Question Number 182875 by HeferH last updated on 15/Dec/22 $$ \\ $$$$\:{Find}\:{the}\:{value}\:{of}\:{S}: \\ $$$$\:{S}\:=\:\mathrm{1}\:+\:\frac{\mathrm{1}}{\mathrm{1}×\mathrm{2}}\:+\:\frac{\mathrm{1}}{\mathrm{2}×\mathrm{3}}\:+\:\frac{\mathrm{1}}{\mathrm{3}×\mathrm{4}}\:… \\ $$$$\: \\ $$ Answered by TheSupreme last updated on 16/Dec/22…
Question Number 182866 by Shrinava last updated on 15/Dec/22 Answered by HeferH last updated on 15/Dec/22 $${if}\:{CD}\:\parallel\:{AB}\:: \\ $$$$\:\mathrm{8}{a}\:−\:\mathrm{60}\:=\:\mathrm{180}\: \\ $$$$\:{a}\:=\:\frac{\mathrm{240}}{\mathrm{8}}\:=\:\mathrm{30}° \\ $$ Terms of…
Question Number 182852 by mnjuly1970 last updated on 16/Dec/22 $$ \\ $$$$\:{challanging}\:{question}.. \\ $$$$\:\:{if}\:\:\:,\:\frac{\mathrm{1}}{\:\mathrm{1}+\sqrt{\mathrm{1}−{x}+{x}^{\:\mathrm{2}} }}\:−\frac{{x}}{\mathrm{1}+\sqrt{\mathrm{1}+{x}+{x}^{\:\mathrm{2}} }}=\mathrm{1} \\ $$$$\:\:\:{find}\:\:{the}\:\:{value}\:{of}\::\:\:\:\frac{{x}}{{x}^{\:\mathrm{2}} +\mathrm{1}} \\ $$$$ \\ $$ Commented by…
Question Number 51772 by Cheyboy last updated on 30/Dec/18 $$\mathrm{x}^{\mathrm{3}} +\mathrm{12x}+\mathrm{12}=\mathrm{0} \\ $$$$ \\ $$$$\mathrm{Express}\:\mathrm{your}\:\mathrm{answer}\:\mathrm{in}\:\mathrm{surd}\:\mathrm{form} \\ $$$$ \\ $$ Answered by ajfour last updated on…
Question Number 182804 by mnjuly1970 last updated on 14/Dec/22 $$ \\ $$$$\:\:\:\:{solve} \\ $$$$\:\:\:\:\lfloor\underset{} {\overset{} {{x}}}\:\rfloor\:−\:\lfloor\:\frac{{x}}{\mathrm{3}}\:\rfloor\:=\:\mathrm{3} \\ $$$$\:\:\:\:\:\:\: \\ $$ Answered by mahdipoor last updated…
Question Number 182794 by mnjuly1970 last updated on 14/Dec/22 $$ \\ $$$$\:\:\:\:\:\:\:\Omega\:=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\:\left(−\underset{} {\overset{} {\mathrm{1}}}\:\right)^{\:{n}} }{\mathrm{2}^{\:{n}} .{n}\left({n}\underset{} {\overset{} {+}}\mathrm{1}\right)\left({n}\underset{} {\overset{} {+}}\mathrm{2}\right)}\:=\:?\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$ \\…
Question Number 182788 by CrispyXYZ last updated on 14/Dec/22 $${y}''+{ay}={b} \\ $$ Answered by qaz last updated on 14/Dec/22 $${y}_{{p}} =\frac{\mathrm{1}}{{D}^{\mathrm{2}} +{a}}{b}=\frac{{b}}{\mathrm{2}{i}\sqrt{{a}}}\left(\frac{\mathrm{1}}{{D}−{i}\sqrt{{a}}}−\frac{\mathrm{1}}{{D}+{i}\sqrt{{a}}}\right) \\ $$$$=\frac{{b}}{\mathrm{2}{i}\sqrt{{a}}}\left({e}^{{i}\sqrt{{a}}{x}} \frac{\mathrm{1}}{{D}}{e}^{−{i}\sqrt{{a}}{x}}…
Question Number 51691 by peter frank last updated on 29/Dec/18 Answered by afachri last updated on 29/Dec/18 $$\mathrm{suppose}\:\gamma\:=\:{U}_{\mathrm{1}} \:;\:\beta\:=\:{U}_{\mathrm{2}} \:;\:\alpha\:=\:{U}_{\mathrm{3}} \\ $$$$\boldsymbol{\gamma}\:,\:\boldsymbol{\beta}\:\:=\:\:\frac{−{a}\:\pm\:\sqrt{{a}^{\mathrm{2}} −\:\mathrm{4}{b}\:}}{\mathrm{2}} \\ $$$$\boldsymbol{\beta}\:−\:\boldsymbol{\gamma}\:\:=\:\:\frac{\left(−{a}\:−\:\sqrt{{a}^{\mathrm{2}}…
Question Number 182721 by mnjuly1970 last updated on 13/Dec/22 Answered by Acem last updated on 13/Dec/22 $$\:{It}\:{period}:\:{T}=\:\frac{\mathrm{1}}{{a}} \\ $$$$\:{f}\left({x}\right)=\mathrm{1}\:\Rightarrow\:{b}=\mathrm{1} \\ $$$$\underset{{x}\rightarrow\frac{\mathrm{15}^{\:−} }{\mathrm{2}}} {\mathrm{lim}}\:{f}\left({x}\right)=\:−\infty\:\Rightarrow\:\mathrm{cos}\:\pi{ax}=\:\mathrm{cos}\:\frac{\pi^{\:+} }{\mathrm{2}}\:\Rightarrow\:{a}<\mathrm{0} \\…
Question Number 182720 by mnjuly1970 last updated on 13/Dec/22 Terms of Service Privacy Policy Contact: info@tinkutara.com