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Category: Algebra

x-1-x-2-x-2-x-3-x-5-x-6-x-6-x-7-solve-for-x-

Question Number 46713 by agniv last updated on 30/Oct/18 $$\frac{{x}−\mathrm{1}}{{x}−\mathrm{2}}−\frac{{x}−\mathrm{2}}{{x}−\mathrm{3}}=\frac{{x}−\mathrm{5}}{{x}−\mathrm{6}}−\frac{{x}−\mathrm{6}}{{x}−\mathrm{7}} \\ $$$$\boldsymbol{{solve}}\:\boldsymbol{{for}}\:\boldsymbol{{x}} \\ $$ Answered by Kunal12588 last updated on 30/Oct/18 $$\frac{\left({x}−\mathrm{1}\right)\left({x}−\mathrm{3}\right)−\left({x}−\mathrm{2}\right)^{\mathrm{2}} }{\left({x}−\mathrm{2}\right)\left({x}−\mathrm{3}\right)}=\frac{\left({x}−\mathrm{5}\right)\left({x}−\mathrm{7}\right)−\left({x}−\mathrm{6}\right)^{\mathrm{2}} }{\left({x}−\mathrm{6}\right)\left({x}−\mathrm{7}\right)} \\…

Question-46680

Question Number 46680 by Sanjarbek last updated on 30/Oct/18 Answered by MJS last updated on 30/Oct/18 $$\mathrm{we}\:\mathrm{had}\:\mathrm{this}\:\mathrm{before}.\:\mathrm{infinite}\:\mathrm{continued}\:\mathrm{fractions} \\ $$$$\mathrm{without}\:\mathrm{cyclic}\:\mathrm{patterns}\:\mathrm{represent}\:\mathrm{transcendental} \\ $$$$\mathrm{numbers}.\:\mathrm{we}\:\mathrm{can}\:\mathrm{only}\:\mathrm{approximate} \\ $$ Terms of…

Question-177741

Question Number 177741 by infinityaction last updated on 08/Oct/22 Answered by Frix last updated on 08/Oct/22 $$\mathrm{let}\:{p}={a}_{{n}} \wedge{q}={a}_{{n}+\mathrm{1}} \\ $$$${q}^{\mathrm{2}} −\mathrm{2}{pq}−{p}=\mathrm{0}\:\Rightarrow\:{q}={p}+\sqrt{{p}\left({p}+\mathrm{1}\right)} \\ $$$$\Rightarrow\:{q}>{p}\:\Leftrightarrow\:{a}_{{n}+\mathrm{1}} >{a}_{{n}} \\…

In-ABC-36-R-r-m-b-2-m-c-2-sin-B-sin-C-9-R-4-2-r-2-

Question Number 177698 by Shrinava last updated on 08/Oct/22 $$\mathrm{In}\:\:\bigtriangleup\mathrm{ABC} \\ $$$$\mathrm{36}\:\mathrm{R}\:\mathrm{r}\:\leqslant\:\Sigma\:\frac{\mathrm{m}_{\boldsymbol{\mathrm{b}}} ^{\mathrm{2}} \:+\:\mathrm{m}_{\boldsymbol{\mathrm{c}}} ^{\mathrm{2}} }{\mathrm{sin}\:\mathrm{B}\:\mathrm{sin}\:\mathrm{C}}\:\leqslant\:\frac{\mathrm{9}\:\mathrm{R}^{\mathrm{4}} }{\mathrm{2}\:\mathrm{r}^{\mathrm{2}} } \\ $$ Terms of Service Privacy Policy…

fog-1-3x-2-gof-x-2x-1-fof-3-

Question Number 177688 by mathlove last updated on 08/Oct/22 $${fog}^{−\mathrm{1}} =\mathrm{3}{x}+\mathrm{2} \\ $$$$\left({gof}\right)_{{x}} =\mathrm{2}{x}−\mathrm{1}\:\:\:\:\:\overset{\left.\begin{matrix}{}\\{}\end{matrix}\right\}\:\:\left({fof}\right)_{\mathrm{3}} =?} {\:} \\ $$ Commented by cortano1 last updated on 08/Oct/22…