Menu Close

Category: None

Three-school-children-share-some-oranges-as-follows-Akwasi-gets-1-3-of-the-total-and-the-remainder-is-shared-between-Abena-and-Juana-in-the-ratio-2-3-If-Abena-gets-24-oranges-how-many-does-Ak

Question Number 67108 by otchereabdullai@gmail.com last updated on 22/Aug/19 $$\mathrm{Three}\:\mathrm{school}\:\mathrm{children}\:\mathrm{share}\:\mathrm{some}\: \\ $$$$\mathrm{oranges}\:\mathrm{as}\:\mathrm{follows}:\:\mathrm{Akwasi}\:\mathrm{gets}\:\frac{\mathrm{1}}{\mathrm{3}}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{total}\:\mathrm{and}\:\mathrm{the}\:\mathrm{remainder}\:\mathrm{is}\:\mathrm{shared} \\ $$$$\mathrm{between}\:\mathrm{Abena}\:\mathrm{and}\:\mathrm{Juana}\:\mathrm{in}\:\mathrm{the}\:\mathrm{ratio} \\ $$$$\mathrm{2}:\:\mathrm{3}\:.\:\mathrm{If}\:\mathrm{Abena}\:\mathrm{gets}\:\mathrm{24}\:\mathrm{oranges}\:,\:\mathrm{how} \\ $$$$\mathrm{many}\:\mathrm{does}\:\mathrm{Akwasi}\:\mathrm{get}. \\ $$ Commented by Prithwish…

Question-132647

Question Number 132647 by aurpeyz last updated on 15/Feb/21 Answered by Dwaipayan Shikari last updated on 15/Feb/21 $$\int\frac{\mathrm{1}}{\:\sqrt{{x}}\left(\sqrt{{x}}+\mathrm{7}\right)}{dx}\:\:\:\:\:\:\:\:{x}={t}^{\mathrm{2}} \\ $$$$=\mathrm{2}\int\frac{\mathrm{1}}{{t}+\mathrm{7}}{dt}=\mathrm{2}{log}\left({t}+\mathrm{7}\right)+{C}={log}\left(\sqrt{{x}}+\mathrm{7}\right)^{\mathrm{2}} +{C} \\ $$ Answered by…

1-e-x-9e-x-dx-

Question Number 132641 by aurpeyz last updated on 15/Feb/21 $$\int\frac{\mathrm{1}}{{e}^{{x}} +\mathrm{9}{e}^{−{x}} }{dx} \\ $$ Answered by MJS_new last updated on 15/Feb/21 $${t}=\mathrm{e}^{{x}} \:\rightarrow\:{dx}=\frac{{dt}}{{t}} \\ $$$$\int\frac{{dt}}{{t}^{\mathrm{2}}…

How-does-one-use-the-following-notation-It-s-new-to-my-understanding-of-using-only-one-sigma-sign-For-e-g-j-1-m-i-1-n-i-1-j-1-

Question Number 1566 by 112358 last updated on 20/Aug/15 $${How}\:{does}\:{one}\:{use}\:{the}\:{following} \\ $$$${notation}?\:{It}'{s}\:{new}\:{to}\:{my}\: \\ $$$${understanding}\:{of}\:{using}\:{only} \\ $$$${one}\:{sigma}\:{sign}. \\ $$$${For}\:{e}.{g}\:\:\:\:\:\:\:\:\underset{{j}=\mathrm{1}} {\overset{{m}} {\sum}}\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\left({i}+\mathrm{1}\right)\left({j}−\mathrm{1}\right) \\ $$ Commented…