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…
Question Number 55954 by Easyman32 last updated on 06/Mar/19 Answered by MJS last updated on 06/Mar/19 $$\mathrm{draw}\:\mathrm{them}:\:\mathrm{2}\:\mathrm{half}\:\mathrm{parabolas}\:\Rightarrow \\ $$$$\Rightarrow\:\mathrm{1}\:\mathrm{intersection} \\ $$$${x}=\mathrm{11}.\mathrm{932};\:{y}=\mathrm{16}.\mathrm{546} \\ $$ Commented by…
Question Number 55942 by Tawa1 last updated on 06/Mar/19 $$\mathrm{Prove}\:\mathrm{that}:\:\:\:\:\:\frac{\mathrm{1}}{\mid\boldsymbol{\mathrm{A}}\mid}\:\:=\:\:\boldsymbol{\mathrm{A}}^{\mathrm{1}} \\ $$ Commented by 121194 last updated on 06/Mar/19 $$\mathrm{what}\:\mathrm{A}? \\ $$ Terms of Service…
Question Number 187008 by mustafazaheen last updated on 12/Feb/23 $$\frac{{a}}{\mathrm{3}}=\frac{{b}}{\mathrm{4}}=\frac{{c}}{\mathrm{5}}\:\:\:\:\:\:\:\:\:\:\:\mathrm{3}{a}+{c}=\mathrm{42}\:\:\:\:\:\:\:\:{b}=? \\ $$$${how}\:{is}\:{solution} \\ $$ Answered by HeferH last updated on 12/Feb/23 $$\frac{{a}}{\mathrm{3}}\:=\:\frac{{c}}{\mathrm{5}} \\ $$$$\:\frac{{a}}{{c}}\:=\:\frac{\mathrm{3}}{\mathrm{5}}\:;\:{a}\:=\:\mathrm{3}{k}\:,\:{c}\:=\:\mathrm{5}{k}\: \\…
Question Number 121446 by mr W last updated on 08/Nov/20 Answered by liberty last updated on 08/Nov/20 $$\mathrm{u}\:=\:\mathrm{x}+\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} } \\ $$$$\mathrm{u}\left(\mathrm{y}+\sqrt{\mathrm{1}+\mathrm{y}^{\mathrm{2}} }\:\right)\:=\:\mathrm{1}\:\Rightarrow\sqrt{\mathrm{1}+\mathrm{y}^{\mathrm{2}} }\:=\:\frac{\mathrm{1}}{\mathrm{u}}−\mathrm{y} \\ $$$$\Rightarrow\mathrm{1}+\mathrm{y}^{\mathrm{2}}…