Question Number 208063 by hardmath last updated on 03/Jun/24 $$\begin{cases}{\mathrm{x}\:+\:\frac{\mathrm{3}}{\mathrm{5}}\:\mathrm{y}\:=\:\mathrm{5}}\\{\frac{\mathrm{3}}{\mathrm{5}}\:\mathrm{x}\:+\:\mathrm{y}\:=\:\mathrm{11}}\end{cases}\:\:\:\:\:\Rightarrow\:\:\mathrm{x}\:+\:\mathrm{y}\:=\:? \\ $$ Answered by A5T last updated on 03/Jun/24 $${x}+{y}+\frac{\mathrm{3}}{\mathrm{5}}\left({x}+{y}\right)=\mathrm{16}=\frac{\mathrm{8}\left({x}+{y}\right)}{\mathrm{5}}\Rightarrow{x}+{y}=\mathrm{10} \\ $$ Terms of Service…
Question Number 208052 by necx122 last updated on 03/Jun/24 $${Sketch}\:{the}\:{curve}\:{y}\:=\:{x}^{\mathrm{3}} . \\ $$$$\left({a}\right)\:{Find}\:{the}\:{equation}\:{of}\:{the}\:{tangent} \\ $$$${to}\:{the}\:{curve}\:{at}\:{A}\left(\mathrm{1},\mathrm{1}\right). \\ $$$$\left({b}\right)\:{Find}\:{the}\:{coordinates}\:{of}\:{point}\:{B}, \\ $$$${where}\:{the}\:{tangent}\:{meets}\:{the}\:{curve}\:{again}. \\ $$$$\left({c}\right)\:{Calculate}\:{the}\:{area}\:{between}\:{the} \\ $$$${tangent}\:{B}\:{and}\:{the}\:{arc}\:{AB}\:{of}\:{the}\:{curve}. \\ $$…
Question Number 208069 by Davidtim last updated on 03/Jun/24 $${F}_{\mathrm{1}} =\mathrm{4}{N},\:\:\:{F}_{\mathrm{2}} =\mathrm{5}{N},\:\:\:{R}=\mathrm{8}.\mathrm{32}{N} \\ $$$$\theta=? \\ $$ Commented by mr W last updated on 03/Jun/24 $$\theta=\mathrm{180}°−\mathrm{cos}^{−\mathrm{1}}…
Question Number 208071 by hardmath last updated on 03/Jun/24 $$\frac{\mathrm{4}\:\centerdot\:\mathrm{10}^{\boldsymbol{\mathrm{n}}} \:−\:\mathrm{52}}{\mathrm{3}} \\ $$$$\mathrm{The}\:\mathrm{numbers}\:\mathrm{are}\:\:\mathrm{38}\:\:\mathrm{in}\:\mathrm{total} \\ $$$$\mathrm{Find}:\:\:\boldsymbol{\mathrm{n}}\:=\:? \\ $$ Commented by mr W last updated on 04/Jun/24…
Question Number 208049 by mr W last updated on 03/Jun/24 Commented by Frix last updated on 03/Jun/24 $${a}=\mathrm{1}\wedge{b}=\mathrm{3} \\ $$$${r}={a}+{b}−\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }=\mathrm{4}−\sqrt{\mathrm{10}} \\ $$ Answered…
Question Number 208066 by Ismoiljon_008 last updated on 03/Jun/24 Answered by Ismoiljon_008 last updated on 03/Jun/24 $${Help}\:{me}\:{please} \\ $$ Answered by A5T last updated on…
Question Number 208031 by Davidtim last updated on 02/Jun/24 $${we}\:{have}\:{y}={f}\left({x}\right)\:{function},\:{if}\:{we}\:{transfer} \\ $$$${its}\:{graph}\:{C}\:{unit}\:{vertically}\:{and}\:{gain} \\ $$$${the}\:{new}\:{function}\:{y}={f}\left({x}\right)\pm{C},\:{it}'{s}\:{meant} \\ $$$${y}\:{is}\:{increased}\:{or}\:{decreased}\:\:{C}\:{units}. \\ $$$${if}\:{we}\:{transfer}\:{the}\:{graph}\:{of}\:{considered} \\ $$$${function}\:{horizontally},\:{we}\:{gain}\:{the}\:{new}\:{function} \\ $$$${y}={f}\left({x}\pm{C}\right),\:{what}\:{does}\:{mean}\:{it}? \\ $$ Answered…
Question Number 207999 by intan_rd last updated on 02/Jun/24 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 208020 by Frix last updated on 02/Jun/24 $$\begin{cases}{\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}=\frac{\mathrm{14}}{\mathrm{625}}}\\{\sqrt{{x}}+\sqrt{{y}}=\mathrm{8}}\end{cases} \\ $$$$\mathrm{Find}\:\mathrm{all}\:\mathrm{solutions}. \\ $$ Answered by efronzo1 last updated on 02/Jun/24 $$\:\:\Rightarrow\mathrm{x}+\mathrm{y}\:+\mathrm{2}\sqrt{\mathrm{xy}}\:=\:\mathrm{64}\: \\ $$$$\:\Rightarrow\:\frac{\mathrm{x}+\mathrm{y}}{\mathrm{xy}}\:=\:\frac{\mathrm{14}}{\mathrm{625}}\:;\:\mathrm{x}+\mathrm{y}\:=\:\frac{\mathrm{14}}{\mathrm{625}}\mathrm{xy}\: \\…
Question Number 208021 by efronzo1 last updated on 02/Jun/24 Answered by A5T last updated on 02/Jun/24 $$\frac{{CD}}{{CD}+{r}}=\frac{{r}}{{r}+{AB}}\Rightarrow\frac{{r}+{AB}}{{r}+{CD}}=\frac{{r}}{{CD}} \\ $$$$\frac{{r}+{AB}}{\mathrm{8}+{r}}=\frac{{r}+{CD}}{\mathrm{3}+{r}}\Rightarrow\frac{{r}+{AB}}{{r}+{CD}}=\frac{\mathrm{8}+{r}}{\mathrm{3}+{r}}=\frac{{r}}{{CD}} \\ $$$$\Rightarrow{CD}=\frac{{r}\left(\mathrm{3}+{r}\right)}{\mathrm{8}+{r}} \\ $$$$\frac{{AB}}{{AB}+{r}}=\frac{{r}}{{CD}+{r}}={AB}\left({CD}+{r}\right)={r}\left({AB}\right)+{r}^{\mathrm{2}} \\ $$$${AB}=\frac{{r}^{\mathrm{2}}…