Question Number 174514 by Best1 last updated on 03/Aug/22 $${a}+{b}+{c}=\mathrm{1} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} =\mathrm{2} \\ $$$${a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} =\mathrm{3} \\ $$$${then}\:{a}^{\mathrm{5}} +{b}^{\mathrm{5}} +{c}^{\mathrm{5}} \:?…
Question Number 174492 by naka3546 last updated on 02/Aug/22 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\:\frac{\mathrm{3}{x}\:\mathrm{tan}\:\frac{\mathrm{2}}{{x}}\:−\:\mathrm{2}{x}\:\mathrm{sin}\:\frac{\mathrm{3}}{{x}}}{\mathrm{cos}\:\frac{\mathrm{1}}{{x}}\:−\:\mathrm{cos}\:\frac{\mathrm{2}}{{x}}}\:=\:\:? \\ $$ Commented by kaivan.ahmadi last updated on 02/Aug/22 $${we}\:{use}\:{Taylor}\:{series} \\ $$$$\sim{li}\underset{{x}\rightarrow\infty} {{m}}\:\frac{\mathrm{3}{x}\left(\frac{\mathrm{2}}{{x}}−\frac{\mathrm{8}}{\mathrm{3}{x}^{\mathrm{3}} }\right)−\mathrm{2}{x}\left(\frac{\mathrm{3}}{{x}}−\frac{\mathrm{9}}{\mathrm{2}{x}^{\mathrm{3}}…
Question Number 108961 by ZiYangLee last updated on 20/Aug/20 $$\mathrm{Solve}\:\frac{\mid{x}−\mathrm{2}\mid+\mathrm{1}}{\mid{x}−\mathrm{2}\mid−\mathrm{1}}<\mathrm{3} \\ $$ Commented by Rasheed.Sindhi last updated on 20/Aug/20 $$\:\frac{\mid{x}−\mathrm{2}\mid+\mathrm{1}}{\mid{x}−\mathrm{2}\mid−\mathrm{1}}<\mathrm{3} \\ $$$$\:\frac{\mid{x}−\mathrm{2}\mid+\mathrm{1}}{\mid{x}−\mathrm{2}\mid−\mathrm{1}}<\frac{\mathrm{2}+\mathrm{1}}{\mathrm{2}−\mathrm{1}} \\ $$$$\:\frac{\mid{x}−\mathrm{2}\mid}{\mid{x}−\mathrm{2}\mid}<\frac{\mathrm{2}}{\mathrm{2}} \\…
Question Number 108950 by ZiYangLee last updated on 20/Aug/20 $$\mathrm{In}\:\bigtriangleup\mathrm{ABC},\: \\ $$$$\mathrm{prove}\:\mathrm{that}\:\frac{\mathrm{1}+\mathrm{cosA}−\mathrm{cosB}+\mathrm{cosC}}{\mathrm{1}+\mathrm{cosA}+\mathrm{cosB}−\mathrm{cosC}}=\mathrm{tan}\frac{\mathrm{B}}{\mathrm{2}}\mathrm{cot}\frac{\mathrm{C}}{\mathrm{2}} \\ $$ Answered by mnjuly1970 last updated on 20/Aug/20 $$\frac{\left(\mathrm{1}+{cos}\left(\mathrm{C}\right)\right)+\left({cos}\left(\mathrm{A}\right)−{cos}\left(\mathrm{B}\right)\right)}{\left(\mathrm{1}−{cos}\left(\mathrm{C}\right)\right)+\left({cos}\left(\mathrm{A}\right)+{cos}\left(\mathrm{B}\right)\right)} \\ $$$$=\frac{\mathrm{2}{cos}^{\mathrm{2}} \left(\frac{\mathrm{C}}{\mathrm{2}}\right)−\mathrm{2}{sin}\left(\frac{\mathrm{A}+\mathrm{B}}{\mathrm{2}}\right){sin}\left(\frac{\mathrm{A}−\mathrm{B}}{\mathrm{2}}\right)}{\mathrm{2}{sin}^{\mathrm{2}}…
