Question Number 201044 by mnjuly1970 last updated on 28/Nov/23 $$ \\ $$$$\:\:\:\:\Omega\:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \int_{\mathrm{0}} ^{\:\mathrm{1}} \left({x}−{y}\:\right)^{\mathrm{2}} {sin}^{\:\mathrm{2}} \:\left(\:{x}+{y}\:\right){dxdy}=? \\ $$ Answered by mathematicsmagic last updated…
Question Number 201047 by Mingma last updated on 28/Nov/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 201041 by hardmath last updated on 28/Nov/23 $$\mathrm{4}\left(\mathrm{33}\right)\mathrm{7} \\ $$$$\mathrm{4}\left(\mathrm{24}\right)\mathrm{6} \\ $$$$\mathrm{5}\left(\:?\:\right)\mathrm{4} \\ $$$$ \\ $$$$\left.{a}\left.\right)\left.\mathrm{9}\left.\:\:\:\:\:{b}\right)\mathrm{18}\:\:\:\:\:{c}\right)\mathrm{27}\:\:\:\:\:{d}\right)\mathrm{36} \\ $$ Answered by Frix last updated…
Question Number 201037 by mr W last updated on 28/Nov/23 Commented by mr W last updated on 28/Nov/23 $${a}\:{triangle}\:{has}\:{sides}\:{a},\:{b},\:{c}.\:{find}\:{the} \\ $$$${fraction}\:{of}\:{its}\:{area}\:{covered}\:{by}\:{all} \\ $$$$\left({infinite}\right)\:{inscribed}\:{circles}\:{as}\:{shown}. \\ $$…
Question Number 201033 by Mingma last updated on 28/Nov/23 Answered by Frix last updated on 28/Nov/23 $$\mathrm{sin}\:\frac{\pi}{\mathrm{14}}\:\mathrm{sin}\:\frac{\mathrm{3}\pi}{\mathrm{14}}\:\mathrm{sin}\:\frac{\mathrm{9}\pi}{\mathrm{14}}\:={x} \\ $$$$\mathrm{0}<{x}<\mathrm{1} \\ $$$$\mathrm{Using}\:\mathrm{trigonometric}\:\mathrm{formulas}\:\Rightarrow \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{1}−\mathrm{cos}\:\frac{\pi}{\mathrm{7}}\:+\mathrm{sin}\:\frac{\mathrm{3}\pi}{\mathrm{14}}\:−\mathrm{sin}\:\frac{\pi}{\mathrm{14}}\right)={x}\:\bigstar \\ $$$$\mathrm{cos}\:\frac{\pi}{\mathrm{7}}\:−\mathrm{sin}\:\frac{\mathrm{3}\pi}{\mathrm{14}}\:+\mathrm{sin}\:\frac{\pi}{\mathrm{14}}\:=\mathrm{1}−\mathrm{4}{x}…
Question Number 201034 by Mingma last updated on 28/Nov/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 201035 by sonukgindia last updated on 28/Nov/23 Commented by mr W last updated on 28/Nov/23 $${the}\:{side}\:{length}\:{of}\:{the}\:{square}\:{can}\:{not} \\ $$$${be}\:{equal}\:{to}\:{the}\:{radius}\:{of}\:{the}\:{smaller} \\ $$$${semi}−{circles}! \\ $$ Commented…
Question Number 201028 by sonukgindia last updated on 28/Nov/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 201029 by sonukgindia last updated on 28/Nov/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 201027 by sonukgindia last updated on 28/Nov/23 Answered by Atomist last updated on 28/Nov/23 $${sechx}=\frac{\mathrm{1}}{{coshx}}\: \\ $$$${coshx}=\frac{{e}^{{x}} +{e}^{−{x}} }{\mathrm{2}} \\ $$$$\int{sechxdx}=\int\frac{\mathrm{2}}{{e}^{{x}} +{e}^{−{x}} }{dx}=…