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Given-E-x-4-10x-2-q-0-U-x-2-so-E-U-U-2-40U-q-0-We-suppose-that-E-U-has-two-roots-such-as-r-1-lt-r-2-We-give-also-that-r-1-r-2-40-and-r-1-r-2-q-1-The-equation-E-has-four-p

Question Number 104191 by mathocean1 last updated on 19/Jul/20 $${Given}\:\left({E}\right):{x}^{\mathrm{4}} −\mathrm{10}{x}^{\mathrm{2}} +{q}=\mathrm{0} \\ $$$${U}={x}^{\mathrm{2}} \:{so}\:\left({E}_{\left({U}\right)} \right)={U}^{\mathrm{2}} −\mathrm{40}{U}+{q}=\mathrm{0}\:. \\ $$$${We}\:{suppose}\:{that}\:{E}_{\left({U}\right)} {has}\:{two}\:{roots} \\ $$$${such}\:{as}\:{r}_{\mathrm{1}} <{r}_{\mathrm{2}\:} . \\…

Given-E-m-1-x-2-m-2-x-1-0-We-suppose-that-it-has-two-roots-x-1-and-x-2-Determinate-m-such-that-x-1-1-x-2-

Question Number 104187 by mathocean1 last updated on 19/Jul/20 $${Given}: \\ $$$$\left({E}\right):\:\left({m}+\mathrm{1}\right){x}^{\mathrm{2}} +\left({m}−\mathrm{2}\right){x}+\mathrm{1}=\mathrm{0} \\ $$$${We}\:{suppose}\:{that}\:{it}\:{has}\:{two}\:{roots}\:{x}_{\mathrm{1}} \\ $$$${and}\:{x}_{\mathrm{2}} . \\ $$$${Determinate}\:{m}\:{such}\:{that}: \\ $$$${x}_{\mathrm{1}} =\mathrm{1}+{x}_{\mathrm{2}} \\ $$…

Given-P-x-x-4-2x-3-41x-2-42x-360-Determinate-Q-x-a-quadratic-poly-nom-such-that-P-x-Q-x-2-42-Q-x-360-

Question Number 104181 by mathocean1 last updated on 19/Jul/20 $${Given} \\ $$$${P}\left({x}\right)={x}^{\mathrm{4}} +\mathrm{2}{x}^{\mathrm{3}} −\mathrm{41}{x}^{\mathrm{2}} +\mathrm{42}{x}+\mathrm{360} \\ $$$${Determinate}\:{Q}\left({x}\right)\:{a}\:{quadratic}\:{poly}− \\ $$$${nom}\:{such}\:{that}: \\ $$$${P}\left({x}\right)=\left({Q}\left({x}\right)\right)^{\mathrm{2}} −\mathrm{42}\left({Q}\left({x}\right)\right)+\mathrm{360} \\ $$ Answered…

Solve-in-R-a-3-x-3-8-x-3-4-0-b-x-2-x-6-x-1-c-x-3-27-6-lt-x-3-

Question Number 104178 by mathocean1 last updated on 19/Jul/20 $${Solve}\:{in}\:\mathbb{R} \\ $$$$\left.{a}\right)\:\mathrm{3}\mid{x}−\sqrt{\mathrm{3}}\mid−\mathrm{8}\sqrt{\left(\mid{x}−\sqrt{\mathrm{3}}\mid\right)+\mathrm{4}}=\mathrm{0} \\ $$$$\left.{b}\left.\right)\sqrt{\left(\mid{x}^{\mathrm{2}} −{x}−\mathrm{6}\right.}\mid\right)={x}+\mathrm{1} \\ $$$$\left.{c}\right)\sqrt{\left({x}^{\mathrm{3}} −\mathrm{27}\right.}+\mathrm{6}<{x}+\mathrm{3} \\ $$ Answered by OlafThorendsen last updated…

App-Updates-v2-110-is-available-now-on-www-tinkutara-com-and-will-be-available-in-playstore-in-a-day-This-version-improves-drawing-sidebar-buttons-are-added-Multiple-shape-selection-for-al

