Question Number 212651 by hardmath last updated on 20/Oct/24 Answered by Ghisom last updated on 20/Oct/24 $${x}+\mathrm{3}{y}+\mathrm{5}{z}\leqslant\mathrm{15} \\ $$$${x}+{y}+{z}\leqslant\mathrm{7} \\ $$$$\mathrm{2}{x}+{y}+\mathrm{4}{z}\leqslant\mathrm{12} \\ $$$$======== \\ $$$${x}+\mathrm{3}{y}+\mathrm{5}{z}\leqslant\mathrm{15}…
Question Number 212601 by Spillover last updated on 18/Oct/24 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 212598 by MATHEMATICSAM last updated on 18/Oct/24 $$\mathrm{Help}\:\mathrm{me}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{pls} \\ $$$$\mathrm{Q}\:\mathrm{212576}\: \\ $$ Answered by A5T last updated on 18/Oct/24 $${Lemma}:\:\frac{{a}}{{b}}=\frac{{c}}{{d}}\Rightarrow\frac{{a}}{{b}}=\frac{{c}}{{d}}=\frac{{a}\underset{−} {+}{c}}{{b}\underset{−} {+}{d}} \\…
Question Number 212602 by Spillover last updated on 18/Oct/24 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 212612 by hardmath last updated on 18/Oct/24 $$\mathrm{Solve}\:\mathrm{the}\:\mathrm{differential}\:\mathrm{equation}: \\ $$$$\frac{\mathrm{dy}}{\mathrm{dx}}\:\:=\:\:\mathrm{cos}\left(\mathrm{x}\:+\:\mathrm{y}\right) \\ $$ Answered by mr W last updated on 19/Oct/24 $${let}\:{t}={x}+{y} \\ $$$$\frac{{dt}}{{dx}}=\mathrm{1}+\frac{{dy}}{{dx}}…
Question Number 212609 by hardmath last updated on 18/Oct/24 $$\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 $${f}\left({x}\right)=\mathrm{2}^{{x}}…
Question Number 212573 by hardmath last updated on 17/Oct/24 $$\mathrm{Prove}\:\mathrm{that}:\:\:\:\frac{\mathrm{5m}^{\mathrm{2}} \:−\:\mathrm{n}}{\mathrm{n}^{\mathrm{2}} \:+\:\mathrm{3m}}\:=\:\mathrm{1} \\ $$ Commented by Frix last updated on 18/Oct/24 $$\mathrm{Prove}\:\mathrm{what}\:\mathrm{exactly}? \\ $$ Commented…
Question Number 212560 by ajfour last updated on 17/Oct/24 $${Exact}\:{solution}\:{just}\:{to}\:{this}\:{please}: \\ $$$$\mathrm{9}{x}^{\mathrm{2}} \left({x}+\mathrm{1}\right)^{\mathrm{3}} =\mathrm{3}\left(\mathrm{3}{x}^{\mathrm{2}} −\mathrm{1}\right)\left(\mathrm{3}{x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{1}\right) \\ $$ Commented by Frix last updated on 18/Oct/24…
Question Number 212552 by mnjuly1970 last updated on 17/Oct/24 $$ \\ $$$$\:\overset{\mathrm{Q}:} {\:}\:\mathrm{In}\:\mathrm{A}\overset{\Delta} {\mathrm{B}C}\::\:\:\:{cos}\left(\mathrm{A}\right)\:+{cos}\left(\mathrm{B}\:\right)+\:\mathrm{2}{cos}\left(\mathrm{C}\:\right)=\:\mathrm{2} \\ $$$$\:\:\: \\ $$$$\:\:\:\:\:\:\mathrm{show}\:\mathrm{that}\::\:\:\:{a}\:+\:{b}\:=\:\mathrm{2}{c}\:\:\:\:\:\:\:\:\:\:\:\blacksquare\: \\ $$$$ \\ $$ Answered by Ghisom…
Question Number 212576 by MATHEMATICSAM last updated on 17/Oct/24 $$\mathrm{If}\:\frac{{x}^{\mathrm{2}} \:−\:{yz}}{{a}^{\mathrm{2}} \:−\:{bc}}\:=\:\frac{{y}^{\mathrm{2}} \:−\:{zx}}{{b}^{\mathrm{2}} \:−\:{ca}}\:=\:\frac{{z}^{\mathrm{2}} \:−\:{xy}}{{c}^{\mathrm{2}} \:−\:{ab}}\:\mathrm{then}\: \\ $$$$\mathrm{prove}\:\mathrm{that}\:\frac{{x}}{{a}}\:=\:\frac{{y}}{{b}}\:=\:\frac{{z}}{{c}}\:. \\ $$ Terms of Service Privacy Policy…