Question Number 203275 by Frix last updated on 13/Jan/24 $$\mathrm{Show}\:\mathrm{this}\:\mathrm{has}\:\mathrm{exactly}\:\mathrm{7}\:\mathrm{solutions}\:\mathrm{for}\:{x}\in\mathbb{C}: \\ $$$${x}^{\mathrm{ln}\:{x}} =\mathrm{1} \\ $$ Commented by aleks041103 last updated on 14/Jan/24 $${When}\:{you}\:{use}\:{ln}\left({x}\right)\:{do}\:{you}\:{imply}\:{the}\:{principle} \\ $$$${branch}\:{of}\:{this}\:{function}?…
Question Number 203264 by hardmath last updated on 13/Jan/24 $$\mathrm{If}\:\:\:\:\:\mathrm{x}\:=\:\frac{\mathrm{1}\:+\:\sqrt{\mathrm{5}}}{\mathrm{2}}\:\:\:\:\:\mathrm{find}:\:\mathrm{x}^{\mathrm{12}} \:=\:? \\ $$ Answered by MM42 last updated on 13/Jan/24 $$\begin{cases}{{x}^{\mathrm{2}} =\frac{\mathrm{3}+\sqrt{\mathrm{5}}}{\mathrm{2}}}\\{{x}^{\mathrm{4}} =\frac{\mathrm{7}+\mathrm{3}\sqrt{\mathrm{5}}}{\mathrm{2}}}\end{cases}\:\:\Rightarrow{x}^{\mathrm{6}} =\mathrm{9}+\mathrm{4}\sqrt{\mathrm{5}} \\…
Question Number 203222 by Tawa11 last updated on 12/Jan/24 XYZ Ltd wants to borrow money to finance a project. They can pay N2500 monthly in…
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Question Number 203219 by hardmath last updated on 12/Jan/24 Answered by witcher3 last updated on 15/Jan/24 $$\mathrm{evident} \\ $$$$\left(\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d}\right)^{\mathrm{2}} =\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} +\mathrm{d}^{\mathrm{2}} +\mathrm{2}\left(\mathrm{ab}+\mathrm{bc}+\mathrm{cd}+\mathrm{da}\right) \\…
Question Number 203208 by hardmath last updated on 12/Jan/24 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 203206 by hardmath last updated on 12/Jan/24 Answered by mr W last updated on 12/Jan/24 $$\underset{{k}=\mathrm{1}} {\overset{\mathrm{100}} {\sum}}\mid{i}−{k}\mid \\ $$$$=\underset{{k}=\mathrm{1}} {\overset{{i}} {\sum}}\mid{i}−{k}\mid+\underset{{k}={i}+\mathrm{1}} {\overset{\mathrm{100}}…
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Question Number 203202 by Abdullahrussell last updated on 12/Jan/24 Answered by cortano12 last updated on 12/Jan/24 $$\:\:\:\underbrace{\leq} \\ $$ Answered by mr W last updated…
Question Number 203157 by navin12345 last updated on 11/Jan/24 Answered by witcher3 last updated on 11/Jan/24 $$\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} +\mathrm{z}^{\mathrm{2}} −\mathrm{2x}−\mathrm{2z}+\mathrm{2}=\mathrm{0} \\ $$$$\left(\mathrm{x}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} +\left(\mathrm{z}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{0}…