Question Number 110173 by ajfour last updated on 27/Aug/20 $${If}\:\:\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}{t}_{{r}} =\frac{{n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)\left({n}+\mathrm{3}\right)}{\mathrm{8}} \\ $$$${then}\:\:\:\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\:\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\:\frac{\mathrm{1}}{{t}_{{r}} }\:=\:? \\ $$$$ \\ $$ Answered by…
Question Number 110156 by I want to learn more last updated on 27/Aug/20 $$\mathrm{If}\:\mathrm{we}\:\mathrm{have}\:\:\mathrm{5}\:\mathrm{people},\:\mathrm{how}\:\mathrm{many}\:\mathrm{ways}\:\mathrm{can}\:\mathrm{they}\:\mathrm{be}\:\mathrm{seated}\:\mathrm{in}\:\mathrm{a}\:\mathrm{row} \\ $$$$\mathrm{on}\:\mathrm{a}\:\mathrm{chair},\:\mathrm{if}\:\mathrm{their}\:\mathrm{are}, \\ $$$$\left(\mathrm{a}\right)\:\:\:\mathrm{7}\:\:\mathrm{chairs}\:\:\:\:\:\:\:\left(\mathrm{b}\right)\:\:\:\:\mathrm{3}\:\:\mathrm{chairs}\:\:\:\:\:\:\:\:\mathrm{available} \\ $$ Answered by mr W last…
Question Number 110157 by I want to learn more last updated on 27/Aug/20 $$\mathrm{If}\:\mathrm{we}\:\mathrm{have}\:\:\mathrm{5}\:\:\mathrm{people},\:\mathrm{how}\:\mathrm{many}\:\mathrm{ways}\:\mathrm{can}\:\mathrm{they}\:\mathrm{be}\:\mathrm{seated} \\ $$$$\mathrm{on}\:\mathrm{a}\:\mathrm{round}\:\mathrm{table},\:\mathrm{if}\:\mathrm{there}\:\mathrm{are}, \\ $$$$\left(\mathrm{a}\right)\:\:\:\mathrm{7}\:\:\mathrm{chairs}\:\:\:\:\:\:\:\:\:\left(\mathrm{b}\right)\:\:\:\:\mathrm{3}\:\:\mathrm{chairs}\:\:\:\:\:\:\:\:\:\:\:\mathrm{available} \\ $$ Answered by bemath last updated…
Question Number 175685 by Linton last updated on 05/Sep/22 $$\mathrm{3}^{\mathrm{5}^{{x}} } =\:\mathrm{5}^{\mathrm{3}^{{x}} } \\ $$$${solve}\:{for}\:{x} \\ $$ Answered by mahdipoor last updated on 05/Sep/22 $${ln}\left({ln}\left(\mathrm{3}^{\mathrm{5}^{{x}}…
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Question Number 110136 by qwerty111 last updated on 27/Aug/20 Answered by $@y@m last updated on 27/Aug/20 $${Let}\:\mathrm{2020}^{{m}} ={a}\: \\ $$$${Given}, \\ $$$${a}+\frac{\mathrm{1}}{{a}}=\mathrm{3}\:…..\left(\mathrm{1}\right) \\ $$$${Squarring}\:{both}\:{sides}, \\…
Question Number 175653 by mnjuly1970 last updated on 04/Sep/22 $$ \\ $$$$\:\:\:{Q}:\:\:{prove}\:{that}\:{the}\:{following} \\ $$$$\:\:\:\:\:{equation}\:{has}\:{no}\:{solution}. \\ $$$$ \\ $$$$\:\:\:\sqrt{{x}\:+\lfloor\:{x}\:\rfloor}\:+\:\sqrt{{x}\:−\sqrt{{x}}\:}\:=\:\mathrm{1} \\ $$$$ \\ $$ Answered by mahdipoor…
Question Number 44570 by deepakkumardas last updated on 01/Oct/18 Answered by tanmay.chaudhury50@gmail.com last updated on 02/Oct/18 Commented by tanmay.chaudhury50@gmail.com last updated on 02/Oct/18 Commented by…
Question Number 175639 by Linton last updated on 04/Sep/22 $${x}^{\mathrm{99}} +{y}^{\mathrm{99}} =\:{x}^{\mathrm{100}} \\ $$$${Interger}\:{solutions}?\: \\ $$ Answered by BaliramKumar last updated on 04/Sep/22 $${x}^{\mathrm{99}} \left[\mathrm{1}\:+\:\left(\frac{{y}}{{x}}\right)^{\mathrm{99}}…
Question Number 110087 by 1549442205PVT last updated on 27/Aug/20 $$\mathrm{Solve}\:\mathrm{the}\:\mathrm{system}\:\mathrm{following}\:\mathrm{of}\:\mathrm{equations} \\ $$$$\begin{cases}{\mathrm{x}+\mathrm{y}+\mathrm{z}=\mathrm{2}}\\{\mathrm{2x}+\mathrm{3y}+\mathrm{z}=\mathrm{1}}\\{\mathrm{x}^{\mathrm{2}} +\left(\mathrm{y}+\mathrm{2}\right)^{\mathrm{2}} +\left(\mathrm{z}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{9}}\end{cases} \\ $$ Commented by bemath last updated on 27/Aug/20 $$\:\:\:\left[\bigtriangleup\frac{{be}}{{math}}\bigtriangledown\right]…