Question Number 198232 by SANOGO last updated on 16/Oct/23 $${find}\: \\ $$$$\underset{{k}={o}} {\overset{{n}} {\sum}}{sin}\left({k}\right) \\ $$ Answered by witcher3 last updated on 14/Oct/23 $$\mathrm{sin}\left(\mathrm{k}\right)\in\mathbb{R} \\…
Question Number 198228 by sulaymonnorboyev140 last updated on 14/Oct/23 $$\left(\mathrm{4}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}\right)^{{x}^{\mathrm{2}} −{x}} >\mathrm{1} \\ $$ Answered by MM42 last updated on 14/Oct/23 $$\begin{cases}{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}>\mathrm{1}}\\{{x}^{\mathrm{2}} −{x}>\mathrm{0}}\end{cases}\Rightarrow\left(−\infty,−\frac{\mathrm{1}}{\mathrm{2}}\right)\cup\left(\mathrm{0},+\infty\right)={A}\:\:\&\:\left(−\infty,\mathrm{0}\right)\cup\left(\mathrm{1},+\infty\right)={B}…
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Question Number 198184 by Blackpanther last updated on 13/Oct/23 Answered by Rasheed.Sindhi last updated on 13/Oct/23 $$\mathrm{Sector}\:\mathrm{DFG}=\mathrm{quarter}\:\mathrm{of}\:\mathrm{the}\:\mathrm{circle}\:\mathrm{of}\:\mathrm{radius}\:\mathrm{3} \\ $$$$=\pi\left(\mathrm{3}\right)^{\mathrm{2}} /\mathrm{4}=\frac{\mathrm{9}\pi}{\mathrm{4}} \\ $$$${Shaded}\:{area}\:{in}\:{one}\:{circle}={Sector}\:\mathrm{DFG}−\blacktriangle\mathrm{DFG} \\ $$$$\:\:\:\:\:\:\:=\frac{\mathrm{9}\pi}{\mathrm{4}}−\frac{\mathrm{3}×\mathrm{3}}{\mathrm{2}}=\frac{\mathrm{9}\pi−\mathrm{18}}{\mathrm{4}} \\…
Question Number 198182 by Blackpanther last updated on 13/Oct/23 Answered by mr W last updated on 14/Oct/23 Commented by mr W last updated on 14/Oct/23…
Question Number 198176 by Blackpanther last updated on 12/Oct/23 Answered by som(math1967) last updated on 13/Oct/23 $${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} \\ $$$$\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{12}×{y}=\mathrm{6}{y} \\ $$$$\:\:\:\:\:\:\:\mathrm{48}\leqslant\mathrm{6}{y}\leqslant\mathrm{60} \\ $$$$\:\:\:\:\:\:\:\mathrm{8}\leqslant{y}\leqslant\mathrm{10} \\ $$$$\because\:{y}\in\mathbb{Z}…
Question Number 198136 by sonukgindia last updated on 11/Oct/23 Answered by Frix last updated on 11/Oct/23 $$\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)}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\left({x}+\mathrm{2}\right)}−\frac{\left(\mathrm{2}−\sqrt{\mathrm{2}}\right){x}+\mathrm{2}\left(\mathrm{2}+\sqrt{\mathrm{2}}\right)}{\mathrm{8}\left({x}^{\mathrm{2}}…
Question Number 198151 by sonukgindia last updated on 11/Oct/23 Answered by Mathspace last updated on 12/Oct/23 $${I}=\int_{\mathrm{0}} ^{\infty} \:\frac{{zlnz}}{\mathrm{1}+{z}^{\mathrm{3}} }{dz}\:\:\:\:\left({z}={t}^{\frac{\mathrm{1}}{\mathrm{3}}} \right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{0}} ^{\infty} \:\:\frac{{t}^{\frac{\mathrm{1}}{\mathrm{3}}}…