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Author: Tinku Tara

we-have-y-f-x-function-if-we-transfer-its-graph-C-unit-vertically-and-gain-the-new-function-y-f-x-C-it-s-meant-y-is-increased-or-decreased-C-units-if-we-transfer-the-graph-of-considered-function

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…

1-x-1-y-14-625-x-y-8-Find-all-solutions-

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-208021

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}}…

y-W-

Question Number 207991 by efronzo1 last updated on 02/Jun/24 $$\:\:\:\:\:\mathrm{y}\:\underline{\underbrace{\mathcal{W}}} \\ $$ Answered by mr W last updated on 02/Jun/24 $${particular}\:{solution}: \\ $$$${y}={A}\:\mathrm{sin}\:{x}+{B}\:\mathrm{cos}\:{x} \\ $$$${y}'={A}\:\mathrm{cos}\:{x}−{B}\:\mathrm{sin}\:{x}…

Question-207984

Question Number 207984 by Thomaseinstein last updated on 02/Jun/24 Answered by Frix last updated on 02/Jun/24 $${p},\:{q}\:>\mathrm{0} \\ $$$$\frac{\mathrm{1}}{{p}^{\mathrm{2}} }−\frac{\mathrm{1}}{{q}^{\mathrm{2}} }=\frac{\mathrm{4}}{\mathrm{3}}\:\Rightarrow\:{q}^{\mathrm{2}} =\frac{\mathrm{3}{p}^{\mathrm{2}} }{\:\mathrm{3}−\mathrm{4}{p}^{\mathrm{2}} } \\…