Question Number 118546 by putriamalia last updated on 18/Oct/20 $${nilai}\:{maksimum}?{fungsi}\:{y}=\:\mathrm{1}+\:{sin}\:\mathrm{2}{x}\:+{cos}\:\mathrm{2}{x} \\ $$ Commented by mr W last updated on 18/Oct/20 $${y}=\mathrm{1}+\mathrm{sin}\:\mathrm{2}{x}+\mathrm{cos}\:\mathrm{2}{x} \\ $$$$=\mathrm{1}+\sqrt{\mathrm{2}}\:\mathrm{sin}\:\left(\mathrm{2}{x}+\frac{\pi}{\mathrm{4}}\right) \\ $$$$\leqslant\mathrm{1}+\sqrt{\mathrm{2}}…
Question Number 52841 by Tawa1 last updated on 13/Jan/19 Commented by maxmathsup by imad last updated on 14/Jan/19 $$\:\pi\:\rightarrow\mathrm{180}^{{o}} \\ $$$$\:\alpha\rightarrow{k}^{{o}} \:\Rightarrow\mathrm{180}^{} .\alpha\:={k}\pi\:\Rightarrow\alpha\:=\frac{{k}\pi}{\mathrm{180}}\:\Rightarrow \\ $$$${S}=\sum_{{k}=\mathrm{0}}…
Question Number 118369 by Samokoji last updated on 17/Oct/20 $$\:\mathrm{657}×\mathrm{10}^{\mathrm{4}} =\frac{\mathrm{5}}{\mathrm{7}}+ \\ $$ Commented by MJS_new last updated on 17/Oct/20 $${a}={b}+{x}\:\Rightarrow\:{x}={a}−{b} \\ $$$${a}−\frac{{p}}{{q}}=\frac{{aq}}{{q}}−\frac{{p}}{{q}}=\frac{{aq}−{p}}{{q}} \\ $$$$\mathrm{where}'\mathrm{s}\:\mathrm{the}\:\mathrm{problem}?…
Question Number 118366 by andilizhaa last updated on 17/Oct/20 $$\mathrm{4}.\:\mathrm{Turunan}\:\mathrm{fungsi}\:\mathrm{f}\left(\mathrm{x}\right)=^{\mathrm{5}} \sqrt{\left(\mathrm{10x}^{\mathrm{2}} −\mathrm{4}\right)^{\mathrm{8}} \:\:} \\ $$$$\mathrm{adalah}\:\mathrm{f}^{\mathrm{1}} \left(\mathrm{x}\right).\:\mathrm{Nilai}\:\mathrm{f}^{\mathrm{1}} \left(\mathrm{1}\right)=… \\ $$$${a}.\:\mathrm{1} \\ $$$${b}.\:\mathrm{8} \\ $$$${c}.\:\mathrm{14} \\ $$$$\mathrm{d}.\:\mathrm{16}…
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Question Number 118209 by bemath last updated on 16/Oct/20 Answered by MJS_new last updated on 16/Oct/20 $${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…