Question Number 213639 by efronzo1 last updated on 12/Nov/24 Answered by golsendro last updated on 12/Nov/24 $$\:\:\:\mathrm{8}^{\mathrm{x}} =\mathrm{25}^{\mathrm{y}} =\:\mathrm{t}\:\Rightarrow\begin{cases}{\mathrm{x}=\:\mathrm{log}\:_{\mathrm{8}} \mathrm{t}}\\{\mathrm{y}=\:\mathrm{log}\:_{\mathrm{25}} \mathrm{t}}\end{cases} \\ $$$$\:\:\Rightarrow\mathrm{8}\left(\frac{\mathrm{1}}{\mathrm{x}}\right)+\mathrm{25}\left(\frac{\mathrm{1}}{\mathrm{y}}\right)=\:\mathrm{1} \\ $$$$\:\:\:\mathrm{8}.\:\mathrm{log}\:_{\mathrm{t}}…
Question Number 213636 by hardmath last updated on 11/Nov/24 $$\mathrm{x}^{\mathrm{2}} \:−\:\mathrm{x}\:+\:\mathrm{1}\:=\:\mathrm{0} \\ $$$$\mathrm{find}:\:\:\:\mathrm{x}^{\mathrm{10}} \:=\:? \\ $$ Answered by A5T last updated on 11/Nov/24 $${x}^{\mathrm{2}} ={x}−\mathrm{1}\Rightarrow{x}^{\mathrm{3}}…
Question Number 213624 by hardmath last updated on 10/Nov/24 $$\mathrm{x}^{\mathrm{3}} \:+\:\mathrm{5x}\:−\:\mathrm{42}\:=\:\left(\mathrm{x}\:−\:\mathrm{3}\right)\centerdot\mathrm{P}\left(\mathrm{x}\right) \\ $$$$\mathrm{Find}:\:\:\:\mathrm{P}\left(\mathrm{3}\right)\:=\:? \\ $$ Answered by A5T last updated on 10/Nov/24 $$\left({x}−\mathrm{3}\right)\left({x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{14}\right)−\left({x}−\mathrm{3}\right){P}\left({x}\right)=\mathrm{0} \\…
Question Number 213621 by hardmath last updated on 10/Nov/24 $$\mathrm{a},\mathrm{b}>\mathrm{0} \\ $$$$\frac{\mathrm{1}}{\mathrm{a}}\:+\:\frac{\mathrm{1}}{\mathrm{b}}\:=\:\mathrm{2}\:\:\:\mathrm{and}\:\:\:\mathrm{a}^{\mathrm{2}} \:+\:\mathrm{b}^{\mathrm{2}} \:=\:\mathrm{12} \\ $$$$\mathrm{a}\:+\:\mathrm{b}\:=\:? \\ $$ Answered by Ghisom last updated on 10/Nov/24…
Question Number 213589 by hardmath last updated on 09/Nov/24 $$\mathrm{Find}: \\ $$$$\frac{\left(\mathrm{3}\:-\:\frac{\mathrm{3}}{\mathrm{4}}\right)\centerdot\left(\mathrm{3}\:-\:\frac{\mathrm{3}}{\mathrm{5}}\right)\centerdot\left(\mathrm{3}\:-\:\frac{\mathrm{1}}{\mathrm{2}}\right)\centerdot\left(\mathrm{3}\:-\:\frac{\mathrm{3}}{\mathrm{7}}\right)\centerdot…\centerdot\left(\mathrm{3}\:-\:\frac{\mathrm{1}}{\mathrm{6}}\right)}{\mathrm{27}^{\mathrm{5}} }\:=\:? \\ $$ Answered by issac last updated on 09/Nov/24 $$\left(\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{15}} \underset{{h}=\mathrm{1}} {\overset{\mathrm{15}}…
