Question Number 41044 by Tawa1 last updated on 31/Jul/18 Answered by candre last updated on 01/Aug/18 $$\mathrm{we}\:\mathrm{have}\:\mathrm{that}\:\mathrm{remainder}\:\mathrm{of}\:\mathrm{sum}\:\mathrm{is}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{remakder} \\ $$$$\mathrm{then},\:\mathrm{making}\:\mathrm{a}\:\mathrm{list} \\ $$$$\mathrm{7},\mathrm{14},\mathrm{21},\mathrm{28},\mathrm{35},\mathrm{42},\mathrm{49}\:\left(\mathrm{7}\:{m}_{\mathrm{0}} \right) \\ $$$$\mathrm{1},\mathrm{8},\mathrm{15},\mathrm{22},\mathrm{29},\mathrm{36},\mathrm{43},\mathrm{50}\:\left(\mathrm{8}\:{m}_{\mathrm{1}} \right)…
Question Number 172111 by Mikenice last updated on 23/Jun/22 $${solve} \\ $$$$\mathrm{2}^{{x}} =\mathrm{4}{x} \\ $$ Commented by mr W last updated on 23/Jun/22 $${you}\:{have}\:{asked}\:\mathrm{2}^{{x}} =\mathrm{10}{x}.\:{it}\:{is}\:{solved}…
Question Number 41032 by adityasin567@gmail.com last updated on 31/Jul/18 $$\frac{{a}}{{x}−{a}}\:+\frac{{b}}{{x}−{b}}=\mathrm{2}.{find}\:{x}. \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 31/Jul/18 $${ax}−{ab}+{bx}−{ab}=\mathrm{2}\left({x}^{\mathrm{2}} −{ax}−{bx}+{ab}\right) \\ $$$$\mathrm{2}{x}^{\mathrm{2}} −\mathrm{2}{ax}−\mathrm{2}{bx}+\mathrm{2}{ab}−{ax}−{bx}+\mathrm{2}{ab}=\mathrm{0} \\…
Question Number 172065 by mnjuly1970 last updated on 23/Jun/22 $$ \\ $$$$\:\:\:\:\:\mathrm{Test}\:: \\ $$$$\:\:\mathrm{Q}\::\:\:\:\mathrm{If}\:,\:\:\mid\overset{\rightarrow} {\mathrm{a}}\:×\:\overset{\rightarrow} {\mathrm{b}}\mid\:=\:\sqrt{\mathrm{11}}\:\:\:,\:\:\:\mathrm{2}\overset{\rightarrow} {\mathrm{a}}\:+\:\mathrm{3}\overset{\rightarrow} {\mathrm{b}}=\:\overset{\rightarrow} {\mathrm{i}}+\:\mathrm{2}\overset{\rightarrow} {\mathrm{j}}+\:\mathrm{3}\overset{\rightarrow\:} {\mathrm{k}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{then}\:\:\::\:\:\:\mid\:\:\mathrm{2}\overset{\rightarrow} {\mathrm{a}}\:+\:\overset{\rightarrow} {\mathrm{a}}×\overset{\rightarrow}…
Question Number 40965 by Glorious Man last updated on 30/Jul/18 Answered by tanmay.chaudhury50@gmail.com last updated on 30/Jul/18 $${t}={x}^{\mathrm{5}} −{x}^{\mathrm{3}} \:\:\:{dt}=\mathrm{5}{x}^{\mathrm{4}} \:−\mathrm{3}{x}^{\mathrm{2}} \:\:{dx} \\ $$$$=−\mathrm{2}\int\frac{\mathrm{5}{x}^{\mathrm{4}} −\mathrm{3}{x}^{\mathrm{2}}…
Question Number 172029 by Mikenice last updated on 23/Jun/22 $${solve}: \\ $$$${x}^{\mathrm{2}} =\mathrm{2}^{{x}} \\ $$ Commented by Rasheed.Sindhi last updated on 23/Jun/22 $${x}=\mathrm{2},\mathrm{4} \\ $$…
Question Number 172030 by Mikenice last updated on 23/Jun/22 $${solve}: \\ $$$$\sqrt{\frac{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{4}}{{x}^{\mathrm{2}} +\mathrm{5}{x}+\mathrm{6}}\:}\:\:+\:\:\mathrm{2}\:\sqrt{\frac{{x}^{\mathrm{2}} +\mathrm{5}{x}+\mathrm{6}}{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{4}}}\:=? \\ $$ Answered by Rasheed.Sindhi last updated on 23/Jun/22…
Question Number 172031 by Mikenice last updated on 23/Jun/22 $${solve}: \\ $$$$\sqrt{{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{4}}\:\:+\sqrt{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{4}}\:=\sqrt{{x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{9}} \\ $$ Answered by puissant last updated on 23/Jun/22 $$\Rightarrow\left({x}−\mathrm{2}\right)+\left({x}+\mathrm{2}\right)=\sqrt{{x}^{\mathrm{2}}…
Question Number 172024 by Mikenice last updated on 23/Jun/22 $${let}\:\alpha\:{and}\:\beta\:{be}\:{the}\:{root}\:{of}\:{the}\:{equation} \\ $$$${ax}^{\mathrm{2}} +{bx}+{c}=\mathrm{0}.\:{find}\:{the}\:{equation}\:{whose}\:{roots} \\ $$$${are}\:\left(\frac{\mathrm{1}}{\alpha}+\frac{\mathrm{1}}{\beta}\right)\:{and}\:\left(\frac{\mathrm{1}}{\alpha}−\frac{\mathrm{1}}{\beta}\right) \\ $$ Answered by Rasheed.Sindhi last updated on 23/Jun/22 $${Given}\:{equation}:…
Question Number 172025 by Mikenice last updated on 23/Jun/22 $${find}\:{the}\:{root}\:{of}\: \\ $$$${x}^{\mathrm{4}} +\mathrm{4}{x}^{\mathrm{3}} +\mathrm{11}{x}^{\mathrm{2}} +\mathrm{14}{x}−\mathrm{30}=\mathrm{0}. \\ $$ Answered by thfchristopher last updated on 23/Jun/22 $${x}^{\mathrm{4}}…