Menu Close

Category: Algebra

f-x-1-1-x-y-f-x-z-f-y-f-z-

Question Number 212711 by RojaTaniya last updated on 21/Oct/24 $$\:{f}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{1}−{x}} \\ $$$$\:{y}={f}\left({x}\right),\:{z}={f}\left({y}\right),\:{f}\left({z}\right)=? \\ $$ Answered by MATHEMATICSAM last updated on 21/Oct/24 $${y}\:=\:{f}\left({x}\right)\:=\:\frac{\mathrm{1}}{\mathrm{1}\:−\:{x}} \\ $$$${f}\left({y}\right)\:=\:{f}\left(\frac{\mathrm{1}}{\mathrm{1}\:−\:{x}}\right)\:=\:\frac{\mathrm{1}}{\mathrm{1}\:−\:\frac{\mathrm{1}}{\mathrm{1}\:−\:{x}}}\:=\:\frac{\mathrm{1}\:−\:{x}}{−\:{x}}\:=\:\frac{{x}\:−\:\mathrm{1}}{{x}} \\…

Question-212673

Question Number 212673 by golsendro last updated on 21/Oct/24 $$\:\:\:\: \\ $$ Answered by Frix last updated on 21/Oct/24 $$\sqrt{\mathrm{2}{x}+\mathrm{3}}={x}^{\mathrm{2}} −{x}−\mathrm{3} \\ $$$$\mathrm{Squaring}\:\&\:\mathrm{transforming} \\ $$$$\:\:\:\:\:\left[\mathrm{introduces}\:\mathrm{false}\:\mathrm{solutions}!\right]…

Question-212651

Question Number 212651 by hardmath last updated on 20/Oct/24 Answered by Ghisom last updated on 20/Oct/24 $${x}+\mathrm{3}{y}+\mathrm{5}{z}\leqslant\mathrm{15} \\ $$$${x}+{y}+{z}\leqslant\mathrm{7} \\ $$$$\mathrm{2}{x}+{y}+\mathrm{4}{z}\leqslant\mathrm{12} \\ $$$$======== \\ $$$${x}+\mathrm{3}{y}+\mathrm{5}{z}\leqslant\mathrm{15}…

Help-me-to-solve-pls-Q-212576-

Question Number 212598 by MATHEMATICSAM last updated on 18/Oct/24 $$\mathrm{Help}\:\mathrm{me}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{pls} \\ $$$$\mathrm{Q}\:\mathrm{212576}\: \\ $$ Answered by A5T last updated on 18/Oct/24 $${Lemma}:\:\frac{{a}}{{b}}=\frac{{c}}{{d}}\Rightarrow\frac{{a}}{{b}}=\frac{{c}}{{d}}=\frac{{a}\underset{−} {+}{c}}{{b}\underset{−} {+}{d}} \\…

If-0-x-2-Prove-that-2-x-1-10-5x-4-0-

Question Number 212609 by hardmath last updated on 18/Oct/24 $$\mathrm{If}\:\:\:\mathrm{0}\leqslant\mathrm{x}\leqslant\mathrm{2} \\ $$$$\mathrm{Prove}\:\mathrm{that}: \\ $$$$\mathrm{2}^{\boldsymbol{\mathrm{x}}} \:+\:\mathrm{1}\:−\:\sqrt{\mathrm{10},\mathrm{5x}\:+\:\mathrm{4}}\:\leqslant\:\mathrm{0} \\ $$ Answered by Frix last updated on 18/Oct/24 $${f}\left({x}\right)=\mathrm{2}^{{x}}…