Question Number 24211 by Physics lover last updated on 14/Nov/17 Commented by Physics lover last updated on 14/Nov/17 $${The}\:{figure}\:{shows}\:{a}\:{mercury} \\ $$$${barometer}. \\ $$$${find}\:{reading}\:{of}\:{the}\:{weighing}\: \\ $$$${machine}.\:{Density}\:{of}\:{mercury}…
Question Number 23508 by math solver last updated on 31/Oct/17 Commented by math solver last updated on 01/Nov/17 $$\mathrm{q}.\mathrm{1}\:\:? \\ $$ Answered by mrW1 last…
Question Number 88940 by jagoll last updated on 14/Apr/20 Answered by john santu last updated on 14/Apr/20 $$\mathrm{3}^{\mathrm{3}} \:+\:\frac{\mathrm{1}}{\mathrm{4}}\:\left[\mathrm{3}×\mathrm{9}+\mathrm{3}×\mathrm{7}\:\right]\:=\: \\ $$$$\mathrm{27}\:+\:\mathrm{12}\:=\:\mathrm{39}\: \\ $$$$\Rightarrow\mathrm{4}^{\mathrm{3}} \:+\:\frac{\mathrm{1}}{\mathrm{4}}\:\left[\:\mathrm{4}×\mathrm{5}+\mathrm{4}×\mathrm{3}\right]\:=\: \\…
Question Number 88943 by jagoll last updated on 14/Apr/20 Commented by john santu last updated on 14/Apr/20 $$\left.\mathrm{1}\right)\:\mathrm{20}+\mathrm{22}−\mathrm{39}\:=\:\mathrm{3} \\ $$$$\Rightarrow\mathrm{3}^{\mathrm{4}−\mathrm{1}} \:=\:\mathrm{3}^{\mathrm{3}} \:=\:\mathrm{27} \\ $$$$\left.\mathrm{2}\right)\mathrm{7}+\mathrm{7}−\mathrm{13}\:=\:\mathrm{1} \\…
Question Number 88690 by jagoll last updated on 12/Apr/20 Answered by john santu last updated on 12/Apr/20 $${i}\:{guess}\:{the}\:{pola}\:{in}\:{question} \\ $$$$\left[\:\left(\mathrm{3}+\mathrm{1}\right)×\mathrm{3}\right]^{\mathrm{2}} \:=\:\mathrm{12}^{\mathrm{2}} \:=\:\mathrm{144} \\ $$$$\left[\:\left(\mathrm{4}+\mathrm{1}\right)×\mathrm{4}\:\right]^{\mathrm{3}} \:=\:\mathrm{20}^{\mathrm{3}}…
Question Number 154194 by chinomso last updated on 15/Sep/21 Commented by alisiao last updated on 15/Sep/21 $$\int_{\mathrm{9}} ^{\:\mathrm{11}} \:\frac{\boldsymbol{{y}}+\mathrm{8}}{\boldsymbol{{c}}}\:\boldsymbol{{dy}}\:=\:\mathrm{1}\:\Rightarrow\:\frac{\mathrm{1}}{\boldsymbol{{c}}}\:\int_{\mathrm{9}} ^{\:\mathrm{11}} \:\left(\boldsymbol{{y}}+\mathrm{8}\right)\boldsymbol{{dy}}\:=\mathrm{1} \\ $$$$ \\ $$$$\boldsymbol{{c}}\:=\:\left(\frac{\boldsymbol{{y}}^{\mathrm{2}}…
Question Number 22872 by Physics lover last updated on 23/Oct/17 Commented by ajfour last updated on 23/Oct/17 Commented by ajfour last updated on 24/Oct/17 $${I}_{{disc}}…
Question Number 151777 by Olaf_Thorendsen last updated on 22/Aug/21 $$\mathrm{Le}\:\mathrm{dernier}\:\mathrm{jour}\:\mathrm{d}'\mathrm{un}\:\mathrm{certain}\:\mathrm{mois}\:\mathrm{au}\: \\ $$$$\mathrm{cours}\:\mathrm{de}\:\mathrm{la}\:\mathrm{premiere}\:\mathrm{guerre}\:\mathrm{mondiale}, \\ $$$$\mathrm{une}\:\mathrm{bombe}\:\mathrm{tombe}\:\mathrm{sur}\:\mathrm{la}\:\mathrm{tombe}\:\mathrm{d}'\mathrm{un} \\ $$$$\mathrm{hallebardier}. \\ $$$$ \\ $$$$\mathrm{Sachant}\:\mathrm{que}\:\mathrm{1}.\mathrm{872}.\mathrm{269}\:\mathrm{est}\:\mathrm{le}\:\mathrm{produit} \\ $$$$\mathrm{de}\:: \\ $$$$ \\…
Question Number 85555 by luxlavanish last updated on 22/Mar/20 $$\mathrm{2}{x}^{\mathrm{2}} +\mathrm{5}{x}+\mathrm{7}=\mathrm{0} \\ $$ Answered by MJS last updated on 23/Mar/20 $${ax}^{\mathrm{2}} +{bx}+{c}=\mathrm{0} \\ $$$$\Rightarrow \\…
Question Number 19516 by Tinkutara last updated on 12/Aug/17 $$\mathrm{Let}\:\mathrm{Akbar}\:\mathrm{and}\:\mathrm{Birbal}\:\mathrm{together}\:\mathrm{have}\:{n} \\ $$$$\mathrm{marbles},\:\mathrm{where}\:{n}\:>\:\mathrm{0}. \\ $$$$\mathrm{Akbar}\:\mathrm{says}\:\mathrm{to}\:\mathrm{Birbal},\:“\mathrm{If}\:\mathrm{I}\:\mathrm{give}\:\mathrm{you}\:\mathrm{some} \\ $$$$\mathrm{marbles}\:\mathrm{then}\:\mathrm{you}\:\mathrm{will}\:\mathrm{have}\:\mathrm{twice}\:\mathrm{as} \\ $$$$\mathrm{many}\:\mathrm{marbles}\:\mathrm{as}\:\mathrm{I}\:\mathrm{will}\:\mathrm{have}.''\:\mathrm{Birbal} \\ $$$$\mathrm{says}\:\mathrm{to}\:\mathrm{Akbar},\:“\mathrm{If}\:\mathrm{I}\:\mathrm{give}\:\mathrm{you}\:\mathrm{some} \\ $$$$\mathrm{marbles}\:\mathrm{then}\:\mathrm{you}\:\mathrm{will}\:\mathrm{have}\:\mathrm{thrice}\:\mathrm{as} \\ $$$$\mathrm{many}\:\mathrm{marbles}\:\mathrm{as}\:\mathrm{I}\:\mathrm{will}\:\mathrm{have}.'' \\…