Question Number 198210 by a.lgnaoui last updated on 14/Oct/23 $$\:\boldsymbol{\mathrm{Red}}\:\boldsymbol{\mathrm{Area}}? \\ $$ Commented by a.lgnaoui last updated on 14/Oct/23 Answered by mr W last updated…
Question Number 198178 by universe last updated on 13/Oct/23 $${f}\left({xf}\left({y}\right)+{x}\right)={xy}+{f}\left({x}\right) \\ $$$${f}:\mathbb{R}\rightarrow\mathbb{R} \\ $$$${f}\left({x}\right)=? \\ $$ Answered by Rasheed.Sindhi last updated on 13/Oct/23 $${Let}\:{f}\left({x}\right)={ax}+{b} \\…
Question Number 198156 by Erico last updated on 12/Oct/23 $$\mathrm{Prove}\:\mathrm{that}\: \\ $$$$\frac{\mathrm{2t}−\mathrm{1}}{\mathrm{lnt}−\mathrm{ln}\left(\mathrm{1}−\mathrm{t}\right)}=\underset{\:\mathrm{0}} {\int}^{\:\mathrm{1}} \mathrm{t}^{\mathrm{x}} \left(\mathrm{1}−\mathrm{t}\right)^{\mathrm{1}−\mathrm{x}} \mathrm{dx} \\ $$$$\mathrm{and}\:\:\:\:\underset{\:\mathrm{0}} {\int}^{\:\mathrm{1}} \frac{\mathrm{2t}−\mathrm{1}}{\mathrm{lnt}−\mathrm{ln}\left(\mathrm{1}−\mathrm{t}\right)}\mathrm{dt}\:\:=\:\:\frac{\pi}{\mathrm{2}}\underset{\:\mathrm{0}} {\int}^{\:\mathrm{1}} \frac{\mathrm{x}\left(\mathrm{1}−\mathrm{x}\right)}{\mathrm{sin}\left(\pi\mathrm{x}\right)}\mathrm{dx} \\ $$ Answered…
Question Number 198158 by mnjuly1970 last updated on 12/Oct/23 Answered by mr W last updated on 12/Oct/23 Commented by mr W last updated on 12/Oct/23…
Question Number 198175 by York12 last updated on 12/Oct/23 $${Prove}\:{The}\:{following}\:{Functional}\:{equation}: \\ $$$$\zeta\left({x},{s}\right)=\frac{\mathrm{2}\Gamma\left(\mathrm{1}−{s}\right)}{\left(\mathrm{2}\pi\right)^{\left(\mathrm{1}−{s}\right)} }\left\{{sin}\left(\frac{\pi{s}}{\mathrm{2}}\right)\underset{{m}=\mathrm{1}} {\overset{\infty} {\sum}}\left[\frac{{cos}\left(\mathrm{2}\pi{mx}\right)}{{m}^{\left(\mathrm{1}−{s}\right)} }\right]+{cos}\left(\frac{\pi{s}}{\mathrm{2}}\right)\underset{{m}=\mathrm{1}} {\overset{\infty} {\sum}}\left[\frac{{sin}\left(\mathrm{2}\pi{mx}\right)}{{m}^{\left(\mathrm{1}−{s}\right)} }\right]\right\} \\ $$ Answered by witcher3 last…
Question Number 198166 by mr W last updated on 12/Oct/23 $${if}\:{f}\left({x}\right)={x}^{\mathrm{2}} +{bx}+{c} \\ $$$${f}\left({f}\left(\mathrm{1}\right)\right)={f}\left({f}\left(\mathrm{2}\right)\right)=\mathrm{0}\:{and}\:{f}\left(\mathrm{1}\right)\neq{f}\left(\mathrm{2}\right) \\ $$$${find}\:{f}\left(\mathrm{0}\right)=? \\ $$ Answered by mr W last updated on…
Question Number 198176 by Blackpanther last updated on 12/Oct/23 Answered by som(math1967) last updated on 13/Oct/23 $${let}\:{side}\:{of}\:\blacksquare{PQRB}={x}\:{unit} \\ $$$${side}\:{of}\:\blacksquare\:{EFGH}\:={y}\:{unit} \\ $$$$\:{AB}={AP}+{PB} \\ $$$$\:\:\:\:\:={xtan}\mathrm{30}\:+{x}\:\left[\:\because\:\frac{{AP}}{{x}}={tan}\mathrm{30}\right] \\ $$$$\:{again}\:{AB}={AH}+{HB}…
Question Number 198161 by Blackpanther last updated on 12/Oct/23 Answered by Rasheed.Sindhi last updated on 12/Oct/23 $$\blacktriangle\mathrm{ABC}=\frac{\mathrm{1}}{\mathrm{2}}.\mathrm{AC}.\mathrm{BC} \\ $$$$\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{12}×{y}=\mathrm{6}{y} \\ $$$$\:\:\:\:\:\:\:\mathrm{48}\leqslant\mathrm{6}{y}\leqslant\mathrm{60} \\ $$$$\:\:\:\:\:\:\:\mathrm{8}\leqslant{y}\leqslant\mathrm{10} \\ $$$$\because\:{y}\in\mathbb{Z}…
Question Number 198141 by Erico last updated on 11/Oct/23 $$\underset{\:\mathrm{0}} {\int}^{\:\mathrm{1}} \:\frac{\mathrm{x}\left(\mathrm{1}−\mathrm{x}\right)}{\mathrm{sin}\left(\pi\mathrm{x}\right)}\mathrm{dx}=??? \\ $$ Answered by witcher3 last updated on 11/Oct/23 $$\frac{\mathrm{1}}{\mathrm{sin}\left(\pi\mathrm{x}\right)}=\frac{\mathrm{2ie}^{−\mathrm{i}\pi\mathrm{x}} }{\mathrm{1}−\mathrm{e}^{−\mathrm{2i}\pi\mathrm{x}} }\: \\…
Question Number 198136 by sonukgindia last updated on 11/Oct/23 Answered by Frix last updated on 11/Oct/23 $$\frac{\mathrm{3}{x}^{\mathrm{3}} +\mathrm{4}{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{16}}{{x}^{\mathrm{5}} +\mathrm{2}{x}^{\mathrm{4}} +\mathrm{16}{x}+\mathrm{32}}=\frac{\left({x}+\mathrm{4}\right)\left({x}^{\mathrm{2}} +\mathrm{4}\right)}{\left({x}+\mathrm{2}\right)\left({x}^{\mathrm{4}} +\mathrm{16}\right)}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\left({x}+\mathrm{2}\right)}−\frac{\left(\mathrm{2}−\sqrt{\mathrm{2}}\right){x}+\mathrm{2}\left(\mathrm{2}+\sqrt{\mathrm{2}}\right)}{\mathrm{8}\left({x}^{\mathrm{2}}…