Question Number 191210 by TUN last updated on 20/Apr/23 $$\mathrm{2}^{{x}} +\mathrm{2}^{{x}+\mathrm{1}} >\mathrm{3}^{{x}} +\mathrm{3}^{{x}+\mathrm{1}} \\ $$ Answered by ARUNG_Brandon_MBU last updated on 20/Apr/23 $$\mathrm{2}^{{x}} +\mathrm{2}^{{x}+\mathrm{1}} >\mathrm{3}^{{x}}…
Question Number 125673 by aurpeyz last updated on 12/Dec/20 Answered by bramlexs22 last updated on 13/Dec/20 $$\left(\mathrm{1}\right){a}+{ar}+{ar}^{\mathrm{2}} \:=\:{p}\:;{a}\left(\mathrm{1}+{r}+{r}^{\mathrm{2}} \right)={p} \\ $$$$\left(\mathrm{2}\right){a}^{\mathrm{2}} +{a}^{\mathrm{2}} {r}^{\mathrm{2}} +{a}^{\mathrm{2}} {r}^{\mathrm{4}}…
Question Number 60132 by Kunal12588 last updated on 18/May/19 Commented by Kunal12588 last updated on 18/May/19 $${is}\:{this}\:{question}\:{correct} \\ $$$${pls}\:{solve} \\ $$ Terms of Service Privacy…
Question Number 191203 by otchereabdullai last updated on 20/Apr/23 $$\:{If}\:\:\mathrm{3}^{{x}} =−\mathrm{1}\:{find}\:{the}\:{value}\:{of}\:{x} \\ $$ Answered by Frix last updated on 20/Apr/23 $$\mathrm{3}^{{x}} =−\mathrm{1} \\ $$$$\mathrm{e}^{{x}\mathrm{ln}\:\mathrm{3}} =\mathrm{e}^{\mathrm{i}\left(\mathrm{2}{n}+\mathrm{1}\right)\pi}…
Question Number 191199 by 073 last updated on 20/Apr/23 Commented by 073 last updated on 20/Apr/23 $$\mathrm{sheed}\:\mathrm{area}=?? \\ $$ Answered by cortano12 last updated on…
Question Number 125663 by Mammadli last updated on 12/Dec/20 $$\frac{\mathrm{1}}{\mathrm{6}^{\mathrm{1}} }+\frac{\mathrm{2}}{\mathrm{6}^{\mathrm{2}} }+\frac{\mathrm{3}}{\mathrm{6}^{\mathrm{3}} }+…+\frac{{n}}{\mathrm{6}^{{n}} }=? \\ $$ Answered by Dwaipayan Shikari last updated on 12/Dec/20 $${S}=\frac{\mathrm{1}}{\mathrm{6}}+\frac{\mathrm{2}}{\mathrm{6}^{\mathrm{2}}…
Question Number 125661 by Mammadli last updated on 12/Dec/20 $$\boldsymbol{{a}}^{\boldsymbol{{x}}} \:=\:\boldsymbol{{b}}^{\boldsymbol{{y}}} \:=\:\boldsymbol{{c}}^{\boldsymbol{{z}}} \\ $$$$\boldsymbol{{x}}^{\boldsymbol{{a}}} \:+\:\boldsymbol{{y}}^{\boldsymbol{{b}}} \:+\:\boldsymbol{{z}}^{\boldsymbol{{c}}} \:=\:? \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 125659 by Mammadli last updated on 12/Dec/20 $$\frac{\mathrm{1}}{\mathrm{2020}^{\mathrm{1}} }+\frac{\mathrm{2}}{\mathrm{2020}^{\mathrm{2}} }+\frac{\mathrm{3}}{\mathrm{2020}^{\mathrm{3}} }+…+\frac{\boldsymbol{{n}}}{\mathrm{2020}^{\boldsymbol{{n}}} }=? \\ $$ Answered by Dwaipayan Shikari last updated on 12/Dec/20 $${S}=\frac{\mathrm{1}}{\mathrm{2020}}+\frac{\mathrm{2}}{\mathrm{2020}^{\mathrm{2}}…
Question Number 125650 by aurpeyz last updated on 12/Dec/20 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 125651 by Algoritm last updated on 12/Dec/20 Answered by Olaf last updated on 12/Dec/20 $$\begin{cases}{{y}_{\mathrm{1}} '\:=\:{y}_{\mathrm{1}} +\mathrm{4}{y}_{\mathrm{2}} \:\left(\mathrm{1}\right)}\\{{y}_{\mathrm{2}} '\:=\:−{y}_{\mathrm{1}} +{y}_{\mathrm{2}} +{e}^{\mathrm{3}{x}} \:\left(\mathrm{2}\right)}\end{cases} \\…