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Category: Algebra

A-hockey-team-plays-in-an-arena-that-has-a-seating-capacity-of-15-000-spectators-With-the-ticket-price-seat-at-14-average-attendance-at-recent-games-has-been-9500-A-market-survey-indicates-that-fo

Question Number 135230 by benjo_mathlover last updated on 11/Mar/21 $${A}\:{hockey}\:{team}\:{plays}\:{in}\:{an}\:{arena}\:{that} \\ $$$${has}\:{a}\:{seating}\:{capacity}\:{of}\:\mathrm{15},\mathrm{000} \\ $$$${spectators}.\:{With}\:{the}\:{ticket}\:{price} \\ $$$${seat}\:{at}\:\$\mathrm{14},\:{average}\:{attendance} \\ $$$${at}\:{recent}\:{games}\:{has}\:{been}\:\mathrm{9500}. \\ $$$${A}\:{market}\:{survey}\:{indicates}\:{that} \\ $$$${for}\:{each}\:{dollar}\:{the}\:{ticket}\:{price} \\ $$$${is}\:{lowered}\:,{the}\:{average}\:{attendance} \\…

prove-that-2x-3-x-2-2x-1-x-3-1-x-3-1-x-4-2x-3-3x-2-2x-1-2-

Question Number 69662 by aliesam last updated on 26/Sep/19 $${prove}\:{that} \\ $$$$ \\ $$$$\frac{\mathrm{2}{x}^{\mathrm{3}} −{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}}{{x}^{\mathrm{3}} +\mathrm{1}}\:+\:\frac{{x}^{\mathrm{3}} +\mathrm{1}}{{x}^{\mathrm{4}} −\mathrm{2}{x}^{\mathrm{3}} +\mathrm{3}{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}}\:=\:\mathrm{2} \\ $$ Answered by…

x-2-yz-3-y-2-zx-5-z-2-xy-1-solve-for-x-y-and-z-

Question Number 135191 by liberty last updated on 11/Mar/21 $$\begin{cases}{\mathrm{x}^{\mathrm{2}} −\mathrm{yz}\:=\:\mathrm{3}}\\{\mathrm{y}^{\mathrm{2}} −\:\mathrm{zx}\:=\:\mathrm{5}}\\{\mathrm{z}^{\mathrm{2}} −\mathrm{xy}\:=\:−\mathrm{1}}\end{cases} \\ $$$$\mathrm{solve}\:\mathrm{for}\:\mathrm{x}\:,\mathrm{y}\:\mathrm{and}\:\mathrm{z}. \\ $$ Answered by MJS_new last updated on 11/Mar/21 $${y}={px}\wedge{z}={qx}…

Question-69644

Question Number 69644 by ahmadshahhimat775@gmail.com last updated on 26/Sep/19 Answered by MJS last updated on 26/Sep/19 $$\mathrm{let} \\ $$$${x}=\alpha \\ $$$${y}=\beta−\sqrt{\gamma} \\ $$$${z}=\beta+\sqrt{\gamma} \\ $$$$\begin{cases}{\alpha+\mathrm{2}\beta−\mathrm{12}=\mathrm{0}}\\{\alpha^{\mathrm{2}}…

Question-69645

Question Number 69645 by ahmadshahhimat775@gmail.com last updated on 26/Sep/19 Answered by Rasheed.Sindhi last updated on 26/Sep/19 $$\sqrt{{x}+\sqrt{{x}+\sqrt{{x}….}}}=\mathrm{2} \\ $$$$\left(\sqrt{{x}+\sqrt{{x}+\sqrt{{x}….}}}\right)=\left(\mathrm{2}\right)^{\mathrm{2}} \\ $$$${x}+\sqrt{{x}+\sqrt{{x}+\sqrt{{x}…}}}=\mathrm{4} \\ $$$${x}+\mathrm{2}=\mathrm{4} \\ $$$${x}=\mathrm{2}…

x-2-9-3x-5-x-3-x-1-x-3-x-1-Find-solution-

Question Number 135172 by bemath last updated on 11/Mar/21 $$\left(\mathrm{x}^{\mathrm{2}} −\mathrm{9}\right)^{\mathrm{3x}+\mathrm{5}} \:=\:\left(\mathrm{x}−\mathrm{3}\right)^{\mathrm{x}−\mathrm{1}} .\left(\mathrm{x}+\mathrm{3}\right)^{\mathrm{x}−\mathrm{1}} \\ $$$$\mathrm{Find}\:\mathrm{solution} \\ $$ Answered by john_santu last updated on 11/Mar/21 $$\Rightarrow\left({x}^{\mathrm{2}}…

1-2-3-3-3-2-4-3-3-5-3-4-6-3-5-

Question Number 135170 by bramlexs22 last updated on 11/Mar/21 $$\mathrm{1}+\frac{\mathrm{2}}{\mathrm{3}}+\frac{\mathrm{3}}{\mathrm{3}^{\mathrm{2}} }+\frac{\mathrm{4}}{\mathrm{3}^{\mathrm{3}} }+\frac{\mathrm{5}}{\mathrm{3}^{\mathrm{4}} }+\frac{\mathrm{6}}{\mathrm{3}^{\mathrm{5}} }+…\:=?\: \\ $$$$ \\ $$ Answered by bemath last updated on 11/Mar/21…