Question Number 126921 by sdfg last updated on 25/Dec/20 Commented by mr W last updated on 25/Dec/20 $$\overset{\rightarrow} {{C}}=\overset{\rightarrow} {{A}}+\overset{\rightarrow} {{B}} \\ $$$$\overset{\rightarrow} {{C}}'=\frac{\mathrm{1}}{\mathrm{2}}\overset{\rightarrow} {{A}}+\frac{\mathrm{1}}{\mathrm{2}}\overset{\rightarrow}…
Question Number 192407 by mnjuly1970 last updated on 17/May/23 $$ \\ $$$$\:\:\:\:\:{find}\:\:{the}\:{value}\:{of}\:{the}\:{following}\:{integral} \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\chi\:=\:\int_{\mathrm{0}} ^{\:\infty} \frac{\:\mathrm{ln}^{\:\mathrm{2}} \left({x}\:\right)}{\mathrm{1}+\:{x}^{\:\mathrm{2}} }\:{dx}\:=\:?\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:−−−−−−−−− \\ $$ Answered…
Question Number 126859 by sdfg last updated on 24/Dec/20 Answered by mr W last updated on 24/Dec/20 $${G}=\mathrm{10}×\mathrm{sin}\:\mathrm{60}°×\mathrm{cos}\:\mathrm{45}°\:{i}+\mathrm{10}×\mathrm{sin}\:\mathrm{60}°×\mathrm{cos}\:\mathrm{45}°\:{j}+\mathrm{10}×\mathrm{cos}\:\mathrm{60}°\:{k} \\ $$$$\Rightarrow\boldsymbol{{G}}=\frac{\mathrm{5}\sqrt{\mathrm{6}}}{\mathrm{2}}\:\boldsymbol{{i}}+\frac{\mathrm{5}\sqrt{\mathrm{6}}}{\mathrm{2}}\:\boldsymbol{{j}}+\mathrm{5}\:\boldsymbol{{k}} \\ $$$$ \\ $$$${H}=−\mathrm{15}×\mathrm{sin}\:\mathrm{45}°×\mathrm{sin}\:\mathrm{30}°\:{i}+\mathrm{15}×\mathrm{sin}\:\mathrm{45}°×\mathrm{cos}\:\mathrm{30}°\:{j}+\mathrm{15}×\mathrm{cos}\:\mathrm{45}°\:{k} \\…
Question Number 126854 by sdfg last updated on 24/Dec/20 Answered by mathmax by abdo last updated on 24/Dec/20 $$\mathrm{h}\rightarrow\mathrm{r}^{\mathrm{2}} \:+\mathrm{k}^{\mathrm{2}} =\mathrm{0}\:\Rightarrow\mathrm{r}^{\mathrm{2}} =−\mathrm{k}^{\mathrm{2}} \:\Rightarrow\mathrm{r}\:=\overset{−} {+}\mathrm{ik}\:\Rightarrow\mathrm{y}_{\mathrm{h}} =\mathrm{ae}^{\mathrm{ikx}}…
Question Number 126813 by BHOOPENDRA last updated on 24/Dec/20 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 126806 by sdfg last updated on 24/Dec/20 Commented by sdfg last updated on 24/Dec/20 $${pleaes}\:{help} \\ $$ Commented by mahdipoor last updated on…
Question Number 192343 by cortano12 last updated on 15/May/23 $$\:\mathrm{Find}\:\mathrm{minimum}\:\mathrm{value}\:\mathrm{of}\: \\ $$$$\:\:\:\:\:\:\:\mathrm{y}=\frac{\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}}{\mathrm{2}\:\mathrm{tan}\:^{\mathrm{2}} \mathrm{x}}\: \\ $$ Answered by manxsol last updated on 15/May/23 $$\:\:{x}\neq{k}\pi\frac{\pi}{\mathrm{2}}\:\:\:\:{y}>\mathrm{0} \\…
Question Number 126776 by snipers237 last updated on 24/Dec/20 $$\:{Nature}\:{and}\:{Sum}\:\Sigma{u}_{{n}} \\ $$$$\:\:{where}\:{u}_{{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{{n}+\mathrm{1}} }{\mathrm{1}+{x}+…+{x}^{{n}} }{dx} \\ $$ Answered by mindispower last updated on…
Question Number 126777 by snipers237 last updated on 24/Dec/20 $$\left.{let}\:{consider}\:\left({u}_{{n}} \right)\:{such}\:{as}\:{u}_{\mathrm{0}} \in\right]\mathrm{0};\mathrm{1}\left[\:{and}\:{u}_{{n}+\mathrm{1}} ={u}_{{n}} −{u}_{{n}} ^{\mathrm{2}} \:\right. \\ $$$$\left.\mathrm{1}\right){Prove}\:{that}\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:^{{n}} \sqrt{{u}_{{n}} }\:=\:\mathrm{1}\:{and}\:{that}\:{the}\:{convergence}\:{domain}\:{of}\:\Sigma{u}_{{n}} {x}^{{n}} \: \\ $$$${is}\:\:{D}=\left[−\mathrm{1};\mathrm{1}\left[\:\right.\right.…
Question Number 126764 by BHOOPENDRA last updated on 24/Dec/20 Terms of Service Privacy Policy Contact: info@tinkutara.com