Question Number 174487 by mnjuly1970 last updated on 02/Aug/22 Answered by dragan91 last updated on 02/Aug/22 $$ \\ $$$$\mathrm{x}^{\mathrm{2}} −\mathrm{25x}=\mathrm{x}+\mathrm{5}\sqrt{\mathrm{x}} \\ $$$$\mathrm{x}\left(\mathrm{x}−\mathrm{25}\right)=\sqrt{\mathrm{x}}\left(\sqrt{\mathrm{x}}+\mathrm{5}\right) \\ $$$$\mathrm{x}\left(\sqrt{\mathrm{x}}−\mathrm{5}\right)\left(\sqrt{\mathrm{x}}+\mathrm{5}\right)=\sqrt{\mathrm{x}}\left(\sqrt{\mathrm{x}}+\mathrm{5}\right) \\…
Question Number 174479 by Michaelfaraday last updated on 02/Aug/22 Commented by mr W last updated on 03/Aug/22 $${see}\:{Q}\mathrm{147791} \\ $$ Terms of Service Privacy Policy…
Question Number 174464 by Shrinava last updated on 01/Aug/22 $$\mathrm{In}\:\:\bigtriangleup\mathrm{ABC} \\ $$$$\mathrm{R}\in\left(\mathrm{AB}\right)\:,\:\mathrm{P}\in\left(\mathrm{BC}\right)\:,\:\mathrm{Q}\in\left(\mathrm{CA}\right) \\ $$$$\mathrm{AR}=\mathrm{3}\:,\:\mathrm{RB}=\mathrm{1}\:,\:\mathrm{BP}=\mathrm{6}\:,\:\mathrm{PC}=\mathrm{2}\:,\:\mathrm{CQ}=\mathrm{5}\:,\:\mathrm{QA}=\mathrm{4} \\ $$$$\mathrm{Prove}\:\mathrm{that}: \\ $$$$\mathrm{PQ}\:+\:\mathrm{QR}\:+\:\mathrm{RP}\:>\:\frac{\mathrm{21}}{\mathrm{2}} \\ $$ Terms of Service Privacy Policy…
Question Number 43391 by gunawan last updated on 10/Sep/18 $${x}^{\mathrm{2}} +{x}={y}^{\mathrm{4}} +{y}^{\mathrm{3}} +{y}^{\mathrm{2}} +{y} \\ $$$${x}^{\mathrm{4}} +\left({x}+\mathrm{1}\right)^{\mathrm{4}} ={y}^{\mathrm{2}} +\left({y}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\mathrm{find}\:{x}\:\mathrm{and}\:{y}\:\:\mathrm{of}\:\mathrm{is}\ldots \\ $$$$ \\ $$…
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Question Number 108917 by pete last updated on 20/Aug/20 $$\mathrm{A}\:\mathrm{woman}\:\mathrm{purchased}\:\mathrm{a}\:\mathrm{number}\:\mathrm{of}\:\mathrm{plates} \\ $$$$\mathrm{for\$150}.\mathrm{00}.\:\mathrm{Four}\:\mathrm{of}\:\mathrm{the}\:\mathrm{plates}\:\mathrm{got}\:\mathrm{broken} \\ $$$$\mathrm{while}\:\mathrm{transporting}\:\mathrm{themto}\:\mathrm{her}\:\mathrm{shop}.\:\mathrm{By}\:\mathrm{selling} \\ $$$$\mathrm{the}\:\mathrm{remaining}\:\mathrm{plates}\:\mathrm{at}\:\mathrm{a}\:\mathrm{profit}\:\mathrm{of}\:\$\:\mathrm{1}.\mathrm{00}\:\mathrm{on}\:\mathrm{each}, \\ $$$$\mathrm{she}\:\mathrm{made}\:\mathrm{a}\:\mathrm{total}\:\mathrm{profit}\:\mathrm{of}\:\$\mathrm{6}.\mathrm{00}.\:\mathrm{How}\:\mathrm{many} \\ $$$$\mathrm{plates}\:\mathrm{did}\:\mathrm{she}\:\mathrm{purchase}? \\ $$ Answered by nimnim…
Question Number 174451 by jahar last updated on 01/Aug/22 $${please}\:{help} \\ $$$${If}\:{x}\:{is}\:{directly}\:{porportional}\:{to}\:{z}\:\:{and}\: \\ $$$${y}\:{is}\:{also}\:{directly}\:{porportional}\:{to}\:{z} \\ $$$$\:{then}\:{what}\:{is} \\ $$$${the}\:{value}\:{of}\:\:{xy}\:{propotional}\:{to}\:\:? \\ $$$$ \\ $$ Terms of Service…
Question Number 108914 by 1549442205PVT last updated on 21/Aug/20 $$ \\ $$$$\mathrm{Q108815}\left(\mathrm{19}/\mathrm{8}/\mathrm{20}\right)\left(\mathrm{unanswer}\right)\mathrm{by}\:\mathrm{1x}.\mathrm{x} \\ $$$$\mathrm{Given}\:\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\mathrm{x}}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\mathrm{a}}}+\frac{\sqrt{\mathrm{ax}}}{\:\sqrt{\mathrm{ax}+\mathrm{8}}} \\ $$$$\mathrm{x},\mathrm{a}\in\mathrm{R};\mathrm{x},\mathrm{a}>\mathrm{0}.\mathrm{Prove}\:\mathrm{that} \\ $$$$\mathrm{1}<\mathrm{f}\left(\mathrm{x}\right)<\mathrm{2} \\ $$$$\mathrm{Solution}:\mathrm{Put}\:\mathrm{x}=\mathrm{tan}^{\mathrm{2}} \mathrm{A},\mathrm{a}=\mathrm{tan}^{\mathrm{2}} \mathrm{B}\left(\mathrm{A},\mathrm{B}\in\left[\mathrm{0},\frac{\pi}{\mathrm{2}}\right)\right. \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{cosA}+\mathrm{cosB}+\frac{\mathrm{tanAtanB}}{\:\sqrt{\mathrm{tan}^{\mathrm{2}} \mathrm{Atan}^{\mathrm{2}}…
Question Number 174440 by mathlove last updated on 01/Aug/22 $$\mathrm{1}+\sqrt{\mathrm{3}^{{x}} }=\mathrm{2}^{{x}} \:\:\:\:\:\:\:\:{x}=? \\ $$ Commented by infinityaction last updated on 01/Aug/22 $$\mathrm{2} \\ $$ Commented…
Question Number 174439 by mnjuly1970 last updated on 01/Aug/22 $$ \\ $$$$\:\:\:\:\:{x}\in\:\left(\:\mathrm{0}\:,\:\mathrm{1}\:\right)\:,\:{k}\:\in\:\mathbb{N} \\ $$$$\:\:\:\:\:\:{prove}\:\:{that}\:: \\ $$$$\:\:{kx}^{\:{k}} \:<\:\frac{{x}}{\mathrm{1}−{x}}\:\:\left({math}\:\:{analysis}\right) \\ $$$$ \\ $$ Answered by behi834171 last…