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Question Number 170143 by otchereabdullai@gmail.com last updated on 17/May/22 Commented by otchereabdullai@gmail.com last updated on 17/May/22 $$\mathrm{please}\:\mathrm{i}\:\mathrm{need}\:\mathrm{help}\:\mathrm{on}\:\mathrm{this}\:\mathrm{prof} \\ $$ Commented by otchereabdullai@gmail.com last updated on…
Question Number 170142 by otchereabdullai@gmail.com last updated on 17/May/22 Commented by otchereabdullai@gmail.com last updated on 17/May/22 $$\mathrm{please}\:\mathrm{i}\:\mathrm{need}\:\mathrm{help}\:\mathrm{on}\:\mathrm{this} \\ $$ Commented by otchereabdullai@gmail.com last updated on…
Question Number 39059 by jasno91 last updated on 02/Jul/18 Answered by $@ty@m last updated on 02/Jul/18 Commented by $@ty@m last updated on 02/Jul/18 Terms of…
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Question Number 170126 by daus last updated on 17/May/22 Commented by daus last updated on 17/May/22 $${pls}\:{show}\:{me}\:{the}\:{way},\:{btw}\:{x}=\mathrm{2} \\ $$ Commented by mr W last updated…
Question Number 170106 by Beginner last updated on 16/May/22 Answered by cortano1 last updated on 17/May/22 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\mathrm{5}^{{x}} \:\mathrm{sin}\:\left(\frac{{a}}{\mathrm{5}^{{x}} }\right)\: \\ $$$$\:{let}\:\frac{\mathrm{1}}{\mathrm{5}^{{x}} }\:=\:{t}\: \\ $$$$\:=\:\underset{{t}\rightarrow\mathrm{0}}…
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Question Number 170084 by Beginner last updated on 16/May/22 Commented by cortano1 last updated on 16/May/22 $$\:\:\:\mathrm{sin}\:{x}=\mathrm{3}\left(\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{cos}\:{x}+\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{3}}\:\mathrm{sin}\:{x}\right) \\ $$$$\:\:\:\mathrm{sin}\:{x}=\:\frac{\mathrm{3}}{\mathrm{2}}\:\mathrm{cos}\:{x}\:+\frac{\mathrm{3}}{\mathrm{2}}\sqrt{\mathrm{3}}\:\mathrm{sin}\:{x} \\ $$$$\:\left(\frac{\mathrm{2}−\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{2}}\right)\mathrm{sin}\:{x}\:=\:\frac{\mathrm{3}}{\mathrm{2}}\:\mathrm{cos}\:{x} \\ $$$$\:\:\:\mathrm{tan}\:{x}\:=\:\frac{\mathrm{3}}{\mathrm{2}−\mathrm{3}\sqrt{\mathrm{3}}}\:=\:\frac{\mathrm{3}\left(\mathrm{2}+\mathrm{3}\sqrt{\mathrm{3}}\right)}{\mathrm{4}−\mathrm{27}} \\ $$$$\:\:\:\mathrm{tan}\:{x}\:=\:−\frac{\mathrm{6}+\mathrm{9}\sqrt{\mathrm{3}}}{\mathrm{23}}…
Question Number 104542 by mohammad17 last updated on 22/Jul/20 Answered by Dwaipayan Shikari last updated on 22/Jul/20 $$\left.\mathrm{2}\right) \\ $$$${tan}^{−\mathrm{1}} \left(\frac{{y}}{{x}}\right)={log}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right) \\ $$$$\frac{\frac{\mathrm{1}}{{x}}.\frac{{dy}}{{dx}}−\frac{{y}}{{x}^{\mathrm{2}}…