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…
Question Number 108930 by mohammad17 last updated on 20/Aug/20 Answered by ajfour last updated on 20/Aug/20 $${z}=\frac{\left(\mathrm{1}+{i}\right)^{\mathrm{15}} }{\mathrm{128}}×\frac{\left(\sqrt{\mathrm{2}}\right)^{\mathrm{15}} }{\left(\sqrt{\mathrm{2}}\right)^{\mathrm{15}} }\:=\sqrt{\mathrm{2}}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}+\frac{{i}}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{15}} \\ $$$$\:\:\:=\:\sqrt{\mathrm{2}}{e}^{\mathrm{15}{i}\pi/\mathrm{4}} \:=\:\sqrt{\mathrm{2}}{e}^{−{i}\pi/\mathrm{4}} \\ $$$$\bar…
Question Number 108931 by mohammad17 last updated on 20/Aug/20 Commented by mohammad17 last updated on 20/Aug/20 $${sir}\:{can}\:{you}\:{help}\:{me}\:{pleas} \\ $$ Answered by Aziztisffola last updated on…
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Question Number 108927 by mohammad17 last updated on 20/Aug/20 Commented by mohammad17 last updated on 20/Aug/20 $${help}\:{me}\:{sir} \\ $$ Answered by 1549442205PVT last updated on…