Question Number 108825 by ajfour last updated on 19/Aug/20 $$\underset{{n}=\mathrm{1}} {\overset{\mathrm{10}} {\sum}}\left({i}^{{n}} +{i}^{{n}+\mathrm{1}} \right)=\:? \\ $$ Commented by ajfour last updated on 19/Aug/20 $${thanks}\:{everyone}\:{who}\:{confirmed}. \\…
Question Number 174356 by behi834171 last updated on 30/Jul/22 $$\:\:\:\:\:\:\:\boldsymbol{{x}}^{\mathrm{3}} +\mathrm{4}\boldsymbol{{x}}^{\mathrm{2}} −\mathrm{9}\boldsymbol{{x}}−\mathrm{14}=\mathrm{0} \\ $$$$\:\:\:\:\:\mathrm{1}.\:\frac{\mathrm{1}}{\boldsymbol{{x}}_{\mathrm{1}} ^{\mathrm{2}} }+\frac{\mathrm{1}}{\boldsymbol{{x}}_{\mathrm{2}} ^{\mathrm{2}} }+\frac{\mathrm{1}}{\boldsymbol{{x}}_{\mathrm{3}} ^{\mathrm{2}} }=? \\ $$$$\:\:\:\:\:\mathrm{2}.\:\Sigma\:\frac{\boldsymbol{{x}}_{\mathrm{1}} ^{\mathrm{2}} +\boldsymbol{{x}}_{\mathrm{2}} ^{\mathrm{2}}…
Question Number 174350 by AbdulAzizabir last updated on 30/Jul/22 Commented by AbdulAzizabir last updated on 30/Jul/22 $$ \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 108813 by Khalmohmmad last updated on 19/Aug/20 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(−\mathrm{ln}\:\mathrm{2}.\mathrm{1}\right)^{\mathrm{2x}} =? \\ $$$$?????? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 108802 by 150505R last updated on 19/Aug/20 Answered by mr W last updated on 19/Aug/20 $$\mathrm{tan}^{−\mathrm{1}} \:\left(\frac{{i}}{{j}}\right)+\mathrm{tan}^{−\mathrm{1}} \left(\frac{{j}}{{i}}\right)=\frac{\pi}{\mathrm{2}} \\ $$$$\mathrm{tan}^{−\mathrm{1}} \:\left(\frac{{i}}{{i}}\right)=\frac{\pi}{\mathrm{4}} \\ $$$$\underset{{i}=\mathrm{1}}…
Question Number 174324 by BaliramKumar last updated on 29/Jul/22 $$\mathrm{F}{actorize}:− \\ $$$$\mathrm{2}{x}^{\mathrm{2}} {y}^{\mathrm{2}} \:+\:\mathrm{2}{y}^{\mathrm{2}} {z}^{\mathrm{2}} \:+\:\mathrm{2}{z}^{\mathrm{2}} {x}^{\mathrm{2}} \:−\:{x}^{\mathrm{4}} \:−\:{y}^{\mathrm{4}} \:−\:{z}^{\mathrm{4}} \\ $$ Commented by kaivan.ahmadi…
Question Number 108781 by 150505R last updated on 19/Aug/20 Answered by 1549442205PVT last updated on 19/Aug/20 $$\mathrm{P}=\left(\mathrm{1}!×\mathrm{1}\right)+\left(\mathrm{2}!×\mathrm{2}\right)+\left(\mathrm{3}!×\mathrm{3}\right)+…+\left(\mathrm{286}!×\mathrm{286}\right) \\ $$$$=\left[\mathrm{1}!×\left(\mathrm{2}−\mathrm{1}\right)\right]+\left[\mathrm{2}!×\left(\mathrm{3}−\mathrm{1}\right)\right]+\left[\mathrm{3}!×\left(\mathrm{4}−\mathrm{1}\right)\right]+\left(\mathrm{4}!×\left(\mathrm{5}−\mathrm{1}\right)\right]+ \\ $$$$…+\left(\mathrm{286}!×\left(\mathrm{287}−\mathrm{1}\right)\right] \\ $$$$=−\mathrm{1}!+\mathrm{2}!−\mathrm{2}!+\mathrm{3}!−\mathrm{3}!+\mathrm{4}!−\mathrm{4}!+\mathrm{5}!−… \\ $$$$−\mathrm{286}!+\mathrm{287}!=\mathrm{287}!…
Question Number 174304 by dragan91 last updated on 29/Jul/22 Answered by mr W last updated on 29/Jul/22 $${x}^{\mathrm{5}} −\mathrm{2}{x}^{\mathrm{4}} −\mathrm{1}=\mathrm{0} \\ $$$$\Sigma{r}=\mathrm{2} \\ $$$$\Sigma{r}_{{i}} {r}_{{j}}…
Question Number 108769 by bemath last updated on 19/Aug/20 $$\:\:\:\frac{\iddots\mathcal{B}{e}\mathcal{M}{ath}\ddots}{\bigstar} \\ $$$$\:\mathrm{G}{iven}\:\frac{\mathrm{2}{x}}{\mathrm{2}{x}+\mathrm{6}}\:=\:\frac{\mathrm{5}{y}}{\mathrm{5}{y}+\mathrm{25}}\:=\:\frac{\mathrm{4}{z}}{\mathrm{4}{z}+\mathrm{16}} \\ $$$${and}\:{xy}\:+\:{yz}\:+\:{xz}\:=\:\mathrm{188}\:.\:\mathrm{F}{ind}\:{the} \\ $$$${solution} \\ $$ Answered by john santu last updated on…
Question Number 174303 by dragan91 last updated on 29/Jul/22 Answered by a.lgnaoui last updated on 29/Jul/22 $${x}^{\mathrm{2}} −\mathrm{10}{x}−{ax}+\mathrm{10}{a}+\mathrm{1}={x}^{\mathrm{2}} +{cx}+{bx}+{bc} \\ $$$${x}^{\mathrm{2}} −\left(\mathrm{10}+{a}\right){x}+\mathrm{10}{a}+\mathrm{1}={x}^{\mathrm{2}} +\left({b}+{c}\right){x}+{bc} \\ $$$$−\left(\mathrm{10}+{a}\right)={b}+{c}…