Question Number 152457 by pete last updated on 28/Aug/21 $$ \\ $$$$\mathrm{Kofi}\:\mathrm{is}\:\mathrm{20\%}\:\:\mathrm{heavier}\:\mathrm{than}\:\mathrm{Afia}.\:\mathrm{If}\:\mathrm{Kofi} \\ $$$$\mathrm{weighs}\:\mathrm{60}\:\mathrm{kg}\:\mathrm{what}\:\mathrm{is}\:\mathrm{Afia}'\mathrm{s}\:\mathrm{weight}? \\ $$ Answered by Rasheed.Sindhi last updated on 28/Aug/21 $${Let}\:{Afia}'{s}\:{weight}\:{is}\:{w} \\…
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Question Number 21355 by Tinkutara last updated on 21/Sep/17 $$\mathrm{Solve}\::\:\mid{x}^{\mathrm{2}} \:+\:\mathrm{3}{x}\mid\:+\:{x}^{\mathrm{2}} \:−\:\mathrm{2}\:\geqslant\:\mathrm{0} \\ $$ Answered by mrW1 last updated on 22/Sep/17 $$\mid\mathrm{x}^{\mathrm{2}} +\mathrm{3x}\mid\geqslant\mathrm{2}−\mathrm{x}^{\mathrm{2}} \\ $$$$…
Question Number 21356 by Tinkutara last updated on 21/Sep/17 $$\mathrm{Solve}\::\:\sqrt[{\mathrm{3}}]{\mathrm{2}^{\frac{\mathrm{3}{x}−\mathrm{1}}{{x}−\mathrm{1}}} }\:<\:\mathrm{8}^{\frac{{x}−\mathrm{3}}{\mathrm{3}{x}−\mathrm{7}}} \\ $$ Answered by dioph last updated on 21/Sep/17 $${x}−\mathrm{1}\:\neq\:\mathrm{0}\:\Rightarrow\:{x}\:\neq\:\mathrm{1} \\ $$$$\mathrm{3}{x}−\mathrm{7}\:\neq\:\mathrm{0}\:\Rightarrow\:{x}\:\neq\:\frac{\mathrm{7}}{\mathrm{3}} \\ $$$$\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{3}}×\frac{\mathrm{3}{x}−\mathrm{1}}{{x}−\mathrm{1}}}…
Question Number 21357 by Tinkutara last updated on 21/Sep/17 $$\mathrm{Solve}\::\:\mathrm{log}_{\mathrm{2}{x}+\mathrm{3}} {x}^{\mathrm{2}} \:<\:\mathrm{1} \\ $$ Answered by dioph last updated on 21/Sep/17 $$\mathrm{2}{x}\:+\:\mathrm{3}\:>\:\mathrm{0}\:\Rightarrow\:{x}\:>\:−\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\mathrm{2}{x}\:+\:\mathrm{3}\:\neq\:\mathrm{1}\:\Rightarrow\:{x}\:\neq\:−\mathrm{1} \\…
Question Number 21354 by Tinkutara last updated on 21/Sep/17 $$\mathrm{Solve}\::\:\sqrt{\mathrm{2}{x}\:+\:\mathrm{5}}\:+\:\sqrt{{x}\:−\:\mathrm{1}}\:>\:\mathrm{8} \\ $$ Commented by Tinkutara last updated on 22/Sep/17 $${Where}\:{is}\:{my}\:{mistake}? \\ $$$$\sqrt{\mathrm{2}{x}+\mathrm{5}}+\sqrt{{x}−\mathrm{1}}>\mathrm{8} \\ $$$$\mathrm{2}{x}+\mathrm{5}+{x}−\mathrm{1}+\mathrm{2}\sqrt{\left(\mathrm{2}{x}+\mathrm{5}\right)\left({x}−\mathrm{1}\right)}>\mathrm{64} \\…
Question Number 152396 by mathdanisur last updated on 28/Aug/21 $$\mathrm{If}\:\:\mathrm{x}^{\mathrm{3}} -\mathrm{x}+\mathrm{3}=\mathrm{0}\:\:\mathrm{has}\:\mathrm{the}\:\mathrm{roots}\:\:\mathrm{a};\mathrm{b}\:\:\mathrm{and}\:\:\mathrm{c} \\ $$$$\mathrm{determine}\:\mathrm{the}\:\mathrm{monic}\:\mathrm{polynomial}\:\mathrm{with} \\ $$$$\mathrm{the}\:\mathrm{roots}\:\:\mathrm{a}^{\mathrm{5}} \:;\:\mathrm{b}^{\mathrm{5}} \:\:\mathrm{and}\:\:\mathrm{c}^{\mathrm{5}} \:. \\ $$ Answered by mindispower last updated…
Question Number 21321 by Tinkutara last updated on 20/Sep/17 $$\mathrm{The}\:\mathrm{number}\:\mathrm{of}\:\mathrm{real}\:\mathrm{solutions}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{equation}\:\mathrm{4}{x}^{\mathrm{99}} \:+\:\mathrm{5}{x}^{\mathrm{98}} \:+\:\mathrm{4}{x}^{\mathrm{97}} \:+\:\mathrm{5}{x}^{\mathrm{96}} \:+ \\ $$$$…..\:+\:\mathrm{4}{x}\:+\:\mathrm{5}\:=\:\mathrm{0}\:\mathrm{is} \\ $$ Answered by dioph last updated…
Question Number 21319 by Tinkutara last updated on 20/Sep/17 $$\mathrm{If}\:{x},\:{y},\:{z}\:\mathrm{are}\:\mathrm{three}\:\mathrm{real}\:\mathrm{numbers}\:\mathrm{such} \\ $$$$\mathrm{that}\:{x}\:+\:{y}\:+\:{z}\:=\:\mathrm{4}\:\mathrm{and}\:{x}^{\mathrm{2}} \:+\:{y}^{\mathrm{2}} \:+\:{z}^{\mathrm{2}} \:=\:\mathrm{6}, \\ $$$$\mathrm{then} \\ $$$$\left(\mathrm{1}\right)\:\frac{\mathrm{2}}{\mathrm{3}}\:\leqslant\:{x},\:{y},\:{z}\:\leqslant\:\mathrm{2} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{0}\:\leqslant\:{x},\:{y},\:{z}\:\leqslant\:\mathrm{2} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{1}\:\leqslant\:{x},\:{y},\:{z}\:\leqslant\:\mathrm{3} \\ $$$$\left(\mathrm{4}\right)\:\mathrm{2}\:\leqslant\:{x},\:{y},\:{z}\:\leqslant\:\mathrm{3}…
Question Number 21316 by Tinkutara last updated on 20/Sep/17 $$\mathrm{Let}\:{p}\:=\:\left({x}_{\mathrm{1}} \:−\:{x}_{\mathrm{2}} \right)^{\mathrm{2}} \:+\:\left({x}_{\mathrm{1}} \:−\:{x}_{\mathrm{3}} \right)^{\mathrm{2}} \:+\:….\:+ \\ $$$$\left({x}_{\mathrm{1}} \:−\:{x}_{\mathrm{6}} \right)^{\mathrm{2}} \:+\:\left({x}_{\mathrm{2}} \:−\:{x}_{\mathrm{3}} \right)^{\mathrm{2}} \:+\:\left({x}_{\mathrm{2}} \:−\:{x}_{\mathrm{4}}…