Question Number 140650 by aurpeyz last updated on 10/May/21 $${find}\:{the}\:{integral}\:{of}\: \\ $$$$\frac{\mathrm{1}}{\pi}\underset{\mathrm{0}} {\overset{\pi} {\int}}{e}^{{ie}^{{ksin}^{\mathrm{2}} \emptyset} } {d}\emptyset \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 9576 by j.masanja06@gmail.com last updated on 18/Dec/16 $$\mathrm{show}\:\mathrm{that}\:\mathrm{x}\:\mathrm{is}\:\mathrm{small}\:\mathrm{enough}\:\mathrm{for}\:\mathrm{its}\:\mathrm{cube}\:\mathrm{and}\:\mathrm{higher}\:\mathrm{power}\:\mathrm{to}\:\mathrm{be}\:\mathrm{neghected} \\ $$$$\sqrt{\frac{\mathrm{1}−\mathrm{x}}{\mathrm{1}+\mathrm{x}}}=\mathrm{1}−\mathrm{x}+\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}^{\mathrm{2}} . \\ $$$$\mathrm{by}\:\mathrm{putting}\:=\frac{\mathrm{1}}{\mathrm{8}},\mathrm{show}\:\mathrm{that}\:\sqrt{\mathrm{7}}\approx\mathrm{2}\frac{\mathrm{83}}{\mathrm{128}}. \\ $$ Commented by prakash jain last updated on 20/Dec/16…
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Question Number 75110 by peter frank last updated on 07/Dec/19 Answered by mind is power last updated on 08/Dec/19 $$\mathrm{ax}+\mathrm{by}+\mathrm{c}=\mathrm{0} \\ $$$$\mathrm{x}=\mathrm{0}\Rightarrow\mathrm{y}_{\mathrm{0}} =−\frac{\mathrm{c}}{\mathrm{b}}…..\mathrm{b}\neq\mathrm{0} \\ $$$$\mathrm{y}=\mathrm{0}\Rightarrow\mathrm{x}_{\mathrm{0}}…
Question Number 9575 by j.masanja06@gmail.com last updated on 18/Dec/16 $$\mathrm{the}\:\mathrm{expression}\:\mathrm{ax}^{\mathrm{2}} \:+\:\mathrm{bx}\:+\:\mathrm{c}\:\mathrm{is}\:\mathrm{divisible}\:\:\mathrm{by}\:\mathrm{x}−\mathrm{1},\mathrm{has}\:\mathrm{reminder}\:\mathrm{2}\:\mathrm{when}\:\mathrm{divided}\:\mathrm{by}\:\mathrm{x}+\mathrm{1},\mathrm{and}\:\mathrm{has}\:\mathrm{reminder}\:\mathrm{8}\:\mathrm{when}\:\mathrm{divided}\:\mathrm{by}\:\mathrm{x}−\mathrm{2}.\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{a},\mathrm{b}\:\mathrm{and}\:\mathrm{c}. \\ $$ Commented by ridwan balatif last updated on 18/Dec/16 $$\mathrm{P}\left(\mathrm{x}\right)\equiv\mathrm{ax}^{\mathrm{2}} +\mathrm{bx}+\mathrm{c} \\ $$$$\left(\mathrm{x}−\mathrm{1}\right)\mathrm{is}\:\mathrm{divisible}\:\mathrm{P}\left(\mathrm{x}\right)\rightarrow\mathrm{P}\left(\mathrm{1}\right)=\mathrm{0}…
Question Number 9573 by lepan last updated on 17/Dec/16 $${If}\:{n}\:{is}\:{positive}\:{integer}\:{prove}\:{that}\: \\ $$$${the}\:{cofficient}\:{of}\:{x}^{\mathrm{2}\:} {and}\:{x}^{\mathrm{3}} \:{in}\:{the}\: \\ $$$${expansion}\:{of}\:\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{2}\right)^{{n}} \:{are}\:\mathrm{2}^{{n}−\mathrm{1}} .{n}^{\mathrm{2}} \\ $$$${and}\:\mathrm{2}^{{n}−\mathrm{1}} {n}\left({n}−\mathrm{1}\right)\frac{\mathrm{1}}{\mathrm{3}}. \\ $$ Commented…
Question Number 9571 by PANKAJ last updated on 16/Dec/16 Answered by ridwan balatif last updated on 17/Dec/16 $$\mathrm{1}.\left(\frac{\mathrm{x}}{\mathrm{3}}+\mathrm{1},\mathrm{y}−\frac{\mathrm{2}}{\mathrm{3}}\right)=\left(\frac{\mathrm{5}}{\mathrm{3}},\frac{\mathrm{1}}{\mathrm{3}}\right) \\ $$$$\frac{\mathrm{x}}{\mathrm{3}}+\mathrm{1}=\frac{\mathrm{5}}{\mathrm{3}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{y}−\frac{\mathrm{2}}{\mathrm{3}}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\frac{\mathrm{x}}{\mathrm{3}}=\frac{\mathrm{2}}{\mathrm{3}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{y}=\mathrm{1} \\ $$$$\mathrm{x}=\mathrm{2} \\…
Question Number 75104 by chess1 last updated on 07/Dec/19 Answered by mr W last updated on 07/Dec/19 Commented by mr W last updated on 08/Dec/19…
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