Question Number 150718 by mathdanisur last updated on 14/Aug/21 $$\mathrm{Calculate}: \\ $$$$\mathrm{Li}_{\mathrm{2}} \left(\boldsymbol{\mathrm{z}}\right)\:+\:\mathrm{Li}_{\mathrm{2}} \left(\mathrm{1}\:-\:\boldsymbol{\mathrm{z}}\right)\:=\:? \\ $$ Answered by Olaf_Thorendsen last updated on 14/Aug/21 $$\mathrm{Li}\left({z}\right)\:=\:−\int_{\mathrm{0}} ^{{z}}…
Question Number 19638 by Tinkutara last updated on 13/Aug/17 $$\mathrm{Let}\:{P}\left({x}\right)\:\mathrm{is}\:\mathrm{a}\:\mathrm{polynomial}\:\mathrm{such}\:\mathrm{that} \\ $$$${P}\left(\mathrm{1}\right)\:=\:\mathrm{1},\:{P}\left(\mathrm{2}\right)\:=\:\mathrm{2},\:{P}\left(\mathrm{3}\right)\:=\:\mathrm{3},\:\mathrm{and} \\ $$$${P}\left(\mathrm{4}\right)\:=\:\mathrm{5}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{P}\left(\mathrm{6}\right). \\ $$ Answered by ajfour last updated on 13/Aug/17 $$\mathrm{let}\:\mathrm{P}\left(\mathrm{x}\right)=\mathrm{ax}^{\mathrm{3}} +\mathrm{bx}^{\mathrm{2}}…
Question Number 85176 by jagoll last updated on 19/Mar/20 $$\mathrm{what}\:\mathrm{is}\:\mathrm{range}\: \\ $$$$\mathrm{function}\:\mathrm{y}\:=\:\sqrt{\mathrm{x}−\mathrm{1}}\:+\:\sqrt{\mathrm{5}−\mathrm{x}} \\ $$ Answered by john santu last updated on 19/Mar/20 $$\mathrm{Domain}\::\:\mathrm{x}\:\geqslant\:\mathrm{1}\:\wedge\:\mathrm{x}\:\leqslant\:\mathrm{5}\:\Rightarrow\:\mathrm{1}\:\leqslant\mathrm{x}\leqslant\mathrm{5} \\ $$$$\mathrm{Range}\::\:\mathrm{y}\:'\:=\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{x}−\mathrm{1}}}\:−\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{5}−\mathrm{x}}}\:=\:\mathrm{0}…
Question Number 150697 by mathdanisur last updated on 14/Aug/21 Answered by dumitrel last updated on 14/Aug/21 $$\mathrm{1011}\leqslant{a}_{{i}} \leqslant\mathrm{2022} \\ $$$$\frac{\mathrm{1}}{\mathrm{2022}}\leqslant\frac{\mathrm{1}}{{b}_{{i}} }\leqslant\frac{\mathrm{1}}{\mathrm{1011}}\Rightarrow \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\leqslant\frac{{a}_{{i}} }{{b}_{{i}} }\leqslant\mathrm{2}\Rightarrow\left(\frac{{a}_{{i}}…
Question Number 19629 by Tinkutara last updated on 13/Aug/17 $$\mathrm{If}\:\mid{z}\mid\:=\:\mathrm{2},\:\mathrm{then}\:\mathrm{the}\:\mathrm{points}\:\mathrm{representing} \\ $$$$\mathrm{the}\:\mathrm{complex}\:\mathrm{numbers}\:−\mathrm{1}\:+\:\mathrm{5}{z}\:\mathrm{will}\:\mathrm{lie} \\ $$$$\mathrm{on}\:\mathrm{a} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{Circle} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{Straight}\:\mathrm{line} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{Parabola} \\ $$$$\left(\mathrm{4}\right)\:\mathrm{Ellipse} \\ $$ Answered…
Question Number 19623 by Tinkutara last updated on 13/Aug/17 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{locus}\:\mathrm{of}\:{z}\:\mathrm{if}\:\mathrm{arg}\left(\frac{{z}\:−\:\mathrm{2}}{{z}\:−\:\mathrm{3}}\right)\:=\:\frac{\pi}{\mathrm{4}} \\ $$ Answered by ajfour last updated on 13/Aug/17 $$\mathrm{let}\:\mathrm{z}=\mathrm{x}+\mathrm{iy} \\ $$$$\Rightarrow\:\:\mathrm{arg}\left[\frac{\mathrm{x}−\mathrm{2}+\mathrm{iy}}{\mathrm{x}−\mathrm{3}+\mathrm{iy}}\right]=\frac{\pi}{\mathrm{4}} \\ $$$$\Rightarrow\:\:\mathrm{arg}\left[\frac{\left(\mathrm{x}−\mathrm{2}+\mathrm{iy}\right)\left(\mathrm{x}−\mathrm{3}−\mathrm{iy}\right)}{\left(\mathrm{x}−\mathrm{3}\right)^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}}…
Question Number 150684 by mathdanisur last updated on 14/Aug/21 $$\mathrm{if}\:\:\mathrm{a};\mathrm{b};\mathrm{c};\mathrm{d}\in\mathbb{Z} \\ $$$$\left(\mathrm{b}-\mathrm{a}\right)\left(\mathrm{c}-\mathrm{a}\right)\left(\mathrm{d}-\mathrm{a}\right)\left(\mathrm{c}-\mathrm{b}\right)\left(\mathrm{d}-\mathrm{b}\right)\left(\mathrm{d}-\mathrm{c}\right) \\ $$$$\mathrm{prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{expression}\:\mathrm{is}\:\mathrm{divide} \\ $$$$\mathrm{into}\:\mathrm{12} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 19609 by ajfour last updated on 13/Aug/17 Answered by ajfour last updated on 13/Aug/17 $$\mathrm{Equation}\:\mathrm{of}\:\mathrm{L}_{\mathrm{1}} : \\ $$$$\:\:\:\:\:\:\bar {\mathrm{z}}_{\mathrm{1}} \mathrm{z}+\mathrm{z}_{\mathrm{1}} \bar {\mathrm{z}}−\mathrm{2}\mid\mathrm{z}_{\mathrm{1}} \mid^{\mathrm{2}}…
Question Number 19604 by ajfour last updated on 13/Aug/17 Commented by ajfour last updated on 13/Aug/17 $$\mathrm{solution}\:\mathrm{to}\:\mathrm{Q}.\mathrm{19508} \\ $$$$\mathrm{To}\:\mathrm{prove}\:\mathrm{p}=\mid\mathrm{z}_{\mathrm{1}} −\mathrm{z}_{\mathrm{0}} \mid=\mid\frac{\bar {\alpha}\mathrm{z}_{\mathrm{1}} +\alpha\bar {\mathrm{z}}_{\mathrm{1}} +\mathrm{2c}}{\mathrm{2}\mid\alpha\mid}\mid…
Question Number 19592 by ajfour last updated on 13/Aug/17 Commented by Tinkutara last updated on 13/Aug/17 $${z}_{{F}} \:=\:\frac{{ab}^{\mathrm{2}} }{{a}^{\mathrm{2}} \:+\:{b}^{\mathrm{2}} }\:+\:{i}\frac{{a}^{\mathrm{2}} {b}}{{a}^{\mathrm{2}} \:+\:{b}^{\mathrm{2}} } \\…