Question Number 98859 by peter frank last updated on 16/Jun/20 Commented by mr W last updated on 16/Jun/20 $${p}\left({x}\right)={q}\left({x}\right)\left({x}−\mathrm{1}\right) \\ $$$${p}\left(\mathrm{1}\right)=\mathrm{0} \\ $$$$\mathrm{1}+{c}−\mathrm{2}{c}+\mathrm{4}=\mathrm{0} \\ $$$$\Rightarrow{c}=\mathrm{5}…
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Question Number 98842 by M±th+et+s last updated on 16/Jun/20 $${let}\:{f}\left({x}\right)\:{be}\:{a}\:{dolvnomial}\:{of}\:{degree}\:\mathrm{4}\: \\ $$$${such}\:{that}\:{f}\left(\mathrm{1}\right)=\mathrm{1}\:,\:{f}\left(\mathrm{2}\right)=\mathrm{2}\:,{f}\left(\mathrm{3}\right)=\mathrm{3},{f}\left(\mathrm{4}\right)=\mathrm{4} \\ $$$${then}\:{f}\left(\mathrm{6}\right)=? \\ $$ Commented by mr W last updated on 16/Jun/20 $${see}\:{also}\:{Q}\mathrm{98268}…
Question Number 98844 by I want to learn more last updated on 16/Jun/20 $$\mathrm{Please}\:\mathrm{explain}:\:\:\:\:\:\:\underset{\mathrm{1}\:\leqslant\:\boldsymbol{\mathrm{i}}\:<\:\boldsymbol{\mathrm{j}}\:\leqslant\:\boldsymbol{\mathrm{n}}} {\sum}\boldsymbol{\mathrm{ij}}\:\:\:\:=\:\:\:\underset{\boldsymbol{\mathrm{j}}\:\:=\:\:\mathrm{2}} {\overset{\boldsymbol{\mathrm{n}}} {\sum}}\frac{\boldsymbol{\mathrm{j}}\left(\boldsymbol{\mathrm{j}}\:−\:\mathrm{1}\right)\boldsymbol{\mathrm{j}}}{\mathrm{2}} \\ $$$$\boldsymbol{\mathrm{I}}\:\mathrm{want}\:\mathrm{to}\:\mathrm{know}\:\mathrm{how}\:\mathrm{L}.\mathrm{H}.\mathrm{S}\:\:=\:\:\mathrm{R}.\mathrm{H}.\mathrm{S} \\ $$ Answered by mr W…
Question Number 33307 by abdo imad last updated on 14/Apr/18 $${let}\:{z}={x}+{iy}\:\:{with}\:{x}\neq\mathrm{0}\:{prove}?{that} \\ $$$$\mid\:\frac{{e}^{{z}} \:−\mathrm{1}}{{z}}\:\mid\leqslant\mid\:\frac{{e}^{{x}} \:−\mathrm{1}}{{x}}\:\mid \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 164372 by mathlove last updated on 16/Jan/22 Commented by mathlove last updated on 16/Jan/22 $${Domain}\:{f}\left({x}\right)=? \\ $$ Answered by mahdipoor last updated on…
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Question Number 164341 by ajfour last updated on 16/Jan/22 $$\:\:{x}^{\mathrm{4}} ={x}^{\mathrm{2}} +{cx} \\ $$$$\left({x}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} ={cx}+\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${let}\:\:\:\:{x}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}={t}\:\:\:\Rightarrow \\ $$$$\left({t}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{2}} ={c}^{\mathrm{2}} \left({t}+\frac{\mathrm{1}}{\mathrm{2}}\right) \\…
Question Number 98796 by Lekhraj last updated on 16/Jun/20 Answered by MJS last updated on 16/Jun/20 $$−\frac{\mathrm{267}}{\mathrm{2}} \\ $$ Commented by Lekhraj last updated on…
Question Number 164331 by HongKing last updated on 16/Jan/22 Commented by mr W last updated on 16/Jan/22 $${see}\:{Q}\mathrm{163656} \\ $$$$\frac{{BC}}{{AD}}=\frac{{PB}^{\mathrm{2}} −{PC}^{\mathrm{2}} }{{PA}^{\mathrm{2}} −{PD}^{\mathrm{2}} }=\mathrm{1}\: \\…