Question Number 212326 by universe last updated on 10/Oct/24 $$\:\:\:\:\:\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{3}\right)^{{n}} \:{n}!}{\mathrm{1}.\mathrm{4}…\left(\mathrm{3}{n}+\mathrm{1}\right)} \\ $$$$\:\left(\mathrm{1}\right)\:{check}\:\:{its}\:{a}\:{absolute}\:{conergent}\:{series} \\ $$$$\:\:\left(\mathrm{2}\right)\:{show}\:{that}\:{its}\:{a}\:{convergent}\:{series} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 212342 by hardmath last updated on 10/Oct/24 $$\mathrm{The}\:\mathrm{number}\:\:\overline {\mathrm{abc}}\:\mathrm{is}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{37}. \\ $$$$\mathrm{Prove}\:\mathrm{that}\:\overline {\mathrm{bca}}\:+\:\overline {\mathrm{cab}}\:\mathrm{is}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{37}.\: \\ $$ Answered by A5T last updated on 10/Oct/24 $$\mathrm{100}{a}+\mathrm{10}{b}+{c}\equiv−\mathrm{11}{a}−\mathrm{27}{b}+{c}=−\left(\mathrm{11}{a}+\mathrm{27}{b}−{c}\right)…
Question Number 212320 by RojaTaniya last updated on 10/Oct/24 $$\:{a}+{b}+{c}+{d}=\mathrm{2},\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{d}^{\mathrm{2}} =\mathrm{2} \\ $$$$\:{a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} +{d}^{\mathrm{3}} =−\mathrm{4},\:{a}^{\mathrm{4}} +{b}^{\mathrm{4}} +{c}^{\mathrm{4}} +{d}^{\mathrm{4}} =−\mathrm{6} \\…
Question Number 212339 by Spillover last updated on 10/Oct/24 Answered by Ghisom last updated on 10/Oct/24 $$\underset{\mathrm{0}} {\overset{\mathrm{3}} {\int}}\:\frac{{x}^{\mathrm{3}/\mathrm{4}} \left(\mathrm{3}−{x}\right)^{\mathrm{1}/\mathrm{4}} }{\left(\mathrm{5}−{x}\right)^{\mathrm{3}} }{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\left(\frac{\mathrm{2}{x}}{\mathrm{5}\left(\mathrm{3}−{x}\right)}\right)^{\mathrm{1}/\mathrm{4}} ;\:{C}=\frac{\mathrm{9}\sqrt[{\mathrm{4}}]{\mathrm{2}}}{\mathrm{5}\sqrt[{\mathrm{4}}]{\mathrm{5}}}\right]…
Question Number 212301 by RojaTaniya last updated on 09/Oct/24 $$\:\:\:\:\mathrm{3}{x}+\frac{\mathrm{2}}{\:\sqrt{{x}}}=\mathrm{1},\:{x}−\sqrt{{x}}\:=? \\ $$$$\:\: \\ $$ Answered by Sutrisno last updated on 09/Oct/24 $${misal}\:\sqrt{{x}}={p}\rightarrow{x}={p}^{\mathrm{2}} \\ $$$$\mathrm{3}{p}^{\mathrm{2}} +\frac{\mathrm{2}}{{p}}=\mathrm{1}…
Question Number 212314 by Spillover last updated on 09/Oct/24 Answered by mehdee7396 last updated on 09/Oct/24 $${f}\left({x}\right)={ln}\left(\frac{\mathrm{2}−{sinx}}{\mathrm{2}+{sinx}}\right) \\ $$$${f}\left(−{x}\right)={ln}\left(\frac{\mathrm{2}+{sinx}}{\mathrm{2}−{sinx}}\right)=−{f}\left({x}\right) \\ $$$$\Rightarrow;{f};\:{is}\:\:;{odd}\Rightarrow\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} {f}\left({x}\right){dx}=\mathrm{0}\:\checkmark \\ $$$$…
Question Number 212291 by hardmath last updated on 08/Oct/24 Commented by hardmath last updated on 08/Oct/24 $$\mathrm{m}\left(\angle\mathrm{BCD}\right)\:=\:\mathrm{90}° \\ $$$$\mid\mathrm{AB}\mid\:=\:\mid\mathrm{AC}\mid \\ $$$$\mathrm{AC}\:\cap\:\mathrm{BD}\:=\:\mathrm{K} \\ $$$$\mathrm{Area}\left(\mathrm{ABCD}\right)\:=\:? \\ $$…
Question Number 212224 by RojaTaniya last updated on 07/Oct/24 $$\:{a},{b},{c}\in{R} \\ $$$$\:{a}+{b}+{c}=\mathrm{1},\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} =\mathrm{1} \\ $$$$\:{a}^{\mathrm{10}} +{b}^{\mathrm{10}} +{c}^{\mathrm{10}} =\mathrm{1},\:{a}^{\mathrm{4}} +{b}^{\mathrm{4}} +{c}^{\mathrm{4}} =? \\ $$…
Question Number 212226 by RojaTaniya last updated on 07/Oct/24 $$\:{Factorize}: \\ $$$$\:{x}^{\mathrm{16}} −{x}^{\mathrm{14}} +{x}^{\mathrm{9}} +{x}^{\mathrm{7}} +{x} \\ $$ Commented by Frix last updated on 08/Oct/24…
Question Number 212219 by hardmath last updated on 06/Oct/24 $$\mathrm{Find}: \\ $$$$\sqrt{\mathrm{21}\centerdot\mathrm{22}\centerdot\mathrm{23}\centerdot\mathrm{24}\:+\:\mathrm{1}}\:=\:? \\ $$ Answered by Rasheed.Sindhi last updated on 06/Oct/24 $$\sqrt{\left(\mathrm{22}.\mathrm{5}−\mathrm{1}.\mathrm{5}\right)\left(\mathrm{22}.\mathrm{5}−.\mathrm{5}\right)\left(\mathrm{22}.\mathrm{5}+.\mathrm{5}\right)\left(\mathrm{22}.\mathrm{5}+\mathrm{1}.\mathrm{5}\right)+\mathrm{1}}\: \\ $$$$\sqrt{\left(\mathrm{22}.\mathrm{5}^{\mathrm{2}} −.\mathrm{5}^{\mathrm{2}}…