Question Number 104171 by Tinku Tara last updated on 19/Jul/20 $$\mathrm{App}\:\mathrm{Updates}: \\ $$$$\mathrm{v2}.\mathrm{110}\:\mathrm{is}\:\mathrm{available}\:\mathrm{now}\:\mathrm{on} \\ $$$$\mathrm{www}.\mathrm{tinkutara}.\mathrm{com}\:\mathrm{and}\:\mathrm{will}\:\mathrm{be} \\ $$$$\mathrm{available}\:\mathrm{in}\:\mathrm{playstore}\:\mathrm{in}\:\mathrm{a}\:\mathrm{day}. \\ $$$$\mathrm{This}\:\mathrm{version}\:\mathrm{improves}\:\mathrm{drawing}: \\ $$$$−\:\mathrm{sidebar}\:\mathrm{buttons}\:\mathrm{are}\:\mathrm{added} \\ $$$$−\:\mathrm{Multiple}\:\mathrm{shape}\:\mathrm{selection}\:\mathrm{for} \\ $$$$\:\:\:\:\:\mathrm{alignment}\:\mathrm{can}\:\mathrm{be}\:\mathrm{done}\:\mathrm{from}…

find-the-domain-of-i-x-x-5-ii-x-2-iii-3-x-2-5-

Question Number 169667 by MathsFan last updated on 05/May/22 $$\boldsymbol{{find}}\:\boldsymbol{{the}}\:\boldsymbol{{domain}}\:\boldsymbol{{of}} \\ $$$$\left(\boldsymbol{{i}}\right)\:\frac{\boldsymbol{{x}}}{\:\sqrt{\boldsymbol{{x}}+\mathrm{5}}} \\ $$$$\left(\boldsymbol{{ii}}\right)\:\sqrt{\boldsymbol{{x}}}+\mathrm{2} \\ $$$$\left(\boldsymbol{{iii}}\right)\:\frac{\mathrm{3}}{\:\sqrt{\boldsymbol{{x}}+\mathrm{2}}+\mathrm{5}} \\ $$ Answered by FelipeLz last updated on 06/May/22…

Given-that-n-A-10-and-n-B-6-i-what-is-the-largest-possible-of-n-A-B-ii-what-is-the-smallest-possible-value-of-n-A-B-iii-what-is-the-smallest-possible-value-of-n-A-B-

Question Number 169664 by MathsFan last updated on 05/May/22 $$\:\boldsymbol{\mathrm{Given}}\:\boldsymbol{\mathrm{that}}\:\boldsymbol{\mathrm{n}}\left(\boldsymbol{\mathrm{A}}\right)=\mathrm{10}\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{n}}\left(\boldsymbol{\mathrm{B}}\right)=\mathrm{6} \\ $$$$\left.\:\boldsymbol{\mathrm{i}}\right)\:\boldsymbol{\mathrm{what}}\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{largest}}\:\boldsymbol{\mathrm{possible}}\:\boldsymbol{\mathrm{of}}\: \\ $$$$\:\:\:\:\:\:\boldsymbol{\mathrm{n}}\left(\boldsymbol{\mathrm{A}}\cup\boldsymbol{\mathrm{B}}\right) \\ $$$$\left.\:\boldsymbol{\mathrm{ii}}\right)\:\boldsymbol{\mathrm{what}}\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{smallest}}\:\boldsymbol{\mathrm{possible}}\:\boldsymbol{\mathrm{value}} \\ $$$$\:\:\:\:\:\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{n}}\left(\boldsymbol{\mathrm{A}}\cup\boldsymbol{\mathrm{B}}\right) \\ $$$$\left.\boldsymbol{\mathrm{iii}}\right)\:\boldsymbol{\mathrm{what}}\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{smallest}}\:\boldsymbol{\mathrm{possible}}\:\boldsymbol{\mathrm{value}} \\ $$$$\:\:\:\:\:\:\:\boldsymbol{\mathrm{of}}\:\:\boldsymbol{\mathrm{n}}\left(\boldsymbol{\mathrm{A}}\cap\boldsymbol{\mathrm{B}}\right) \\ $$$$ \\…