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 $${x}={ka};{y}={kb};{z}={kc} \\ $$$$\left(\frac{{a}+{b}+{c}}{{x}+{y}+{z}}\right)^{\mathrm{3}}…
Question Number 202306 by hardmath last updated on 24/Dec/23 $$ \\ $$Find the 2023rd term in the sequence 2,3,5,6,7,8,10,11,12,13,14,15,17,18,… obtained by subtracting integer squares from…
Question Number 202307 by MATHEMATICSAM last updated on 24/Dec/23 $$\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}} \\ $$ Answered by aleks041103 last updated on 24/Dec/23…
Question Number 202258 by MATHEMATICSAM last updated on 23/Dec/23 $$\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 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…
Question Number 202198 by MATHEMATICSAM last updated on 22/Dec/23 $$\mathrm{If}\:\alpha,\:\beta\:\mathrm{are}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:{x}^{\mathrm{2}} \:+\:{ax}\:−\:{b}\:=\:\mathrm{0}\: \\ $$$$\mathrm{and}\:\gamma,\:\delta\:\mathrm{are}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:{x}^{\mathrm{2}} \:+\:{ax}\:+\:{c}\:=\:\mathrm{0}\: \\ $$$$\mathrm{then}\:\mathrm{show}\:\mathrm{that}\:\frac{\alpha\:−\:\gamma}{\beta\:−\:\gamma}\:=\:\frac{\beta\:−\:\delta}{\alpha\:−\:\delta}\:\:. \\ $$ Answered by MM42 last updated on 22/Dec/23…
Question Number 202193 by MATHEMATICSAM last updated on 22/Dec/23 $$\mathrm{If}\:\alpha,\:\beta\:\mathrm{are}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:{x}^{\mathrm{2}} \:+\:{ax}\:+\:{b}\:=\:\:\mathrm{0}\: \\ $$$$\mathrm{and}\:\alpha\:+\:\delta,\:\beta\:+\:\delta\:\mathrm{are}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\: \\ $$$${x}^{\mathrm{2}} \:+\:{px}\:+\:{q}\:=\:\mathrm{0}\:\mathrm{then}\:\mathrm{show}\:\mathrm{that}\: \\ $$$${a}^{\mathrm{2}} \:−\:{p}^{\mathrm{2}} \:=\:\mathrm{4}\left({b}\:−\:{q}\right). \\ $$ Answered by Rasheed.Sindhi…
Question Number 202184 by mr W last updated on 22/Dec/23 $${with}\:{f}\left({x}\right)={x}^{\mathrm{2}} +\mathrm{12}{x}+\mathrm{30}\:{and}\:{x}\in{R} \\ $$$${solve}\:{f}\left({f}\left({f}\left({f}\left({f}\left({x}\right)\right)\right)\right)\right)=\mathrm{0} \\ $$ Answered by cortano12 last updated on 22/Dec/23 $$\mathrm{f}\left(\mathrm{x}\right)=\left(\mathrm{x}+\mathrm{6}\right)^{\mathrm{2}} −\mathrm{6}\:\Rightarrow\mathrm{x}=\sqrt{\mathrm{f}\left(\mathrm{x}\right)+\mathrm{6}}−\mathrm{6}…
Question Number 202208 by hardmath last updated on 22/Dec/23 Commented by mr W last updated on 22/Dec/23 $${which}\:{country}\:{are}\:{you}\:{from}? \\ $$$${internationally}\:{the}\:{ABO}\:{blood} \\ $$$${group}\:{system}\:{is}\:{common},\:{see}\:{below}. \\ $$$${but}\:{you}\:{are}\:{talking}\:{about}\:{blood}\: \\…
Question Number 202129 by MATHEMATICSAM last updated on 21/Dec/23 $$\mathrm{If}\:{p}\:=\:\mathrm{9999}\:\mathrm{then}\:\frac{\mathrm{4}{p}^{\mathrm{3}} \:−\:{p}}{\left(\mathrm{2}{p}\:+\:\mathrm{1}\right)\left(\mathrm{6}{p}\:−\:\mathrm{3}\right)}\:=\:? \\ $$ Answered by AST last updated on 21/Dec/23 $$\frac{{p}\left(\mathrm{2}{p}−\mathrm{1}\right)\left(\mathrm{2}{p}+\mathrm{1}\right)}{\mathrm{3}\left(\mathrm{2}{p}+\mathrm{1}\right)\left(\mathrm{2}{p}−\mathrm{1}\right)}=\frac{{p}=\mathrm{9999}}{\mathrm{3}}=\mathrm{3333} \\ $$ Terms of…