Question Number 126987 by sdfg last updated on 25/Dec/20 Answered by mr W last updated on 26/Dec/20 $${Q}\mathrm{4} \\ $$$$\boldsymbol{{C}}=\alpha\boldsymbol{{A}}+\beta\boldsymbol{{B}} \\ $$$$\mid\boldsymbol{{C}}\mid=\sqrt{\alpha^{\mathrm{2}} \mid\boldsymbol{{A}}\mid^{\mathrm{2}} +\beta^{\mathrm{2}} \mid\boldsymbol{{B}}\mid^{\mathrm{2}}…
Question Number 126971 by mnjuly1970 last updated on 25/Dec/20 $$\:\:\:\:\:\:\:\:\:\:\:\:…{nice}\:\:{calculus}… \\ $$$$\:\:\:\:{evaluate}\:': \\ $$$$\:\:\:\:\:\:\Omega=\int_{\mathrm{0}} ^{\:\mathrm{1}} \left({arctan}\left({x}\right)\right)^{\mathrm{2}} {dx}=? \\ $$$$\:\:\:\: \\ $$ Answered by Dwaipayan Shikari…
Question Number 126969 by mnjuly1970 last updated on 25/Dec/20 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{calculus} \\ $$$$\:\:\:\:\:\:{please}\:\:{evaluate}\::: \\ $$$$\:\:\:\:\:\:\:\:\phi\:=\int_{\mathrm{0}} ^{\:\mathrm{1}} \lfloor{ln}\left({x}\right)\rfloor{dx}=? \\ $$$$ \\ $$ Answered by mindispower last updated…
Question Number 192466 by Engr_Jidda last updated on 18/May/23 Commented by Frix last updated on 19/May/23 $$\left(\mathrm{sin}^{\mathrm{4}} \:\theta\:−\mathrm{cos}^{\mathrm{4}} \:\theta\right)\mathrm{csc}^{\mathrm{2}} \:\theta\:=\mathrm{2} \\ $$$$\frac{\mathrm{2sin}^{\mathrm{2}} \:\theta\:−\mathrm{1}}{\mathrm{sin}^{\mathrm{2}} \:\theta}=\mathrm{2} \\…
Question Number 126921 by sdfg last updated on 25/Dec/20 Commented by mr W last updated on 25/Dec/20 $$\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}}…
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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…