Question Number 121631 by Study last updated on 10/Nov/20 Commented by MJS_new last updated on 10/Nov/20 $$\mathrm{see}\:\mathrm{question}\:\mathrm{121612} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 56092 by Mikael_Marshall last updated on 10/Mar/19 $${How}\:{to}\:{rationalize}\:{a}\:{denominator}\:{in} \\ $$$${a}\:{fraction}?\:{Like}\:{this}. \\ $$$$\frac{\sqrt[{\mathrm{3}}]{{x}}+\mathrm{2}}{\:\sqrt[{\mathrm{3}}]{{x}}−\mathrm{2}}\: \\ $$ Answered by math1967 last updated on 10/Mar/19 $$\frac{\left(^{\mathrm{3}} \sqrt{{x}}+\mathrm{2}\right)\left({x}^{\frac{\mathrm{2}}{\mathrm{3}}}…
Question Number 187166 by ajfour last updated on 14/Feb/23 $${x}^{\mathrm{3}} ={x}+{c} \\ $$$${let}\:\:{x}=\frac{{mt}}{\mathrm{1}−{t}} \\ $$$${m}^{\mathrm{3}} {t}^{\mathrm{3}} ={mt}\left(\mathrm{1}−{t}\right)^{\mathrm{2}} +{c}\left(\mathrm{1}−{t}\right)^{\mathrm{3}} \\ $$$$\Rightarrow \\ $$$$\left({m}^{\mathrm{3}} −{m}−{c}\right){t}^{\mathrm{3}} +\left(\mathrm{2}{m}−\mathrm{3}{c}\right){t}^{\mathrm{2}} \\…
Question Number 187124 by Rasheed.Sindhi last updated on 13/Feb/23 $$\frac{{a}}{{x}}=\frac{{b}}{{y}}=\frac{{c}}{{z}}=\frac{\mathrm{1}}{\mathrm{3}}\:, \\ $$$${a}−\mathrm{2}{b}+{c}=\mathrm{2}\:\:{and}\:\:−\mathrm{2}{y}+{z}=\mathrm{1}\:\:\:\: \\ $$$${x}=? \\ $$$${An}\:{altered}\:{form}\:{of}\:\:{q}#\mathrm{187020} \\ $$$$\left({In}\:{this}\:{case}\:{solveable}\right) \\ $$ Answered by HeferH last updated…
Question Number 121590 by pooooop last updated on 09/Nov/20 $$\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{2}\boldsymbol{{x}}=\mathrm{8}\sqrt{\mathrm{6}} \\ $$$$\mathrm{2}\boldsymbol{{x}}−\frac{\mathrm{12}}{\boldsymbol{{x}}}=? \\ $$ Answered by mathmax by abdo last updated on 09/Nov/20 $$\mathrm{x}^{\mathrm{2}}…
Question Number 56052 by mr W last updated on 09/Mar/19 $${Solve}\:{following}\:{equation}\:{for}\:{x}: \\ $$$$\boldsymbol{{x}}^{\boldsymbol{{x}}^{\boldsymbol{{x}}^{\boldsymbol{{n}}} } } =\boldsymbol{{n}}\:{with}\:{n}\in\mathbb{N} \\ $$ Answered by MJS last updated on 09/Mar/19…
Question Number 56042 by Tawa1 last updated on 09/Mar/19 Answered by math1967 last updated on 09/Mar/19 $$\overset{{n}} {{c}}_{{r}} =\frac{{n}!}{{r}!\left({n}−{r}\right)!}×\frac{\left({n}−{r}+\mathrm{1}\right)}{\left({n}−{r}+\mathrm{1}\right)}=\frac{{n}!×\left({n}−{r}+\mathrm{1}\right)}{{r}×\left({r}−\mathrm{1}\right)!\left({n}−{r}+\mathrm{1}\right)!}\: \\ $$$$=\frac{{n}!}{\left({r}−\mathrm{1}\right)!\left({n}−{r}+\mathrm{1}\right)!}×\left[\frac{{n}−{r}+\mathrm{1}}{{r}}\right] \\ $$$$=\overset{{n}} {\:}{c}_{{r}−\mathrm{1}} \left[\frac{\left({n}−{r}+\mathrm{1}\right)}{{r}}\right]…
Question Number 121552 by I want to learn more last updated on 09/Nov/20 Commented by mr W last updated on 10/Nov/20 $$\mathrm{4}\:{ways}: \\ $$$$\mathrm{8}+\mathrm{7}+\mathrm{2}+\mathrm{1}=\mathrm{6}+\mathrm{5}+\mathrm{4}+\mathrm{3} \\…
Question Number 121538 by Khalmohmmad last updated on 09/Nov/20 Answered by benjo_mathlover last updated on 09/Nov/20 $$\left(\mathrm{i}\right)\:\mid\mathrm{2x}^{\mathrm{2}} +\mathrm{3x}−\mathrm{2}\mid\:=\:\mid\left(\mathrm{2x}−\mathrm{1}\right)\left(\mathrm{x}+\mathrm{2}\right)\mid \\ $$$$\:\:\:\rightarrow\begin{cases}{\mathrm{2x}^{\mathrm{2}} +\mathrm{3x}−\mathrm{2}\:;\:\mathrm{if}\:\mathrm{x}\leqslant−\mathrm{2}\:\cup\:\mathrm{x}\geqslant\frac{\mathrm{1}}{\mathrm{2}}}\\{−\mathrm{2x}^{\mathrm{2}} −\mathrm{3x}+\mathrm{2}\:;\:\mathrm{if}\:−\mathrm{2}\leqslant\mathrm{x}\leqslant\frac{\mathrm{1}}{\mathrm{2}}}\end{cases} \\ $$$$\left(\mathrm{ii}\right)\:\mid\mathrm{x}−\mathrm{2}\mid\:\rightarrow\begin{cases}{\mathrm{x}−\mathrm{2}\:;\:\mathrm{if}\:\mathrm{x}>\mathrm{2}}\\{−\mathrm{x}+\mathrm{2}\:;\:\mathrm{if}\:\mathrm{x}<\mathrm{2}}\end{cases} \\…
Question Number 55969 by Muhammad bello last updated on 07/Mar/19 $$\:\:\:\boldsymbol{\mathrm{solve}}\:\:\:\boldsymbol{\mathrm{for}}\:\:\:\boldsymbol{\mathrm{x}}\:\:\:\&\:\:\boldsymbol{\mathrm{y}} \\ $$$$\:\:\:\: \\ $$$$\:\:\:\:\boldsymbol{\mathrm{x}}^{\boldsymbol{\mathrm{logy}}} \:\:\:=\:\:\mathrm{4} \\ $$$$\:\: \\ $$$$\:\:\:\:\boldsymbol{\mathrm{xy}}\:\:\:\:=\:\:\:\:\mathrm{40}\: \\ $$ Answered by MJS…