Question Number 202315 by MATHEMATICSAM last updated on 24/Dec/23 $$\mathrm{If}\:{x}\::\:{y}\::\:{z}\:=\:{a}\::\:{b}\::\:{c}\:\mathrm{then}\:\mathrm{show}\:\mathrm{that} \\ $$$$\left(\frac{{a}\:+\:{b}\:+\:{c}}{{x}\:+\:{y}\:+\:{z}}\right)^{\mathrm{3}} \:=\:\frac{{abc}}{{xyz}}\:. \\ $$ Answered by AST last updated on 24/Dec/23 $${x}={ka};{y}={kb};{z}={kc} \\ $$$$\left(\frac{{a}+{b}+{c}}{{x}+{y}+{z}}\right)^{\mathrm{3}}…
Question Number 202308 by 2024 last updated on 24/Dec/23 Commented by mr W last updated on 24/Dec/23 $$\frac{\mathrm{4}^{{x}} }{\mathrm{9}^{{x}} }=\mathrm{9} \\ $$$$\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{2}{x}} =\mathrm{3}^{\mathrm{2}} \\ $$$$\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{{x}}…
Question Number 202306 by hardmath last updated on 24/Dec/23 $$ \\ $$Find the 2023rd term in the sequence 2,3,5,6,7,8,10,11,12,13,14,15,17,18,… obtained by subtracting integer squares from…
Question Number 202307 by MATHEMATICSAM last updated on 24/Dec/23 $$\mathrm{Show}\:\mathrm{that}\:\frac{{a}^{\mathrm{3}} \:+\:{b}^{\mathrm{3}} }{{a}^{\mathrm{2}} \:+\:{b}^{\mathrm{2}} }\:>\:\frac{{a}^{\mathrm{2}} \:+\:{b}^{\mathrm{2}} }{{a}\:+\:{b}} \\ $$ Answered by aleks041103 last updated on 24/Dec/23…
Question Number 202224 by sonukgindia last updated on 23/Dec/23 Answered by witcher3 last updated on 23/Dec/23 $$\mathrm{x}\rightarrow\frac{\mathrm{1}}{\mathrm{a}+\mathrm{bcos}^{\mathrm{2}} \left(\mathrm{x}\right)}=\mathrm{f}\left(\mathrm{x}\right),\mathrm{is}\:\mathrm{p}\:\mathrm{peridic} \\ $$$$\Rightarrow\int_{\mathrm{k}\pi} ^{\left(\mathrm{k}+\mathrm{1}\right)\pi} \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=\int_{\mathrm{0}} ^{\pi} \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx};\mathrm{by}\:\mathrm{k}\pi+\mathrm{y}=\mathrm{x} \\…
Question Number 202257 by cortano12 last updated on 23/Dec/23 Commented by Frix last updated on 23/Dec/23 Each wife is specific �� Commented by mr W last updated on 22/Apr/24…
Question Number 202258 by MATHEMATICSAM last updated on 23/Dec/23 $$\mathrm{If}\:\frac{{x}}{{a}}\:=\:\frac{{y}}{{b}}\:\mathrm{then}\:\mathrm{show}\:\mathrm{that}\: \\ $$$$\frac{{x}^{\mathrm{3}} \:+\:\mathrm{3}{xy}^{\mathrm{2}} }{{a}^{\mathrm{3}} \:+\:\mathrm{3}{ab}^{\mathrm{2}} }\:=\:\frac{\:{y}^{\mathrm{3}} \:+\:\mathrm{3}{x}^{\mathrm{2}} {y}}{{b}^{\mathrm{3}} \:+\:\mathrm{3}{a}^{\mathrm{2}} {b}}\:. \\ $$ Answered by AST…
Question Number 202255 by SANOGO last updated on 23/Dec/23 $${calcul}\:{f}_{{n}} '\left({x}\right) \\ $$$${f}_{{n}} \left({x}\right)={n}^{\alpha} {x}\left(\mathrm{1}−{x}\right)^{{n}\:} \: \\ $$ Answered by SANOGO last updated on 23/Dec/23…
Question Number 202249 by cortano12 last updated on 23/Dec/23 $$\:\:\:\:\mathrm{C}\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{x}\:\sqrt[{\mathrm{3}}]{\mathrm{cos}\:\mathrm{x}}\:−\:\mathrm{sin}\:\mathrm{x}}{\mathrm{x}^{\mathrm{5}} }\:=? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 202250 by MATHEMATICSAM last updated on 23/Dec/23 $$\left({a}^{\mathrm{2}} \:−\:{bc}\right){x}^{\mathrm{2}} \:+\:\mathrm{2}\left({b}^{\mathrm{2}} \:−\:{ca}\right){x}\:+\:\left({c}^{\mathrm{2}} \:−\:{ab}\right)\:=\:\mathrm{0} \\ $$$$\mathrm{has}\:\mathrm{two}\:\mathrm{equal}\:\mathrm{roots}.\:\mathrm{Show}\:\mathrm{that}\:\mathrm{either}\: \\ $$$${b}\:=\:\mathrm{0}\:\mathrm{or}\:\frac{{a}^{\mathrm{2}} }{{bc}}\:+\:\frac{{b}^{\mathrm{2}} }{{ca}}\:+\:\frac{{c}^{\mathrm{2}} }{{ab}}\:=\:\mathrm{3}. \\ $$ Answered by…