Question Number 77565 by TawaTawa last updated on 07/Jan/20 Commented by MJS last updated on 07/Jan/20 $$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{think}\:\mathrm{that}\:{A}^{{B}} \:\mathrm{for}\:\mathrm{matrices}\:{A},\:{B}\:\mathrm{is} \\ $$$$\mathrm{defined}.\:\mathrm{I}\:\mathrm{only}\:\mathrm{know}\:{A}^{{n}} \:\mathrm{with}\:{n}\in\mathbb{N};\:{A}^{−{n}} \:\mathrm{is} \\ $$$$\mathrm{only}\:\mathrm{possible}\:\mathrm{if}\:\mathrm{the}\:\mathrm{inverse}\:\mathrm{matrix}\:{A}^{−\mathrm{1}} \:\mathrm{exists}…
Question Number 12028 by tawa last updated on 09/Apr/17 $$\mathrm{Express}\::\:\mathrm{sin}\left(\mathrm{33}\right)\:\mathrm{in}\:\mathrm{surd}\:\mathrm{form}. \\ $$ Answered by sandy_suhendra last updated on 10/Apr/17 $$\mathrm{let}\:\mathrm{A}=\mathrm{18}° \\ $$$$\mathrm{5A}=\mathrm{90}° \\ $$$$\mathrm{3A}=\mathrm{90}°−\mathrm{2A} \\…
Question Number 143097 by ZiYangLee last updated on 10/Jun/21 $$\mathrm{If}\:{z}=\mathrm{cos}\:\theta+{i}\mathrm{sin}\:\theta,\: \\ $$$$\mathrm{prove}\:\mathrm{that}\:\frac{\mathrm{1}−{z}^{\mathrm{2}} }{\mathrm{1}+{z}^{\mathrm{2}} }=−{i}\mathrm{tan}\:\theta \\ $$ Answered by MJS_new last updated on 10/Jun/21 $$\frac{\mathrm{1}−\left(\mathrm{c}+\mathrm{is}\right)^{\mathrm{2}} }{\mathrm{1}+\left(\mathrm{c}+\mathrm{is}\right)^{\mathrm{2}}…
Question Number 143098 by mnjuly1970 last updated on 10/Jun/21 $$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…..{Prove}….\: \\ $$$$\:\:\:\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{s}{inh}\left(\pi{n}\right)}\right)^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{6}}\:−\frac{\mathrm{1}}{\mathrm{2}\pi}\:\:\:… \\ $$$$\:\:\:\:\:\:\:…… \\ $$ Answered by Dwaipayan Shikari…
Question Number 12023 by tawa last updated on 09/Apr/17 $$\mathrm{Solve}\:\mathrm{simultaneously} \\ $$$$\mathrm{x}\:+\:\mathrm{y}\:−\:\sqrt{\mathrm{xy}}\:=\:\mathrm{3}\:\:\:\:\:……….\:\left(\mathrm{i}\right) \\ $$$$\sqrt{\mathrm{x}\:+\:\mathrm{1}}\:+\:\sqrt{\mathrm{y}\:+\:\mathrm{1}}\:=\:\mathrm{4}\:\:\:\:\:……….\:\left(\mathrm{ii}\right) \\ $$ Answered by Mr Chheang Chantria last updated on 10/Apr/17…
Question Number 77556 by behi83417@gmail.com last updated on 07/Jan/20 $$\mathrm{1}.\:\:\:\mathrm{1}×\mathrm{1}!×\mathrm{2}!+\mathrm{2}×\mathrm{2}!×\mathrm{3}!+\mathrm{3}×\mathrm{3}!×\mathrm{4}!+….=? \\ $$$$\mathrm{2}.\:\:\:\:\frac{\mathrm{1}+\sqrt{\mathrm{2}}}{\mathrm{2}+\sqrt{\mathrm{3}}}+\frac{\mathrm{2}+\sqrt{\mathrm{3}}}{\mathrm{3}+\sqrt{\mathrm{4}}}+\frac{\mathrm{3}+\sqrt{\mathrm{4}}}{\mathrm{4}+\sqrt{\mathrm{5}}}+……=? \\ $$$$\mathrm{3}.\:\:\:\frac{\mathrm{1}}{\mathrm{1}+\sqrt{\mathrm{2}}+\sqrt{\mathrm{3}}}+\frac{\mathrm{2}}{\:\sqrt{\mathrm{2}}+\sqrt{\mathrm{3}}+\sqrt{\mathrm{4}}}+\frac{\mathrm{3}}{\:\sqrt{\mathrm{3}}+\sqrt{\mathrm{4}}+\sqrt{\mathrm{5}}}+….=?\: \\ $$ Commented by mr W last updated on 07/Jan/20 $$\mathrm{1}.\:\:\rightarrow\infty,\:{clear}…
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Question Number 77554 by TawaTawa last updated on 07/Jan/20 Commented by TawaTawa last updated on 07/Jan/20 $$\mathrm{Please}\:\mathrm{help}\:\mathrm{with}\:\:\:\mathrm{1},\:\mathrm{2},\:\mathrm{3} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 12019 by Nayon last updated on 09/Apr/17 Answered by ajfour last updated on 10/Apr/17 Commented by ajfour last updated on 10/Apr/17 $${AC}^{\:\mathrm{2}} +{BD}^{\mathrm{2}}…
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