Question Number 80037 by Aashishyadavgangaria last updated on 30/Jan/20 $$\mathrm{A}\:\mathrm{matrix}\:{A}=\begin{bmatrix}{{a}_{{ij}} }\end{bmatrix}\:\mathrm{is}\:\mathrm{an}\:\mathrm{upper}\:\mathrm{triangular} \\ $$$$\mathrm{matrix}\:\mathrm{if} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 79950 by mhmd last updated on 29/Jan/20 $$\mathrm{If}\:\underset{{a}} {\overset{{b}} {\int}}\:\:\frac{{x}^{{n}} }{{x}^{{n}} +\left(\mathrm{16}−{x}\right)^{{n}} }\:{dx}\:=\:\mathrm{6},\:\mathrm{then}\: \\ $$ Commented by mr W last updated on 29/Jan/20…
Question Number 14128 by chux last updated on 28/May/17 $$\mathrm{If}\:\:\frac{\mathrm{P}{x}}{\left({b}−{c}\right)}\:=\:\frac{\mathrm{Q}{y}}{\left({c}−{a}\right)}\:=\:\frac{\mathrm{R}{z}}{\left({a}−{b}\right)}\:,\:\mathrm{then}\:\mathrm{find} \\ $$$$\mathrm{P}{ax}\:+\:\mathrm{Q}{by}\:+\:\mathrm{R}{cz}. \\ $$ Answered by ajfour last updated on 28/May/17 $${let}\:\:\frac{{Px}}{{b}−{c}}=\frac{{Qy}}{{c}−{a}}=\frac{{Rz}}{{a}−{b}}={m}\:\:\left({say}\right) \\ $$$${to}\:{find}:\:\:\: \\…
Question Number 79456 by Vishal Sharma last updated on 25/Jan/20 $$\mathrm{If}\:{A}\:\mathrm{and}\:{B}\:\mathrm{are}\:\mathrm{acute}\:\mathrm{positive}\:\mathrm{angles} \\ $$$$\mathrm{satisfying}\:\mathrm{the}\:\mathrm{equations}\: \\ $$$$\mathrm{3}\:\mathrm{sin}^{\mathrm{2}} {A}+\mathrm{2}\:\mathrm{sin}^{\mathrm{2}} {B}=\mathrm{1}\:\mathrm{and}\: \\ $$$$\mathrm{3}\:\mathrm{sin}\:\mathrm{2}{A}−\mathrm{2}\:\mathrm{sin}\:\mathrm{2}{B}=\mathrm{0},\:\mathrm{then}\:{A}+\mathrm{2}{B}= \\ $$ Commented by john santu…
Question Number 79452 by Vishal Sharma last updated on 25/Jan/20 $$\:\:\underset{\:\mathrm{1}} {\overset{\sqrt[{\mathrm{7}}]{\mathrm{2}}} {\int}}\:\frac{\mathrm{1}}{{x}\left(\mathrm{2}{x}^{\mathrm{7}} +\:\mathrm{1}\right)}\:{dx}\:= \\ $$ Commented by john santu last updated on 25/Jan/20 $${look}\:\int\:\frac{{dx}}{\mathrm{2}{x}^{\mathrm{8}}…
Question Number 79455 by Vishal Sharma last updated on 25/Jan/20 $$\mathrm{If}\:\:\mathrm{sin}\:\theta_{\mathrm{1}} +\mathrm{sin}\:\theta_{\mathrm{2}} +\mathrm{sin}\:\theta_{\mathrm{3}} \:=\:\mathrm{3},\:\mathrm{then} \\ $$$$\mathrm{cos}\:\theta_{\mathrm{1}} +\mathrm{cos}\:\theta_{\mathrm{2}} +\mathrm{cos}\:\theta_{\mathrm{3}} \:= \\ $$ Answered by $@ty@m123 last…
Question Number 79453 by Vishal Sharma last updated on 25/Jan/20 $$\mathrm{If}\:\mathrm{in}\:\mathrm{a}\:\mathrm{triangle}\:{ABC} \\ $$$$\mathrm{2}\:\frac{\mathrm{cos}\:{A}}{{a}}+\frac{\mathrm{cos}\:{B}}{{b}}+\mathrm{2}\frac{\mathrm{cos}\:{C}}{{c}}\:=\:\frac{{a}}{{bc}}\:+\:\frac{{b}}{{ca}} \\ $$$$\mathrm{then}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{the}\:\mathrm{angle}\:{A}\:\mathrm{is} \\ $$ Answered by $@ty@m123 last updated on 25/Jan/20 $$\mathrm{2}\:\frac{\mathrm{cos}\:{A}}{{a}}+\frac{\mathrm{cos}\:{B}}{{b}}+\mathrm{2}\frac{\mathrm{cos}\:{C}}{{c}}\:=\:\frac{{a}}{{bc}}\:+\:\frac{{b}}{{ca}}…
Question Number 79437 by Vishal Sharma last updated on 25/Jan/20 $$\mathrm{The}\:\mathrm{value}\:\mathrm{of}\:\mathrm{tan}\:\mathrm{1}°\:\mathrm{tan}\:\mathrm{2}°\:\mathrm{tan}\:\mathrm{3}°…\mathrm{tan}\:\mathrm{89}° \\ $$$$\mathrm{is} \\ $$ Commented by mr W last updated on 25/Jan/20 $$=\mathrm{1} \\…
Question Number 79435 by Vishal Sharma last updated on 25/Jan/20 $$\mathrm{16}\:\mathrm{cos}\:\frac{\mathrm{2}\pi}{\mathrm{15}}\:\mathrm{cos}\:\frac{\mathrm{4}\pi}{\mathrm{15}}\:\mathrm{cos}\:\frac{\mathrm{8}\pi}{\mathrm{15}}\:\mathrm{cos}\:\frac{\mathrm{14}\pi}{\mathrm{15}}\:=\_\_\_\_. \\ $$ Commented by jagoll last updated on 25/Jan/20 $$\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{m}} {\prod}}\mathrm{cos}\:\left(\frac{\mathrm{k}\pi}{\mathrm{2m}+\mathrm{1}}\right)=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}^{\mathrm{m}} } \\…
Question Number 13490 by aamaguestt last updated on 20/May/17 $$\mathrm{The}\:\mathrm{area}\:\mathrm{of}\:\mathrm{a}\:\mathrm{rectangle}\:\mathrm{is}\:\mathrm{255}\:\mathrm{m}^{\mathrm{2}} .\:\mathrm{If} \\ $$$$\mathrm{its}\:\mathrm{length}\:\mathrm{is}\:\mathrm{decreased}\:\mathrm{by}\:\mathrm{1}\:\mathrm{m},\:\mathrm{it}\: \\ $$$$\mathrm{becomes}\:\mathrm{a}\:\mathrm{square}.\:\mathrm{The}\:\mathrm{perimeter}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{square}\:\mathrm{is}\:\_\_\_\_\:\mathrm{m}. \\ $$ Answered by ajfour last updated on…