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Category: Algebra

Question-42719

Question Number 42719 by jbm last updated on 01/Sep/18 Commented by Joel578 last updated on 02/Sep/18 $$\mathrm{Sophie}\:\mathrm{Germain} \\ $$$${p}^{\mathrm{4}} \:+\:\mathrm{4}{q}^{\mathrm{4}} \:=\:\left({p}^{\mathrm{2}} \:+\:\mathrm{2}{pq}\:+\:\mathrm{2}{q}^{\mathrm{2}} \right)\left({p}^{\mathrm{2}} \:−\:\mathrm{2}{pq}\:+\:\mathrm{2}{q}^{\mathrm{2}} \right)…

4sin-2-x-4sin-x-cos-2-x-2cos-x-0-lt-x-lt-pi-2-Find-sin-x-

Question Number 173787 by mnjuly1970 last updated on 18/Jul/22 $$\:\mathrm{4}{sin}^{\:\mathrm{2}} \left({x}\right)−\:\mathrm{4}{sin}\left({x}\right)={cos}^{\:\mathrm{2}} \left({x}\right)−\mathrm{2}{cos}\left({x}\right) \\ $$$$\:\:\:,\:\mathrm{0}<{x}<\frac{\pi}{\mathrm{2}}\:.\:\:\:{Find}\:\:\:\:\:\:\:{sin}\left({x}\right)=? \\ $$ Commented by infinityaction last updated on 18/Jul/22 $$\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}} \\…

Question-173776

Question Number 173776 by AgniMath last updated on 18/Jul/22 Answered by thfchristopher last updated on 18/Jul/22 $$\mathrm{When}\:{n}=\mathrm{1}, \\ $$$$\mathrm{R}.\mathrm{H}.\mathrm{S}.=\frac{\mathrm{1}×\mathrm{2}}{\mathrm{2}} \\ $$$$=\mathrm{1} \\ $$$$=\mathrm{L}.\mathrm{H}.\mathrm{S}. \\ $$$$\mathrm{Assume}\:{n}={k}\:\mathrm{is}\:\mathrm{true},…

y-e-x-ln-sin2x-dy-dx-

Question Number 108238 by Study last updated on 15/Aug/20 $${y}={e}^{{x}} {ln}\left({sin}\mathrm{2}{x}\right)\:\:\:\:\:\:\:\:\:\:\:\frac{{dy}}{{dx}}=?? \\ $$ Answered by Dwaipayan Shikari last updated on 15/Aug/20 $$\frac{{dy}}{{dx}}={e}^{{x}} {log}\left({sin}\mathrm{2}{x}\right)+\mathrm{2}{e}^{{x}} \frac{{cos}\mathrm{2}{x}}{{sin}\mathrm{2}{x}} \\…

If-pqr-1-Hence-evaluate-1-1-e-f-1-1-1-f-g-1-1-1-g-e-1-

Question Number 42671 by Tawa1 last updated on 31/Aug/18 $$\mathrm{If}\:\mathrm{pqr}\:=\:\mathrm{1} \\ $$$$\mathrm{Hence}\:\mathrm{evaluate}:\:\:\:\:\frac{\mathrm{1}}{\mathrm{1}\:+\:\mathrm{e}\:+\:\mathrm{f}^{−\mathrm{1}} }\:\:+\:\:\frac{\mathrm{1}}{\mathrm{1}\:+\:\mathrm{f}\:+\:\mathrm{g}^{−\mathrm{1}} }\:\:+\:\:\frac{\mathrm{1}}{\mathrm{1}\:+\:\mathrm{g}\:+\:\mathrm{e}^{−\mathrm{1}} } \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com