Question Number 222733 by gabthemathguy25 last updated on 06/Jul/25 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 222730 by fantastic last updated on 06/Jul/25 Commented by fantastic last updated on 06/Jul/25 $${find}\:{x}\:{in}\:{terms}\:{of}\: \\ $$$$\alpha,\beta,\gamma, \\ $$ Answered by mr W…
Question Number 222756 by Osefavour last updated on 06/Jul/25 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}−\sqrt{\boldsymbol{\mathrm{cos}}\left(\boldsymbol{\mathrm{x}}\right)}}{\:\boldsymbol{\mathrm{x}}−\boldsymbol{\mathrm{xcos}}\left(\sqrt{\boldsymbol{\mathrm{x}}}\right)} \\ $$ Answered by gregori last updated on 07/Jul/25 $$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\mathrm{cos}\:{x}}{{x}\left(\mathrm{1}−\mathrm{cos}\:\sqrt{{x}}\:\right)\left(\mathrm{1}+\sqrt{\mathrm{cos}\:{x}}\right)} \\ $$$$\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2sin}\:^{\mathrm{2}}…
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Question Number 222758 by chidera last updated on 06/Jul/25 Commented by chidera last updated on 06/Jul/25 $${pls} \\ $$$$ \\ $$ Answered by fantastic last…
Question Number 222753 by mr W last updated on 06/Jul/25 Answered by A5T last updated on 07/Jul/25 $$\mathrm{Let}\:\theta\:\mathrm{be}\:\mathrm{the}\:\mathrm{base}\:\mathrm{angle}\:\mathrm{of}\:\mathrm{any}\:\mathrm{of}\:\mathrm{the}\:\mathrm{isosceles}\:\bigtriangleup \\ $$$$\mathrm{180}−\mathrm{2}\theta+\mathrm{180}−\left(\mathrm{2}\left(\mathrm{180}−\theta−?\right)\right)+\mathrm{90}=\mathrm{180} \\ $$$$\Rightarrow\mathrm{2}?=\mathrm{90}\Rightarrow?=\mathrm{45}° \\ $$ Commented…
Question Number 222754 by hardmath last updated on 06/Jul/25 Commented by hardmath last updated on 06/Jul/25 $$\int_{\mathrm{0}} ^{\:\mathrm{8}} \left(\mathrm{f}^{\:−\mathrm{1}} \left(\mathrm{x}\right)\:−\:\mathrm{f}\left(\mathrm{x}\right)\right)\:\mathrm{dx}\:=\:\mathrm{12} \\ $$$$\int_{\mathrm{0}} ^{\:\mathrm{8}} \:\mathrm{f}\left(\mathrm{x}\right)\:\mathrm{dx}\:=\:? \\…
Question Number 222684 by alvan545 last updated on 05/Jul/25 Commented by alvan545 last updated on 05/Jul/25 Answered by mr W last updated on 05/Jul/25 Commented…
Question Number 222685 by gabthemathguy25 last updated on 05/Jul/25 Answered by mr W last updated on 05/Jul/25 $$\mathrm{tan}\:\left({A}−\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}}\right)=\mathrm{tan}\:\left(\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{3}}+\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{7}}\right) \\ $$$$\frac{\mathrm{tan}\:{A}−\frac{\mathrm{1}}{\mathrm{2}}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{tan}\:{A}}=\frac{\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{7}}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}×\frac{\mathrm{1}}{\mathrm{7}}} \\ $$$$\Rightarrow\mathrm{tan}\:{A}=\frac{\mathrm{4}}{\mathrm{3}}\:\checkmark…
Question Number 222718 by Mingma last updated on 05/Jul/25 Commented by gabthemathguy25 last updated on 06/Jul/25 $${what}? \\ $$ Commented by AntonCWX8 last updated on…