Question Number 57222 by Tawa1 last updated on 31/Mar/19 $$\mathrm{Express}\:\:\:\mathrm{5}.\mathrm{27}\:\:\mathrm{in}\:\mathrm{form}\:\mathrm{of}\:\mathrm{a}\:\mathrm{series}\:\mathrm{and}\:\mathrm{show}\:\mathrm{that}\:\mathrm{is}\:\mathrm{equal} \\ $$$$\mathrm{to}\:\:\:\mathrm{5}\:\frac{\mathrm{5}}{\mathrm{18}} \\ $$ Answered by Joel578 last updated on 01/Apr/19 $$\mathrm{5}.\mathrm{2}\overset{\_} {\mathrm{7}}\:=\:\mathrm{5}.\mathrm{2}\:+\:\frac{\mathrm{7}}{\mathrm{10}^{\mathrm{2}} }\:+\:\frac{\mathrm{7}}{\mathrm{10}^{\mathrm{3}} }\:+\:\frac{\mathrm{7}}{\mathrm{10}^{\mathrm{4}}…
Question Number 188286 by mustafazaheen last updated on 27/Feb/23 $${when}\:\:\:\:\:\:{sin}\left({x}\right)+{cos}\left({x}\right)={a} \\ $$$${find}\:\:\:\:\:\:\:\:\:{sec}\left({x}\right)+{csc}\left({x}\right)=? \\ $$ Answered by ARUNG_Brandon_MBU last updated on 27/Feb/23 $$\mathrm{sin}{x}+\mathrm{cos}{x}={a}\:\Rightarrow\mathrm{sin}{x}\mathrm{cos}{x}=\frac{{a}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{sec}{x}+\mathrm{cosec}{x}=\frac{\mathrm{sin}{x}+\mathrm{cos}{x}}{\mathrm{sin}{x}\mathrm{cos}{x}}=\frac{\mathrm{2}{a}}{{a}^{\mathrm{2}}…
Question Number 188280 by MATHEMATICSAM last updated on 27/Feb/23 $$\mathrm{Prove}\:\mathrm{that}\:\mathrm{1}+\mathrm{2}+\mathrm{3}+\mathrm{4}+…..\:=\:−\frac{\mathrm{1}}{\mathrm{12}} \\ $$ Commented by normans last updated on 27/Feb/23 $$\:{I}\:{think}\:{is}\:{the}\:{old}\:{problem} \\ $$ Commented by JDamian…
Question Number 188270 by mnjuly1970 last updated on 27/Feb/23 Commented by mnjuly1970 last updated on 27/Feb/23 $$\:\:{in}\:\:{A}\overset{\Delta} {{B}C}\:{prove}\::\: \\ $$$$\:\:\:\frac{\mathrm{1}}{{cos}\left({A}\right)}\:+\frac{\mathrm{1}}{{cos}\left({B}\right)}\:+\frac{\mathrm{1}}{{cos}\left({C}\right)}\:\geqslant\:\mathrm{6} \\ $$$$\: \\ $$$$\:\:\:\:\:{note}\::\:\mathrm{0}\:<\:{A}\:,\:{B}\:,\:{C}\:<\:\mathrm{90}^{°} \\…
Question Number 188263 by mathlove last updated on 27/Feb/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 188262 by normans last updated on 27/Feb/23 $$ \\ $$$$\:\:\:\:\:\:\:\boldsymbol{{solve}}\:\boldsymbol{{the}}\:\boldsymbol{{equation}};\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\left.\begin{matrix}{\boldsymbol{{x}}\:+\:\boldsymbol{{y}}\:+\boldsymbol{{z}}\:=\:\:\mathrm{30}\sqrt{\mathrm{2}}}\\{\boldsymbol{{x}}\:−\:\boldsymbol{{y}}\:−\:\boldsymbol{{z}}\:=\:\mathrm{7},\mathrm{5}}\\{\boldsymbol{{x}}\:+\:\boldsymbol{{y}}\:−\:\boldsymbol{{z}}\:=\:\sqrt{\mathrm{22}}}\end{matrix}\right\} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{x}}\:;\:\boldsymbol{{y}}\:;\:\boldsymbol{{z}}\:=\:?? \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:{they}\:{form}\:{funny}\:{positions}\: \\ $$$$ \\ $$ Answered…
Question Number 188248 by mathlove last updated on 27/Feb/23 $$\left({x}^{\mathrm{3}} −{y}−\mathrm{3}{x}\right)\left[\left({x}^{\mathrm{3}} −\mathrm{3}{x}\right)^{\mathrm{2}} −{y}^{\mathrm{2}} \right]=\mathrm{200} \\ $$$$\left({x}^{\mathrm{3}} +{y}−\mathrm{3}{x}\right)\left[\left({x}^{\mathrm{3}} −\mathrm{3}{x}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} \right]=\mathrm{600} \\ $$$${solved}\:{in}\:{R} \\ $$ Terms…
Question Number 188247 by cortano12 last updated on 27/Feb/23 $$\mathrm{Prove}\:\mathrm{that}\: \\ $$$$\left(\mathrm{1}\right)\:\mathrm{5555}^{\mathrm{2222}} +\mathrm{2222}^{\mathrm{5555}} \:\mathrm{divisible}\:\mathrm{by}\:\mathrm{7} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{3}^{\mathrm{105}} +\mathrm{4}^{\mathrm{105}} \:\mathrm{divisible}\:\mathrm{by}\:\mathrm{7}\: \\ $$ Answered by Rasheed.Sindhi last updated…
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Question Number 188224 by Shrinava last updated on 26/Feb/23 $$\mathrm{If}\:\:\:\Omega\:=\:\underset{\boldsymbol{\mathrm{n}}=\mathrm{1}} {\overset{\infty} {\sum}}\:\left(\underset{\boldsymbol{\mathrm{k}}=\mathrm{2}} {\overset{\infty} {\prod}}\:\frac{\mathrm{k}^{\mathrm{3}} \:−\:\mathrm{1}}{\mathrm{k}^{\mathrm{3}} \:+\:\mathrm{1}}\right)^{\boldsymbol{\mathrm{n}}} \\ $$$$\mathrm{Solve}\:\mathrm{for}\:\mathrm{complex}\:\mathrm{numbees}: \\ $$$$\mathrm{z}^{\mathrm{4}} \:+\:\mathrm{3z}^{\mathrm{3}} \:+\:\Omega\mathrm{z}^{\mathrm{2}} \:+\:\mathrm{3z}\:+\:\mathrm{1}\:=\:\mathrm{0} \\ $$…