Question Number 205657 by mnjuly1970 last updated on 26/Mar/24 Answered by mr W last updated on 26/Mar/24 $${small}\:{square}:\:{a}×{a}=\mathrm{6}×\mathrm{6} \\ $$$${big}\:{square}:\:{b}×{b} \\ $$$${area}\:{of}\:{triangle}: \\ $$$${A}_{\mathrm{1}} +{A}_{\mathrm{2}}…
Question Number 205639 by mr W last updated on 26/Mar/24 Answered by mahdipoor last updated on 26/Mar/24 $${Trapezium}−\mathrm{2}\:{Right}\:{triangle}={area}\:\Rightarrow \\ $$$$\left[\frac{\left({r}\right)+\left(\mathrm{4}+{r}\right)}{\mathrm{2}}×\mathrm{4}\right]−\left[\frac{\left(\mathrm{4}+{r}\right)\left({r}\right)}{\mathrm{2}}+\frac{\left(\mathrm{4}−{r}\right)\left({r}\right)}{\mathrm{2}}\right]=\mathrm{8} \\ $$ Commented by mr…
Question Number 205614 by mr W last updated on 25/Mar/24 Answered by A5T last updated on 25/Mar/24 Commented by A5T last updated on 25/Mar/24 $$\frac{{sin}\left(\angle{CFB}\right)}{\mathrm{1}}=\frac{{sin}\mathrm{90}=\mathrm{1}}{\:\sqrt{\mathrm{0}.\mathrm{5}^{\mathrm{2}}…
Question Number 205562 by mr W last updated on 24/Mar/24 Answered by A5T last updated on 24/Mar/24 Commented by A5T last updated on 24/Mar/24 $${CE}=\sqrt{\mathrm{2}{r}^{\mathrm{2}}…
Question Number 205530 by cherokeesay last updated on 23/Mar/24 Answered by mr W last updated on 24/Mar/24 $${P}\left({a}\:\mathrm{cos}\:\theta,\:{b}\:\mathrm{sin}\:\theta\right) \\ $$$$\mathrm{tan}\:\varphi=−\frac{{dy}}{{dx}}=\frac{{b}}{{a}\:\mathrm{tan}\:\theta} \\ $$$${r}={a}\:\mathrm{cos}\:\theta−{r}\:\mathrm{sin}\:\varphi\:\Rightarrow{r}\:\mathrm{sin}\:\varphi={a}\:\mathrm{cos}\:\theta−{r} \\ $$$${r}={b}\:\mathrm{sin}\:\theta−{r}\:\mathrm{cos}\:\varphi\:\Rightarrow{r}\:\mathrm{cos}\:\varphi={b}\:\mathrm{sin}\:\theta−{r} \\…
Question Number 205507 by mr W last updated on 23/Mar/24 Commented by mr W last updated on 23/Mar/24 $${an}\:{unsolved}\:{old}\:{question} \\ $$ Answered by mr W…
Question Number 205428 by mr W last updated on 21/Mar/24 Answered by Rasheed.Sindhi last updated on 21/Mar/24 Commented by Rasheed.Sindhi last updated on 21/Mar/24 $${White}\:{area}\:{of}\:\Box{ABCD}\:{is}\:{composed}…
Question Number 205461 by mr W last updated on 21/Mar/24 Commented by mr W last updated on 21/Mar/24 $${find}\:{the}\:{area}\:{of}\:{the}\:{third}\:{part}\: \\ $$$${between}\:{two}\:{squares}. \\ $$ Answered by…
Question Number 205406 by mr W last updated on 20/Mar/24 Commented by lepuissantcedricjunior last updated on 21/Mar/24 $$\boldsymbol{{d}}'\boldsymbol{{apre}}'\boldsymbol{{s}}\:\boldsymbol{{pythagore}}\: \\ $$$$\boldsymbol{{d}}^{\mathrm{2}} =\mathrm{7}^{\mathrm{2}} +\boldsymbol{{k}}^{\mathrm{2}} \:\left(\mathrm{1}\right) \\ $$$$\boldsymbol{{d}}'\boldsymbol{{apres}}\:\boldsymbol{{alkashil}}…
Question Number 205334 by cortano12 last updated on 17/Mar/24 Commented by Ghisom last updated on 17/Mar/24 $${x}^{\mathrm{2}} −\frac{\mathrm{5}}{\mathrm{2}}{x}^{\mathrm{3}/\mathrm{2}} +\mathrm{2}{x}−\frac{\mathrm{5}}{\mathrm{4}}{x}^{\mathrm{1}/\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{4}}=\mathrm{0} \\ $$$$\left({x}−\mathrm{2}{x}^{\mathrm{1}/\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}\right)\left({x}−\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{1}/\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}\right)=\mathrm{0} \\…