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Category: Algebra

Question-104369

Question Number 104369 by quvonchbek3737 last updated on 21/Jul/20 Answered by mr W last updated on 21/Jul/20 $$\mathrm{6}\left(\sqrt[{{x}}]{\mathrm{3}}\right)^{\mathrm{2}} −\mathrm{13}\left(\sqrt[{{x}}]{\mathrm{3}}\right)\left(\sqrt[{{x}}]{\mathrm{2}}\right)+\mathrm{6}\left(\sqrt[{\mathrm{3}}]{\mathrm{2}}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{6}\left(\sqrt[{{x}}]{\frac{\mathrm{3}}{\mathrm{2}}}\right)^{\mathrm{2}} −\mathrm{13}\left(\sqrt[{{x}}]{\frac{\mathrm{3}}{\mathrm{2}}}\right)+\mathrm{6}=\mathrm{0} \\ $$$$\sqrt[{{x}}]{\frac{\mathrm{3}}{\mathrm{2}}}=\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{{x}}…

Suppose-you-had-x-rupees-Your-father-gave-you-more-6-rupees-You-gave-y-rupees-to-your-brother-Now-how-many-rupees-you-have-You-will-buy-three-shirts-with-the-rupees-that-you-have-now-Write-the-co

Question Number 104326 by Anindita last updated on 20/Jul/20 $$\boldsymbol{\mathrm{Suppose}}\:\boldsymbol{\mathrm{you}}\:\boldsymbol{\mathrm{had}}\:\boldsymbol{{x}}\:\boldsymbol{\mathrm{rupees}}.\:\boldsymbol{\mathrm{Your}} \\ $$$$\boldsymbol{\mathrm{father}}\:\boldsymbol{\mathrm{gave}}\:\boldsymbol{\mathrm{you}}\:\boldsymbol{\mathrm{more}}\:\mathrm{6}\:\boldsymbol{\mathrm{rupees}}.\:\boldsymbol{\mathrm{You}} \\ $$$$\boldsymbol{\mathrm{gave}}\:\boldsymbol{{y}}\:\boldsymbol{\mathrm{rupees}}\:\boldsymbol{\mathrm{to}}\:\boldsymbol{\mathrm{your}}\:\boldsymbol{\mathrm{brother}}.\:\boldsymbol{\mathrm{Now}} \\ $$$$\boldsymbol{\mathrm{how}}\:\boldsymbol{\mathrm{many}}\:\boldsymbol{\mathrm{rupees}}\:\boldsymbol{\mathrm{you}}\:\boldsymbol{\mathrm{have}}?\:\boldsymbol{\mathrm{You}}\:\boldsymbol{\mathrm{will}} \\ $$$$\boldsymbol{\mathrm{buy}}\:\boldsymbol{\mathrm{three}}\:\boldsymbol{\mathrm{shirts}}\:\boldsymbol{\mathrm{with}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{rupees}}\:\boldsymbol{\mathrm{that}} \\ $$$$\boldsymbol{\mathrm{you}}\:\boldsymbol{\mathrm{have}}\:\boldsymbol{\mathrm{now}}.\:\boldsymbol{\mathrm{Write}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{cost}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{each}} \\ $$$$\boldsymbol{\mathrm{shirt}}. \\ $$ Commented…

y-x-2-1-y-0-x-1-x-2-find-the-volume-of-the-object-obtained-by-rotating-the-figure-bounded-by-lines-around-the-abscissa-axis-

Question Number 169860 by Shrinava last updated on 11/May/22 $$\mathrm{y}\:=\:\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{1} \\ $$$$\mathrm{y}\:=\:\mathrm{0} \\ $$$$\mathrm{x}\:=\:-\:\mathrm{1} \\ $$$$\mathrm{x}\:=\:\mathrm{2} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{volume}\:\mathrm{of}\:\mathrm{the}\:\mathrm{object}\:\mathrm{obtained} \\ $$$$\mathrm{by}\:\mathrm{rotating}\:\mathrm{the}\:\mathrm{figure}\:\mathrm{bounded}\:\mathrm{by}\:\mathrm{lines} \\ $$$$\mathrm{around}\:\mathrm{the}\:\mathrm{abscissa}\:\mathrm{axis} \\ $$…

y-x-2-2-y-x-x-0-x-1-find-the-area-of-the-figure-bounded-by-lines-

Question Number 169852 by Shrinava last updated on 10/May/22 $$\mathrm{y}\:=\:\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{2} \\ $$$$\mathrm{y}\:=\:-\:\mathrm{x} \\ $$$$\mathrm{x}\:=\:\mathrm{0} \\ $$$$\mathrm{x}\:=\:\mathrm{1} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{figure}\:\mathrm{bounded}\:\mathrm{by} \\ $$$$\mathrm{lines} \\ $$ Answered by…

Question-38786

Question Number 38786 by Rio Mike last updated on 29/Jun/18 Commented by Rio Mike last updated on 29/Jun/18 $${Three}\:{triangles}\:{are}\:{colined}\:{or}\:{lined} \\ $$$${on}\:{a}\:{line}\:{with}\:{equation}\: \\ $$$${y}\:=\:\mathrm{3}{x}\:+\:\mathrm{1}\:{as}\:{shown}\:{above}. \\ $$$${the}\:{height}\:{of}\:{the}\:{triangles}\:{are}…

Question-169855

Question Number 169855 by Shrinava last updated on 11/May/22 Answered by thfchristopher last updated on 11/May/22 $$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{sin}\:\left(\mathrm{3}{x}−{x}\right){dx} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{sin}\:\mathrm{2}{xdx} \\ $$$$−\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{cos}\:\mathrm{2}{x}\right]_{\mathrm{0}}…

Question-169842

Question Number 169842 by mathlove last updated on 10/May/22 Answered by Rasheed.Sindhi last updated on 10/May/22 $${f}^{\:\mathrm{2}} \left({x}\right).{f}\left(\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\right)={x}^{\mathrm{2}} ………….\left(\mathrm{A}\right) \\ $$$$\left(\mathrm{A}\right)^{\mathrm{2}} :\:\:{f}^{\:\:\mathrm{4}} \left({x}\right).{f}^{\:\mathrm{2}} \left(\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\right)={x}^{\mathrm{4}} \\…