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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…
Question Number 43363 by rahul 19 last updated on 10/Sep/18 $$\mathrm{If}\:\mathrm{z}=\:\mathrm{cos}\:\theta\:+\:\mathrm{isin}\:\theta\:,\:\mathrm{0}<\theta<\frac{\pi}{\mathrm{6}}\:,\:\mathrm{then}\:\mathrm{prove} \\ $$$$\mathrm{that}\:\mathrm{argument}\:\mathrm{of}\:\:\mathrm{1}−\mathrm{z}^{\mathrm{4}} \:=\:\mathrm{2}\theta\:−\:\frac{\pi}{\mathrm{2}}\:. \\ $$ Commented by maxmathsup by imad last updated on 10/Sep/18…
Question Number 174424 by mnjuly1970 last updated on 31/Jul/22 $$ \\ $$$$\:\:\:\:#\:\:{prove}\:\:{that}\:\:# \\ $$$$\:\:\:{tan}^{\:\mathrm{2}} \left(\:\mathrm{10}^{\:\mathrm{o}} \right)\:+\:{tan}^{\:\mathrm{2}} \left(\mathrm{50}^{\:\mathrm{o}} \:\right)+{tan}^{\:\mathrm{2}} \left(\mathrm{70}^{\:\mathrm{o}} \right)\overset{?} {=}\:\mathrm{9} \\ $$$$\:\:\:\:\:−−−−− \\ $$$$…
Question Number 174427 by mnjuly1970 last updated on 31/Jul/22 $$ \\ $$$$\:\:\:{solve}\:\:{for}\:\:\:{x}\:\in\:\mathbb{R}\:: \\ $$$$ \\ $$$$\:\sqrt[{\mathrm{4}}]{\mathrm{12}+\mathrm{4}{x}^{\:} −{x}^{\:\mathrm{2}} }\:+\sqrt{\mathrm{1}+\mathrm{8}{x}−\mathrm{2}{x}^{\:\mathrm{2}} }\:=\:\mathrm{2}{x}^{\:\mathrm{2}} −\mathrm{8}{x}\:+\mathrm{13}\:\:\blacksquare \\ $$$$ \\ $$$$ \\…
Question Number 43353 by pieroo last updated on 10/Sep/18 $$\mathrm{Given}\:\mathrm{the}\:\mathrm{functions}\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{2x}−\mathrm{1}\:\mathrm{and}\:\mathrm{f}\bullet\mathrm{g}\left(\mathrm{x}\right)=\mathrm{x}^{\mathrm{2}} −\mathrm{x}+\mathrm{2}, \\ $$$$\mathrm{find}\:\mathrm{g}\left(\mathrm{x}\right) \\ $$ Commented by maxmathsup by imad last updated on 10/Sep/18 $${we}\:{have}\:{f}\left({x}\right)={y}\:\Rightarrow\mathrm{2}{x}−\mathrm{1}={y}\:\Rightarrow{x}\:=\frac{{y}+\mathrm{1}}{\mathrm{2}}\:\Rightarrow{f}^{−\mathrm{1}}…