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Category: Algebra

6-2-2-1-

Question Number 16840 by Sai dadon. last updated on 27/Jun/17 $$\mathrm{6}/\mathrm{2}\left(\mathrm{2}+\mathrm{1}\right) \\ $$ Commented by RasheedSoomro last updated on 27/Jun/17 $$\mathrm{6}/\mathrm{2}\left(\mathrm{2}+\mathrm{1}\right) \\ $$$$=\mathrm{6}/\left\{\mathrm{2}\left(\mathrm{2}+\mathrm{1}\right)\right\} \\ $$$$=\mathrm{6}/\left(\mathrm{2}×\mathrm{3}\right)…

Question-16835

Question Number 16835 by tawa tawa last updated on 26/Jun/17 Answered by sandy_suhendra last updated on 27/Jun/17 $$\left.\mathrm{1}\right) \\ $$$$\mathrm{5C3}×\mathrm{5C3}×\mathrm{4C2} \\ $$$$=\:\mathrm{10}×\mathrm{10}×\mathrm{6} \\ $$$$=\:\mathrm{600}\:\mathrm{ways} \\…

Solve-for-x-6-4-x-9-x-13-6-x-

Question Number 16829 by tawa tawa last updated on 26/Jun/17 $$\mathrm{Solve}\:\mathrm{for}\:\mathrm{x} \\ $$$$\mathrm{6}\left(\mathrm{4}^{\mathrm{x}} \:+\:\mathrm{9}^{\mathrm{x}} \right)\:=\:\left(\mathrm{13}.\mathrm{6}\right)^{\mathrm{x}} \\ $$ Commented by mrW1 last updated on 26/Jun/17 $$\mathrm{please}\:\mathrm{check}\:\mathrm{the}\:\mathrm{question}.…

x-1-1-x-4-x-1-1-x-

Question Number 147893 by mathdanisur last updated on 24/Jul/21 $${x}\:+\:\mathrm{1}\:+\:\frac{\mathrm{1}}{{x}}\:=\:\mathrm{4}\:\:\Rightarrow\:\:{x}\:+\:\mathrm{1}\:-\:\frac{\mathrm{1}}{{x}}\:=\:? \\ $$ Commented by aliibrahim1 last updated on 24/Jul/21 $${x}+\frac{\mathrm{1}}{{x}}=\mathrm{3}\Rrightarrow\:\:\:\:\:\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} =\mathrm{3}^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }+\mathrm{2}=\mathrm{9}…

if-x-y-z-gt-1-then-x-1-y-1-z-1-x-1-y-1-z-1-lt-xyz-8-

Question Number 147892 by mathdanisur last updated on 24/Jul/21 $${if}\:\:\:{x};{y};{z}>\mathrm{1}\:\:\:{then}: \\ $$$$\sqrt{\frac{\left({x}−\mathrm{1}\right)\left({y}−\mathrm{1}\right)\left({z}−\mathrm{1}\right)}{\left({x}+\mathrm{1}\right)\left({y}+\mathrm{1}\right)\left({z}+\mathrm{1}\right)}\:}\:<\:\frac{{xyz}}{\mathrm{8}} \\ $$ Answered by mindispower last updated on 24/Jul/21 $$\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}<\frac{{x}^{\mathrm{2}} }{\mathrm{4}}…?\:\forall{x}>\mathrm{1} \\ $$$$\Leftrightarrow\mathrm{4}\left({x}−\mathrm{1}\right)<{x}^{\mathrm{2}}…

x-2-3x-4-0-x-2-16-x-2-

Question Number 147894 by mathdanisur last updated on 24/Jul/21 $${x}^{\mathrm{2}} \:-\:\mathrm{3}{x}\:+\:\mathrm{4}\:=\:\mathrm{0}\:\:\Rightarrow\:\:{x}^{\mathrm{2}} \:+\:\frac{\mathrm{16}}{{x}^{\mathrm{2}} }\:=\:? \\ $$ Answered by mindispower last updated on 24/Jul/21 $$\Leftrightarrow{x}+\frac{\mathrm{4}}{{x}}=\mathrm{3} \\ $$$$\Rightarrow{x}^{\mathrm{2}}…

Find-the-sum-to-close-form-1-1-2-3-1-4-5-6-1-7-8-9-

Question Number 147877 by Tawa11 last updated on 24/Jul/21 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{to}\:\mathrm{close}\:\mathrm{form}. \\ $$$$\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{1}.\mathrm{2}.\mathrm{3}}\:\:+\:\:\frac{\mathrm{1}}{\mathrm{4}.\mathrm{5}.\mathrm{6}}\:\:\:+\:\:\frac{\mathrm{1}}{\mathrm{7}.\mathrm{8}.\mathrm{9}}\:\:\:+\:\:\:…\:\:\: \\ $$ Commented by Tinku Tara last updated on 24/Jul/21 $$\mathrm{Welcome}\:\mathrm{back}.\:\mathrm{Tawa}. \\ $$🙂🙂…