Question Number 140827 by mathsuji last updated on 13/May/21 $${Find} \\ $$$$\mathrm{1}.\:\int\frac{{dx}}{\left[{x}\right]^{\mathrm{2}} }\:,\:{x}\geqslant\mathrm{1} \\ $$$$\mathrm{2}.\:\int\frac{\left[{x}\right]^{\lambda} }{{x}^{\lambda+\mathrm{1}} }\:,\:{x}\geqslant\mathrm{1} \\ $$$${Where}\:\left[\ast\right]\:{denote}\:{the}\:{integer}\:{part} \\ $$ Answered by mathmax by…
Question Number 9753 by j.masanja06@gmail.com last updated on 31/Dec/16 $${find}\:{the}\:{value}\:{of}\:{x} \\ $$$$\left({a}\right)\:\:\mathrm{4}^{\mathrm{2}{x}+\mathrm{1}} .\:\mathrm{5}^{{x}−\mathrm{2}} =\mathrm{6}^{\mathrm{1}−{x}} \\ $$$$\left({b}\right)\:\:\:\mathrm{4}\left(\mathrm{3}^{\mathrm{2}{x}+\mathrm{1}} \right)+\mathrm{17}\left(\mathrm{3}^{{x}} \right)−\mathrm{7}=\mathrm{0} \\ $$ Commented by ridwan balatif last…
Question Number 140816 by liberty last updated on 13/May/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
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Question Number 75272 by peter frank last updated on 09/Dec/19 $$\left.{a}\right)\:{If}\:{z}=\mathrm{1}+{i}\sqrt{\mathrm{3}}\:{prove}\:{that} \\ $$$${prove}\:{that} \\ $$$${z}^{\mathrm{14}} =\mathrm{2}^{\mathrm{13}} \left(−\mathrm{1}+{i}\sqrt{\mathrm{3}}\:\right) \\ $$$$ \\ $$$$\left.{b}\right){prove}\:{that}\:{in}\:{triangle}\:{ABC} \\ $$$${a}^{\mathrm{2}} −\left({b}−{c}\right)^{\mathrm{2}} \mathrm{cos}\:^{\mathrm{2}}…
Question Number 140809 by mathdanisur last updated on 12/May/21 $$\boldsymbol{{z}}^{\mathrm{5}} \:+\:\frac{\mathrm{1}}{\boldsymbol{{z}}^{\mathrm{5}} }\:=\:\frac{\mathrm{205}}{\mathrm{16}}\:\centerdot\:\left(\boldsymbol{{z}}\:+\:\frac{\mathrm{1}}{\boldsymbol{{z}}}\right) \\ $$ Answered by Rasheed.Sindhi last updated on 19/May/21 $$\boldsymbol{{z}}^{\mathrm{5}} \:+\:\frac{\mathrm{1}}{\boldsymbol{{z}}^{\mathrm{5}} }\:−\:\frac{\mathrm{205}}{\mathrm{16}}\:\centerdot\:\left(\boldsymbol{{z}}\:+\:\frac{\mathrm{1}}{\boldsymbol{{z}}}\right)=\mathrm{0} \\…
Question Number 140810 by mathsuji last updated on 12/May/21 $${Find}\:{the}\:{natural}\:{value}\:{of}\:\boldsymbol{{x}}\:{that} \\ $$$${satisfies}\:{the}\:{equation}: \\ $$$$\frac{\mathrm{1}}{\mathrm{6}\centerdot\mathrm{7}}\:+\:\frac{\mathrm{1}}{\mathrm{7}\centerdot\mathrm{8}}\:+\:\frac{\mathrm{1}}{\mathrm{8}\centerdot\mathrm{9}}\:+…+\:\frac{\mathrm{1}}{{x}\centerdot\left({x}+\mathrm{1}\right)}\:=\:\frac{\mathrm{1}}{\mathrm{12}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 75267 by aliesam last updated on 09/Dec/19 Commented by mind is power last updated on 09/Dec/19 $$\mathrm{let}\:\mathrm{f}\left(\mathrm{x}\right)=\frac{\sqrt{\mathrm{1}+\mathrm{sin}\left(\mathrm{x}\right)}+\sqrt{\mathrm{1}−\mathrm{sin}\left(\mathrm{x}\right)}}{\:\sqrt{\mathrm{1}+\mathrm{sin}\left(\mathrm{x}\right)}−\sqrt{\mathrm{1}−\mathrm{sin}\left(\mathrm{x}\right)}} \\ $$$$\mathrm{sin}\left(\mathrm{x}\right)=\mathrm{2sin}\left(\frac{\mathrm{x}}{\mathrm{2}}\right)\mathrm{cos}\left(\frac{\mathrm{x}}{\mathrm{2}}\right) \\ $$$$\mathrm{1}\underset{−} {+}\mathrm{sin}\left(\mathrm{x}\right)=\left(\mathrm{cos}\left(\frac{\mathrm{x}}{\mathrm{2}}\right)\underset{−} {+}\mathrm{sin}\left(\frac{\mathrm{x}}{\mathrm{2}}\right)\right)^{\mathrm{2}}…
Question Number 75262 by chess1 last updated on 09/Dec/19 Commented by chess1 last updated on 09/Dec/19 $$\mathrm{Sir}\:\boldsymbol{\mathrm{mind}}\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{power}} \\ $$$$\mathrm{Sir}\:\boldsymbol{\mathrm{mathmax}}\:\mathrm{and}\:\boldsymbol{\mathrm{Mjs}}\:\:\mathrm{and}\:\:\boldsymbol{\mathrm{W}}\:\:\:\boldsymbol{\mathrm{solution}}\:\boldsymbol{\mathrm{please}} \\ $$ Commented by chess1 last…
Question Number 140785 by mathdanisur last updated on 12/May/21 $${P}-{intersection}\:{point}\:{of}\:{bimedians}\:{in} \\ $$$${ABCD}-{convexe}\:{quadrilateral}\:{with} \\ $$$${a};{b};{c};{d}-{sides},\:{e};{f}-{diagonals}, \\ $$$${E}-{point}\:{in}\:{plane},\:{x};{y};{z};{t}-{distances} \\ $$$${from},\:{E}-{to}\:{A};{B};{C};{D}.\:{Prove}\:{that}… \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{e}^{\mathrm{2}} +{f}^{\mathrm{2}} \right)+\mathrm{4}{PE}^{\mathrm{2}}…