Question Number 108949 by mohammad17 last updated on 20/Aug/20 Commented by kaivan.ahmadi last updated on 20/Aug/20 $${z}_{{x}} =\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−\left({x}+{e}^{\sqrt{{y}−\mathrm{1}}} \right)^{\mathrm{2}} }}.{tan}\left({xy}\right)+{y}\left(\mathrm{1}+{tan}^{\mathrm{2}} \left({xy}\right)\right).{sin}^{−\mathrm{1}} \left({x}+{e}^{\sqrt{{y}−\mathrm{1}}} \right) \\ $$$$\Rightarrow{z}_{{x}}…
Question Number 174477 by Best1 last updated on 02/Aug/22 Answered by Rasheed.Sindhi last updated on 02/Aug/22 $${Q}\mathrm{3}\:{monthly}\:{salary}\:\:\:\:{Birr}\:\mathrm{2000} \\ $$$${Commission}\:{on}\:\mathrm{30000}\left(\mathrm{45000}−\mathrm{15000}\right) \\ $$$$\:\:\:\:\:\:\:\:\:=\mathrm{10\%}\:{of}\:\mathrm{30000} \\ $$$$\:\:\:\:\:\:\:\:\:=\frac{\mathrm{10}}{\mathrm{100}}×\mathrm{30000}=\mathrm{3000} \\ $$$$\mathcal{T}{otal}\:{income}={monthly}\:{salary}+{commission}…
Question Number 108947 by mohammad17 last updated on 20/Aug/20 Answered by 1549442205PVT last updated on 21/Aug/20 $$\mathrm{We}\:\mathrm{find}\:\mathrm{the}\:\mathrm{angle}\:\alpha,\beta,\gamma\:\:\mathrm{between}\:\mathrm{the}\:\mathrm{vector} \\ $$$$\overset{\rightarrow} {\mathrm{m}}=\left(\mathrm{1},\mathrm{2},\mathrm{1}\right)\:\mathrm{and}\:\mathrm{the}\:\mathrm{axises}\:\:\mathrm{Ox},\mathrm{Oy},\mathrm{Oz} \\ $$$$\mathrm{cos}\alpha=\mathrm{cos}\left\{\left[\widehat {\left(\mathrm{1},\mathrm{2},\mathrm{1}\right),\left(\mathrm{1},\mathrm{0},\mathrm{0}\right)\right]}\right\}= \\ $$$$\frac{\mid\mathrm{1}.\mathrm{1}+\mathrm{2}.\mathrm{0}+\mathrm{1}.\mathrm{0}\mid}{\:\sqrt{\mathrm{1}^{\mathrm{2}}…
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Question Number 108936 by mohammad17 last updated on 20/Aug/20 Commented by mohammad17 last updated on 20/Aug/20 $${sir}\:{can}\:{you}\:{help}\:{me}\:\left({true}\:{or}\:{false}\right) \\ $$ Answered by Aziztisffola last updated on…
Question Number 174472 by floor(10²Eta[1]) last updated on 01/Aug/22 $$\mathrm{let}\:\mathrm{f},\:\mathrm{g}\:\mathrm{be}\:\mathrm{continuous}\:\mathrm{at}\:\left[\mathrm{a},\mathrm{b}\right],\:\mathrm{with}\:\mathrm{f}\left(\mathrm{x}\right)\geqslant\mathrm{0}\:\mathrm{at}\:\left[\mathrm{a},\mathrm{b}\right]. \\ $$$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{exists}\:\mathrm{some}\:\theta\in\left[\mathrm{a},\mathrm{b}\right]\:\mathrm{such}\:\mathrm{that} \\ $$$$\int_{\mathrm{a}} ^{\mathrm{b}} \mathrm{f}\left(\mathrm{x}\right)\mathrm{g}\left(\mathrm{x}\right)\mathrm{dx}=\mathrm{g}\left(\theta\right)\int_{\mathrm{a}} ^{\mathrm{b}} \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx} \\ $$ Terms of Service Privacy Policy…