Question Number 209974 by Abdullahrussell last updated on 27/Jul/24 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 209975 by Ismoiljon_008 last updated on 27/Jul/24 Answered by mr W last updated on 27/Jul/24 Commented by mr W last updated on 27/Jul/24…
Question Number 209932 by cherokeesay last updated on 26/Jul/24 Answered by mahdipoor last updated on 26/Jul/24 $${BD}^{\mathrm{2}} =\mathrm{100}+\mathrm{49}−\mathrm{140}{cosC}=\mathrm{121}+\mathrm{64}−\mathrm{176}{cosA} \\ $$$$\frac{{BD}}{{sinC}}=\frac{{BD}}{{sinA}}=\mathrm{2}{R}\:\Rightarrow{C}+{A}=\mathrm{180} \\ $$$$\Rightarrow\mathrm{149}−\mathrm{140}{cosC}=\mathrm{185}−\mathrm{176}{cosA} \\ $$$$\Rightarrow{cosC}=\frac{\mathrm{185}−\mathrm{149}}{−\mathrm{176}−\mathrm{140}}=−\frac{\mathrm{9}}{\mathrm{79}} \\…
Question Number 209944 by Tawa11 last updated on 26/Jul/24 Commented by mr W last updated on 26/Jul/24 $${f}=\frac{\mathrm{100}×\mathrm{6}}{\mathrm{2}\pi×\mathrm{2}.\mathrm{5}}=\mathrm{38}.\mathrm{2}\:{Hz} \\ $$ Terms of Service Privacy Policy…
Question Number 209929 by Abdullahrussell last updated on 26/Jul/24 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 209956 by Jubr last updated on 26/Jul/24 Find the maximum value of 7cosA + 24sinA + 32 Answered by Frix last updated on…
Question Number 209924 by OmoloyeMichael last updated on 26/Jul/24 $$\boldsymbol{{If}}\:\:\boldsymbol{{f}}\left(\boldsymbol{{x}}\right)=\left(\boldsymbol{{x}}!\right)\centerdot\left(\boldsymbol{{x}}!!\right)\centerdot\left(\boldsymbol{{x}}!!!\right)\:\: \\ $$$$\boldsymbol{{find}}\:\:\frac{\boldsymbol{{d}}}{\boldsymbol{{dx}}}\left(\boldsymbol{{f}}\left(\boldsymbol{{x}}\right)\right)=? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 209957 by Jubr last updated on 26/Jul/24 If x, y are contain in natural numbers and x² + y² = 613² Find the…
Question Number 209926 by depressiveshrek last updated on 26/Jul/24 $$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{1}}{\mathrm{3}{n}+\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{3}{n}+\mathrm{2}}+…+\frac{\mathrm{1}}{\mathrm{4}{n}} \\ $$ Commented by Frix last updated on 26/Jul/24 $$\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\:\frac{\mathrm{1}}{\mathrm{3}{n}+{k}}\:={H}_{\mathrm{4}{n}} −{H}_{\mathrm{3}{n}} \\…
Question Number 209937 by Ari last updated on 26/Jul/24 Answered by mr W last updated on 26/Jul/24 Commented by mr W last updated on 26/Jul/24…