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Question Number 118209 by bemath last updated on 16/Oct/20 Answered by MJS_new last updated on 16/Oct/20 $${AB}={y} \\ $$$$\left(\mathrm{1}\right)\:\:\:\:\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{20}^{\mathrm{2}} \\ $$$$\left(\mathrm{2}\right)\:\:\:\:\:\left({x}−\mathrm{7}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{15}^{\mathrm{2}}…
Question Number 52629 by scientist last updated on 10/Jan/19 $${show}\:{that}\:\frac{{sin}\alpha+{sin}\mathrm{3}\alpha+{sin}\mathrm{5}\alpha}{{cos}\alpha+{cos}\mathrm{3}\alpha+{cos}\mathrm{5}\alpha}=\:{tan}\mathrm{3}\alpha \\ $$ Answered by math1967 last updated on 10/Jan/19 $$\frac{{sin}\mathrm{3}\alpha+\mathrm{2sin}\:\mathrm{3}\alpha.\mathrm{cos}\:\mathrm{2}\alpha}{\mathrm{cos}\:\mathrm{3}\alpha+\mathrm{2cos}\:\mathrm{3}\alpha\mathrm{cos}\:\mathrm{2}\alpha} \\ $$$$=\frac{\mathrm{sin}\:\mathrm{3}\alpha\left(\mathrm{1}+\mathrm{cos}\:\mathrm{2}\alpha\right)}{\mathrm{cos}\:\mathrm{3}\alpha\left(\mathrm{1}+\mathrm{cos}\:\mathrm{2}\alpha\right)}={tan}\mathrm{3}\alpha \\ $$ Answered…
Question Number 118133 by andilizhaa last updated on 15/Oct/20 $$\mathrm{2}.\:\mathrm{Turunan}\:\mathrm{pertama}\:\mathrm{dari}\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{5x}^{\mathrm{4}} +\mathrm{x}\sqrt{\mathrm{x}}+\frac{\mathrm{6}}{\:\sqrt{×}}−\mathrm{sin}\:\mathrm{x}−\mathrm{2cos}\:\mathrm{x}+\mathrm{5}\:{adalah}… \\ $$$$\mathrm{a}.\:\mathrm{f}^{\mathrm{1}} \left(\mathrm{x}\right)=\mathrm{20x}^{\mathrm{3}} +\frac{\mathrm{3}}{\mathrm{2}}\sqrt{\mathrm{x}}+\frac{\mathrm{6}}{×\sqrt{×}}+\mathrm{cos}\:\mathrm{x}−\mathrm{2sin}\:\mathrm{x} \\ $$$$\mathrm{b}.\:\mathrm{f}^{\mathrm{1}} \left(\mathrm{x}\right)=\mathrm{20x}^{\mathrm{3}} +\frac{\mathrm{3}}{\mathrm{2}}\sqrt{\mathrm{x}}−\frac{\mathrm{3}}{\mathrm{x}\sqrt{\mathrm{x}}}−\mathrm{cos}\:\mathrm{x}+\mathrm{2sin}\:\mathrm{x} \\ $$$$\mathrm{c}.\:\mathrm{f}^{\mathrm{1}} \left(\mathrm{x}\right)=\mathrm{20x}^{\mathrm{3}} +\frac{\mathrm{3}}{\mathrm{2}}\sqrt{\mathrm{x}}−\frac{\mathrm{1}}{\mathrm{x}\sqrt{\mathrm{x}}}−\mathrm{cos}\:\mathrm{x}+\mathrm{2sin}\:\mathrm{x} \\ $$$$\mathrm{d}.\:\mathrm{f}^{\mathrm{1}}…
Question Number 118120 by bemath last updated on 15/Oct/20 $$\mathrm{tan}\:\left(\mathrm{tan}\:{x}\right)\:+\:\mathrm{tan}\:\left(\mathrm{2}{x}\right)=\mathrm{tan}\:\left(\mathrm{3}{x}\right) \\ $$ Commented by MJS_new last updated on 15/Oct/20 $$\mathrm{not}\:\mathrm{possible}\:\mathrm{to}\:\mathrm{completely}\:\mathrm{solve}\:\mathrm{it} \\ $$ Answered by TANMAY…
Question Number 183633 by a.lgnaoui last updated on 27/Dec/22 $${Montrer}\:{que} \\ $$$$\left(\frac{{a}}{{b}}\right)^{\mathrm{2}} =\frac{{c}}{{d}}+\mathrm{1} \\ $$ Commented by a.lgnaoui last updated on 27/Dec/22 Answered by mr…
Question Number 118078 by bemath last updated on 15/Oct/20 $$\mathrm{3}\left(\mathrm{cos}\:\mathrm{x}+\mathrm{sin}\:\mathrm{x}\right)^{\mathrm{4}} +\mathrm{6}\left(\mathrm{sin}\:\mathrm{x}−\mathrm{cos}\:\mathrm{x}\right)^{\mathrm{2}} − \\ $$$$\mathrm{3}\left(\mathrm{sin}\:^{\mathrm{6}} \mathrm{x}+\mathrm{cos}\:^{\mathrm{6}} \mathrm{x}\right)\:=\:?\: \\ $$ Answered by bobhans last updated on 15/Oct/20…
Question Number 118070 by andilizhaa last updated on 15/Oct/20 Commented by bemath last updated on 15/Oct/20 $$\mathrm{see}\:\mathrm{qn}\:\mathrm{118064} \\ $$ Commented by andilizhaa last updated on…
Question Number 118065 by andilizhaa last updated on 15/Oct/20 Answered by bobhans last updated on 15/Oct/20 $$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{1}−\mathrm{cot}\:\mathrm{x} \\ $$$$\frac{\mathrm{df}\left(\mathrm{x}\right)}{\mathrm{dx}}\:=\:\mathrm{cosec}\:^{\mathrm{2}} \left(\mathrm{x}\right) \\ $$$$\frac{\mathrm{df}\left(\mathrm{x}\right)}{\mathrm{dx}}\mid_{\frac{\pi}{\mathrm{6}}} \:=\:\frac{\mathrm{1}}{\mathrm{sin}\:^{\mathrm{2}} \left(\frac{\pi}{\mathrm{6}}\right)}\:=\:\mathrm{4} \\…
Question Number 118064 by andilizhaa last updated on 15/Oct/20 Answered by bobhans last updated on 15/Oct/20 $$\mathrm{f}\left(\mathrm{x}\right)=\sqrt[{\mathrm{5}}]{\left(\mathrm{10x}^{\mathrm{2}} −\mathrm{8}\right)^{\mathrm{4}} } \\ $$$$\mathrm{by}\:\mathrm{chain}\:\mathrm{rule} \\ $$$$\mathrm{letting}\:\mathrm{u}\:=\:\mathrm{10x}^{\mathrm{2}} −\mathrm{8}\:\rightarrow\frac{\mathrm{du}}{\mathrm{dx}}\:=\:\mathrm{20x} \\…