Question Number 213535 by justenspi last updated on 07/Nov/24 $$\mathrm{Are}\:\mathrm{not}\:\mathrm{there}\:\mathrm{simple}\:\mathrm{means}\:\mathrm{of}\:\mathrm{solving}\:\mathrm{this} \\ $$$$\mathrm{system}\:\mathrm{of}\:\mathrm{equations}\:\mathrm{in}\:\mathrm{non}-\mathrm{negative}\:\mathrm{reals} \\ $$ Commented by justenspi last updated on 07/Nov/24 Terms of Service Privacy…
Question Number 213530 by Ari last updated on 07/Nov/24 Commented by Ari last updated on 07/Nov/24 $${C}\left({x},{y}\right)=\mathrm{4}{x}+\mathrm{6}{y} \\ $$$${where}\:{x}−{one}\:{type}\:{of}\:{tables} \\ $$$${y}−{other}\:{type} \\ $$$$ \\ $$…
Question Number 213467 by efronzo1 last updated on 06/Nov/24 $$\:\:\:\underbrace{\boldsymbol{{B}}} \\ $$ Commented by mr W last updated on 06/Nov/24 �� Commented by mr W…
Question Number 213463 by golsendro last updated on 06/Nov/24 $$\:\:\mathrm{For}\:\mathrm{p},\mathrm{q}\:\mathrm{and}\:\mathrm{r}\:\mathrm{prime}\:\mathrm{numbers}\: \\ $$$$\:\:\mathrm{satisfying}\:\begin{cases}{\mathrm{p}\left(\mathrm{q}+\mathrm{1}\right)\left(\mathrm{r}+\mathrm{1}\right)=\mathrm{1064}}\\{\mathrm{r}\left(\mathrm{p}+\mathrm{1}\right)\left(\mathrm{q}+\mathrm{1}\right)=\mathrm{1554}}\end{cases} \\ $$$$\:\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{p}\left(\mathrm{q}+\mathrm{1}\right)\mathrm{r}\: \\ $$ Answered by A5T last updated on 06/Nov/24 $${p}\mid\mathrm{1064}=\mathrm{2}^{\mathrm{3}} ×\mathrm{7}×\mathrm{19};\:{p}+\mathrm{1}\mid\mathrm{1554}=\mathrm{2}×\mathrm{3}×\mathrm{7}×\mathrm{37}…
Question Number 213459 by golsendro last updated on 06/Nov/24 $$\:\:\mathrm{Find}\:\mathrm{tupple}\:\mathrm{natural}\:\mathrm{numbers}\:\left(\mathrm{a},\mathrm{b},\mathrm{c}\right) \\ $$$$\:\mathrm{such}\:\mathrm{that}\: \\ $$$$\:\:\:\:\begin{cases}{\mathrm{max}\left\{\frac{\mathrm{a}+\mathrm{b}}{\mathrm{2}}+\frac{\mid\mathrm{a}−\mathrm{b}\mid}{\mathrm{2}}\:,\:\frac{\mathrm{b}+\mathrm{c}}{\mathrm{2}}+\frac{\mid\mathrm{b}−\mathrm{c}\mid}{\mathrm{2}}\:,\frac{\mathrm{c}+\mathrm{a}}{\mathrm{2}}+\frac{\mid\mathrm{c}−\mathrm{a}\mid}{\mathrm{2}}\right\}=\mathrm{a}}\\{\mathrm{min}\left\{\frac{\mathrm{a}+\mathrm{b}}{\mathrm{2}}−\frac{\mid\mathrm{a}−\mathrm{b}\mid}{\mathrm{2}}\:,\:\frac{\mathrm{b}+\mathrm{c}}{\mathrm{2}}−\frac{\mid\mathrm{b}−\mathrm{c}\mid}{\mathrm{2}}\:,\:\frac{\mathrm{c}+\mathrm{a}}{\mathrm{2}}−\frac{\mid\mathrm{c}−\mathrm{a}\mid}{\mathrm{2}}\right\}=\mathrm{b}}\end{cases} \\ $$$$\:\:\mathrm{where}\:\mathrm{a}+\mathrm{b}+\mathrm{c}\:=\:\mathrm{10} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com