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} \\ $$$$\left({x}−\mathrm{2}{x}^{\mathrm{1}/\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}\right)\left({x}−\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{1}/\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}\right)=\mathrm{0} \\…
Question Number 205302 by cortano12 last updated on 15/Mar/24 Answered by Skabetix last updated on 15/Mar/24 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 205289 by mr W last updated on 14/Mar/24 Answered by A5T last updated on 16/Mar/24 Commented by A5T last updated on 16/Mar/24 $${General}\:{idea},\:{there}\:{could}\:{be}\:{a}\:{far}\:{more}\:{simpler}…
Question Number 205280 by cherokeesay last updated on 14/Mar/24 Commented by mr W last updated on 14/Mar/24 $${figure}\:{is}\:{not}\:{uniquely}\:{defined}. \\ $$$${or}\:{you}\:{mean}\:{that}\:{the}\:{hypotenuse}\:{of} \\ $$$${both}\:{triangles}\:{is}\:{of}\:{same}\:{length}? \\ $$ Commented…
Question Number 205227 by Ari last updated on 13/Mar/24 Commented by Ari last updated on 13/Mar/24 $${colored}\:{surface}? \\ $$ Commented by cherokeesay last updated on…
Question Number 205160 by cortano12 last updated on 11/Mar/24 Answered by A5T last updated on 11/Mar/24 Commented by A5T last updated on 11/Mar/24 $${AG}=\sqrt{\mathrm{5}^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}}…
Question Number 205161 by York12 last updated on 11/Mar/24 $$\mathrm{Calculate}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{green}\:\mathrm{shaded}\:\mathrm{portions} \\ $$ Commented by York12 last updated on 11/Mar/24 Answered by mr W last updated…
Question Number 205135 by mr W last updated on 09/Mar/24 Commented by mr W last updated on 09/Mar/24 $${ABCD}\:{is}\:{square}. \\ $$$${find}\:\frac{{red}\:{area}}{{blue}\:{area}}=? \\ $$ Answered by…