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Question Number 40366 by mondodotto@gmail.com last updated on 20/Jul/18 $$\boldsymbol{\mathrm{use}}\:\boldsymbol{\mathrm{newton}}\:\boldsymbol{\mathrm{raphson}}\:\boldsymbol{\mathrm{method}} \\ $$$$\boldsymbol{\mathrm{to}}\:\boldsymbol{\mathrm{approximate}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{positive}}\:\boldsymbol{\mathrm{root}} \\ $$$$\boldsymbol{{x}}^{\mathrm{2}} −\mathrm{1}=\mathrm{0}\:\boldsymbol{\mathrm{correct}}\:\boldsymbol{\mathrm{to}}\:\mathrm{4}\:\boldsymbol{\mathrm{decimal}}\:\boldsymbol{\mathrm{places}} \\ $$$$\boldsymbol{\mathrm{perform}}\:\mathrm{3}\:\boldsymbol{\mathrm{iteration}}\:\boldsymbol{\mathrm{only}} \\ $$$$\boldsymbol{\mathrm{setting}}\:\boldsymbol{\mathrm{with}}\:\boldsymbol{{x}}=\mathrm{2} \\ $$ Commented by math khazana…
Question Number 105897 by Tinku Tara last updated on 01/Aug/20 $$\mathrm{App}\:\mathrm{Updates}: \\ $$$$\mathrm{cyrillic}\:\mathrm{alohabets}\:\mathrm{are}\:\mathrm{now}\:\mathrm{available} \\ $$$$\mathrm{in}\:\mathrm{app}.\:\mathrm{Email}\:\mathrm{us}\:\mathrm{for}\:\mathrm{any}\:\mathrm{missing} \\ $$$$\mathrm{conjunctions}. \\ $$ Commented by mr W last updated…
Question Number 105892 by mohammad17 last updated on 01/Aug/20 $${prove}\:{that}\:{sin}\left({x}\right)+{cos}\left({x}\right)=\sqrt{\mathrm{2}}{cos}\left({x}−\frac{\pi}{\mathrm{4}}\right)\:\:? \\ $$ Answered by john santu last updated on 01/Aug/20 $$\mathrm{sin}\:{x}+\mathrm{cos}\:{x}\:=\:{p} \\ $$$$\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\mathrm{sin}\:{x}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\mathrm{cos}\:{x}\:=\:\frac{{p}}{\:\sqrt{\mathrm{2}}} \\ $$$$\Rightarrow\:\mathrm{sin}\:\left(\frac{\pi}{\mathrm{4}}\right)\mathrm{sin}\:{x}+\:\mathrm{cos}\:\left(\frac{\pi}{\mathrm{4}}\right)\mathrm{cos}\:{x}…
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Question Number 171421 by mokys last updated on 14/Jun/22 $${Prove}\:{that}\:{arg}\left({e}^{{z}} \right)=\:{Im}\left({z}\right)\:+\:\mathrm{2}{k}\pi\:,{k}=\mathrm{0},\pm\mathrm{1},\pm\mathrm{2},…\sqrt{} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 105885 by mohammad17 last updated on 01/Aug/20 Answered by bemath last updated on 01/Aug/20 $$\mathrm{sin}\:\left({x}\right)+\mathrm{cos}\:\left({x}\right)\:=\:\sqrt{\mathrm{2}}\:\mathrm{cos}\:\left({x}−\frac{\pi}{\mathrm{4}}\right) \\ $$$$\int\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\mathrm{sec}\:\left({x}−\frac{\pi}{\mathrm{4}}\right)\:{dx}\:=\: \\ $$$$\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}\:}\:\mathrm{ln}\:\mid\mathrm{sec}\:\left({x}−\frac{\pi}{\mathrm{4}}\right)+\mathrm{tan}\:\left({x}−\frac{\pi}{\mathrm{4}}\right)\mid+{C} \\ $$$$\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}\:}\mathrm{ln}\:\mid\frac{\sqrt{\mathrm{2}}\:+\mathrm{sin}\:{x}−\mathrm{cos}\:{x}}{\mathrm{sin}\:{x}+\mathrm{cos}\:{x}}\mid\:+{C} \\ $$…
Question Number 171406 by floor(10²Eta[1]) last updated on 14/Jun/22 $$\mathrm{1}=\sqrt{\mathrm{1}}=\sqrt{\left(−\mathrm{1}\right)\left(−\mathrm{1}\right)}=\sqrt{\left(−\mathrm{1}\right)}×\sqrt{\left(−\mathrm{1}\right)}=\mathrm{i}.\mathrm{i}=\mathrm{i}^{\mathrm{2}} =−\mathrm{1} \\ $$$$\mathrm{where}\:\mathrm{is}\:\mathrm{the}\:\mathrm{mistake}? \\ $$ Commented by infinityaction last updated on 14/Jun/22 $${if}\:{a}\:{and}\:{b}\:{are}\:{positive}\:{number} \\ $$$${then}\:\sqrt{{ab}}\:=\:\sqrt{{a}}\:\sqrt{{b}}…
Question Number 105850 by PRITHWISH SEN 2 last updated on 01/Aug/20 $$\mathrm{A}\:\mathrm{bit}\:\mathrm{modification}\:\mathrm{of}\:\mathrm{Mr}.\:\mathrm{W}'\mathrm{s}\:\mathrm{question} \\ $$$$\mathrm{How}\:\mathrm{many}\:\mathrm{numbers}\:\mathrm{are}\:\mathrm{there}\:\mathrm{with}\:\mathrm{atleast}\:\mathrm{one} \\ $$$$\mathrm{zero}\:\mathrm{between}\:\mathrm{1}\:\mathrm{to}\:\mathrm{12345}\:?\:\mathrm{For}\:\mathrm{instance}\:\mathrm{consider} \\ $$$$\mathrm{10020}\:\mathrm{as}\:\mathrm{1}\:\mathrm{number}\:?\:\mathrm{Can}\:\mathrm{the}\:\mathrm{answer}\:\mathrm{ofQ105791}\: \\ $$$$\mathrm{be}\:\mathrm{derived}\:\mathrm{from}\:\mathrm{the}\:\mathrm{answer}\:\mathrm{of}\:\mathrm{this}\:\mathrm{question}\:? \\ $$ Terms of Service…
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