Question Number 217159 by ArshadS last updated on 03/Mar/25 $${Solve}\:{for}\:{x}: \\ $$$$\frac{{x}+\mathrm{3}}{{x}−\mathrm{2}}+\frac{\mathrm{2}{x}−\mathrm{5}}{{x}+\mathrm{4}}=\frac{\mathrm{4}{x}+\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{8}} \\ $$ Answered by som(math1967) last updated on 03/Mar/25 $$\:\frac{{x}+\mathrm{3}}{{x}−\mathrm{2}}+\frac{\mathrm{2}{x}−\mathrm{5}}{{x}+\mathrm{4}}=\frac{\mathrm{4}{x}+\mathrm{1}}{\left({x}−\mathrm{2}\right)\left({x}+\mathrm{4}\right)} \\ $$$$\Rightarrow\frac{{x}+\mathrm{3}}{{x}−\mathrm{2}}\:−\frac{\mathrm{4}{x}+\mathrm{1}}{\left({x}−\mathrm{2}\right)\left({x}+\mathrm{4}\right)}+\frac{\mathrm{2}{x}−\mathrm{5}}{{x}+\mathrm{4}}=\mathrm{0}…
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Question Number 217140 by peter frank last updated on 02/Mar/25 if a, b, c are three digits, abc and bca are two numbers. where abc…
Question Number 217101 by MathematicalUser2357 last updated on 01/Mar/25 $$\mathrm{is}\:\mathrm{this}\:\mathrm{right}\:\mathrm{when}\:\left({a}+{bi}\right)^{{c}+{di}} =\mid{a}+{bi}\mid^{{c}+{di}} {e}^{{i}\left({c}+{di}\right)\mathrm{arg}\left({a}+{bi}\right)} ? \\ $$$$\mathrm{I}\:\mathrm{had}\:\mathrm{let}\:\mathrm{arg}\left({a}+{bi}\right)=\begin{cases}{\mathrm{tan}^{−\mathrm{1}} \left(\frac{{b}}{{a}}\right)}&{{a}\geqslant\mathrm{0}\:\mathrm{and}\:{b}\geqslant\mathrm{0}}\\{\pi−\mathrm{tan}^{−\mathrm{1}} \left(−\frac{{b}}{{a}}\right)}&{{a}<\mathrm{0}\:\mathrm{and}\:{b}\geqslant\mathrm{0}}\\{−\left(\pi−\mathrm{tan}^{−\mathrm{1}} \left(\frac{{b}}{{a}}\right)\right)}&{{a}<\mathrm{0}\:\mathrm{and}\:{b}<\mathrm{0}}\\{−\mathrm{tan}^{−\mathrm{1}} \left(\frac{{b}}{{a}}\right)}&{{a}\geqslant\mathrm{0}\:\mathrm{and}\:{b}<\mathrm{0}}\end{cases}\:\mathrm{before}\:\mathrm{I}\:\mathrm{solved}\:\mathrm{it} \\ $$$$\left({a}+{bi}\right)^{{c}+{di}} =\mid{a}+{bi}\mid^{{c}+{di}} {e}^{{i}\left({c}+{di}\right)\mathrm{arg}\left({a}+{bi}\right)} \\ $$$$=\mid{a}+{bi}\mid^{{c}}…
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Question Number 217080 by hardmath last updated on 28/Feb/25 Commented by mr W last updated on 01/Mar/25 $${also}\:{Area}\left({ABCD}\right)\leqslant\sqrt{{abcd}} \\ $$ Answered by mr W last…
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Question Number 217064 by ArshadS last updated on 28/Feb/25 $$\mathrm{Two}\:\mathrm{numbers}\:\mathrm{differ}\:\mathrm{by}\:\mathrm{6}.\:\mathrm{The}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{their}\:\mathrm{reciprocals}\:\mathrm{is}\:\frac{\mathrm{2}}{\mathrm{15}}\:. \\ $$$$\mathcal{D}{etermine}\:{the}\:{numbers}. \\ $$ Answered by A5T last updated on 28/Feb/25 $$\frac{\mathrm{1}}{\mathrm{a}}+\frac{\mathrm{1}}{\mathrm{b}}=\frac{\mathrm{2}}{\mathrm{15}}\Rightarrow\mathrm{15}\left(\mathrm{a}+\mathrm{b}\right)=\mathrm{2ab} \\ $$$$\mathrm{a}−\mathrm{b}=\underset{−} {+}\mathrm{6}\Rightarrow\mathrm{a}=\mathrm{b}\underset{−}…
Question Number 217058 by ArshadS last updated on 28/Feb/25 $$\mathrm{The}\:\mathrm{quadratic}\:\mathrm{equation}\:\:\mathrm{has}\:\mathrm{two}\:\mathrm{equal}\:\mathrm{roots}: \\ $$$$\:\:\:\:\:{x}^{\mathrm{2}} +\left({k}−\mathrm{3}\right){x}+{k}=\mathrm{0} \\ $$$$\left(\mathrm{a}\right)\:\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\:{k}. \\ $$$$\left(\mathrm{b}\right)\:\mathrm{For}\:\mathrm{this}\:\mathrm{value}\:\mathrm{of}\:\:{k},\:\mathrm{solve}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{for}\:{x} \\ $$$$\left(\mathrm{c}\right)\mathrm{If}\:\:{x}\:\mathrm{is}\:\mathrm{the}\:\mathrm{length}\:\mathrm{of}\:\mathrm{a}\:\mathrm{rectangle}\:\mathrm{and}\:\mathrm{its}\:\mathrm{width}\:\mathrm{is}\:{x}−\mathrm{2},\:\:\mathrm{find}\:\mathrm{the}\:\mathrm{area}\: \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{rectangle}. \\ $$ Answered by…
Question Number 217047 by hardmath last updated on 27/Feb/25 $$\mathrm{Parties}\:-\:\mathrm{a},\mathrm{b},\mathrm{c},\mathrm{d} \\ $$$$\mathrm{For}\:\mathrm{a}\:\mathrm{convex}\:\mathrm{quadrilateral} \\ $$$$\mathrm{Prove}\:\mathrm{that}: \\ $$$$\mathrm{Area}\left(\mathrm{ABCD}\right)\:\leqslant\:\frac{\mathrm{a}^{\mathrm{2}} \:+\:\mathrm{b}^{\mathrm{2}} \:+\:\mathrm{c}^{\mathrm{2}} \:+\:\mathrm{d}^{\mathrm{2}} }{\mathrm{4}} \\ $$ Terms of Service…
Question Number 217015 by hardmath last updated on 26/Feb/25 $$\mathrm{If} \\ $$$$\mathrm{f}\left(\mathrm{2x}\:+\:\mathrm{1}\right)\:=\:\mathrm{3x}\:+\:\mathrm{5} \\ $$$$\mathrm{Find} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)\:=\:? \\ $$ Answered by Rasheed.Sindhi last updated on 26/Feb/25…