrm{6}}}{\mathrm{3}}…
Question Number 157644 by Ar Brandon last updated on 26/Oct/21 $$\mathrm{Prove}\:\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}}=\frac{\sqrt{\mathrm{6}}−\sqrt{\mathrm{2}}}{\mathrm{4}} \\ $$ Answered by ajfour last updated on 26/Oct/21 $${v}=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}} \\ $$$${v}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}=\frac{\mathrm{2}−\sqrt{\mathrm{3}}}{\mathrm{4}} \\…
Question Number 92016 by peter frank last updated on 04/May/20 Answered by Rio Michael last updated on 04/May/20 $$\left(\mathrm{ii}\right)\:\mathrm{C}_{\mathrm{4}} \mathrm{H}_{\mathrm{9}} \mathrm{Br}\:+\:\mathrm{NaOH}\:\rightarrow\:\mathrm{C}_{\mathrm{4}} \mathrm{H}_{\mathrm{9}} \mathrm{OH}\:+\:\mathrm{NaBr} \\ $$$$\left(\mathrm{iii}\right)\:{CH}_{\mathrm{3}}…
Question Number 157508 by cortano last updated on 24/Oct/21 Commented by MJS_new last updated on 24/Oct/21 $$\mathrm{I}\:\mathrm{think}\:\mathrm{we}\:\mathrm{can}\:\mathrm{only}\:\mathrm{approximate}.\:\mathrm{I}\:\mathrm{get} \\ $$$${x}\approx.\mathrm{575926677918} \\ $$ Terms of Service Privacy…
Question Number 91912 by jagoll last updated on 03/May/20 $$\mathrm{if}\:\mid\mathrm{x}\mid,\:\mid\mathrm{x}−\mathrm{1}\mid,\:\mid\mathrm{x}+\mathrm{1}\mid\:\mathrm{are}\:\mathrm{first} \\ $$$$\mathrm{three}\:\mathrm{terms}\:\mathrm{of}\:\mathrm{an}\:\mathrm{AP}.\:\mathrm{then}\: \\ $$$$\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{it}'\mathrm{s}\:\mathrm{first} \\ $$$$\mathrm{10}\:\mathrm{terms}\:\mathrm{equal}\:\mathrm{to}\: \\ $$ Commented by jagoll last updated on 03/May/20…
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Question Number 157074 by amin96 last updated on 19/Oct/21 $$\sqrt{\mathrm{4}+\sqrt{\mathrm{4}^{\mathrm{2}} +\sqrt{\mathrm{4}^{\mathrm{3}} +\sqrt{\mathrm{4}^{\mathrm{4}} +\ldots}}}}=? \\ $$ Answered by MJS_new last updated on 19/Oct/21 $$\mathrm{3} \\ $$…
Question Number 157046 by ajfour last updated on 18/Oct/21 $${q}=\left\{{a}+\left(\frac{{a}}{\mathrm{9}}−\frac{\mathrm{1}}{\mathrm{108}}−{a}^{\mathrm{3}} \right)^{\mathrm{1}/\mathrm{3}} \right\}^{\mathrm{1}/\mathrm{3}} \\ $$$$\:\:+\left\{{b}+\left(\frac{{b}}{\mathrm{9}}−\frac{\mathrm{1}}{\mathrm{108}}−{b}^{\mathrm{3}} \right)^{\mathrm{1}/\mathrm{3}} \right\}^{\mathrm{1}/\mathrm{3}} \\ $$$$\:\:\:{a}=\frac{\mathrm{9}+\sqrt{\mathrm{37}}}{\mathrm{72}}\:\:\:,\:\:{b}=\frac{\mathrm{9}−\sqrt{\mathrm{37}}}{\mathrm{72}} \\ $$$$\:{find}\:\boldsymbol{{q}}\:{correct}\:{to}\:\mathrm{5}\:{decimal} \\ $$$$\:{places}. \\ $$ Commented…