Question Number 80515 by M±th+et£s last updated on 03/Feb/20 $$\int\frac{{dx}}{\left(\mathrm{1}+{x}^{\phi} \right)^{\phi} } \\ $$ Commented by john santu last updated on 04/Feb/20 $${t}\:=\:\mathrm{1}+{x}^{\phi} \:\Rightarrow{x}=\left({t}−\mathrm{1}\right)^{\frac{\mathrm{1}}{\phi}} \\…
Question Number 14970 by uchechukwu okorie favour last updated on 06/Jun/17 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 80504 by Rio Michael last updated on 03/Feb/20 $$\mathrm{Solve}\:\mathrm{the}\:\mathrm{system}\:\mathrm{of}\:\mathrm{congruences} \\ $$$${x}\:\equiv\:\mathrm{2}\:\left(\mathrm{mod}\:\mathrm{3}\right) \\ $$$${x}\:\equiv\:\mathrm{5}\left(\:\mathrm{mod}\:\mathrm{7}\right) \\ $$$$\: \\ $$ Commented by mr W last updated…
Question Number 80505 by Rio Michael last updated on 03/Feb/20 $$\mathrm{Given}\:\mathrm{that}\:\:\mathrm{7}^{{k}} \:\equiv\mathrm{1}\:\left(\mathrm{mod}\:\mathrm{15}\right) \\ $$$$\left.\mathrm{a}\right)\:\mathrm{Write}\:\mathrm{down}\:\mathrm{three}\:\mathrm{values}\:\mathrm{of}\:{k}. \\ $$$$\left.\mathrm{b}\right)\:\mathrm{Find}\:\mathrm{the}\:\mathrm{general}\:\mathrm{solution}\:\mathrm{of}\: \\ $$$$\mathrm{the}\:\mathrm{equation}\:\:\mathrm{7}^{{k}} \:\equiv\:\mathrm{1}\:\left({mod}\:\mathrm{15}\right) \\ $$ Commented by mr W…
Question Number 14962 by tawa tawa last updated on 06/Jun/17 $$\mathrm{Two}\:\mathrm{30}\:\mathrm{ohms}\:\mathrm{resistor}\:\mathrm{are}\:\mathrm{connected}\:\mathrm{in}\:\mathrm{parallel},\:\mathrm{what}\:\mathrm{should}\:\mathrm{be}\:\mathrm{the}\:\mathrm{resistance} \\ $$$$\mathrm{to}\:\mathrm{be}\:\mathrm{connected}\:\mathrm{in}\:\mathrm{series}\:\mathrm{with}\:\mathrm{this}\:\mathrm{parallel}\:\mathrm{combination}\:\mathrm{such}\:\mathrm{that}\:\mathrm{the}\:\mathrm{power} \\ $$$$\mathrm{in}\:\mathrm{each}\:\mathrm{30}\:\mathrm{ohms}\:\mathrm{is}\:\frac{\mathrm{1}}{\mathrm{4}}\:\mathrm{th}\:\mathrm{of}\:\mathrm{total}\:\mathrm{power}. \\ $$ Answered by ajfour last updated on 06/Jun/17 $$\frac{{P}}{\mathrm{4}}={I}^{\mathrm{2}}…
Question Number 14963 by tawa tawa last updated on 06/Jun/17 $$\mathrm{A}\:\mathrm{resistor}\:\mathrm{R}\:\mathrm{is}\:\mathrm{connected}\:\mathrm{in}\:\mathrm{series}\:\mathrm{with}\:\mathrm{a}\:\mathrm{parallel}\:\mathrm{combination}\:\mathrm{of}\:\mathrm{two}\:\mathrm{resistors} \\ $$$$\mathrm{of}\:\mathrm{24}\:\mathrm{and}\:\mathrm{8}\:\mathrm{ohms}\:.\:\mathrm{The}\:\mathrm{total}\:\mathrm{power}\:\mathrm{disipated}\:\mathrm{in}\:\mathrm{the}\:\mathrm{circuit}\:\mathrm{is}\:\mathrm{64}\:\mathrm{watt}\:\mathrm{when}\:\mathrm{the} \\ $$$$\mathrm{applied}\:\mathrm{voltage}\:\mathrm{is}\:\mathrm{24}\:\mathrm{volt}.\mathrm{Find}\:\mathrm{R} \\ $$ Answered by ajfour last updated on 06/Jun/17 $$\:{P}=\frac{{V}^{\mathrm{2}}…
Question Number 14853 by tawa tawa last updated on 04/Jun/17 $$\mathrm{Solve}\:\mathrm{for}\:\mathrm{x} \\ $$$$\mathrm{3}^{\mathrm{2x}} \:=\:\mathrm{18x} \\ $$ Answered by mrW1 last updated on 04/Jun/17 $${e}^{\mathrm{2}{x}\mathrm{ln}\:\mathrm{3}} =\mathrm{18}{x}…
Question Number 14816 by tawa tawa last updated on 04/Jun/17 Answered by arnabpapu550@gmail.com last updated on 08/Jun/17 $$\mathrm{Answer}\:\mathrm{to}\:\mathrm{part}\:\mathrm{1} \\ $$$$\mathrm{Given}\:,\:\mathrm{x}=\frac{\mathrm{5}}{\mathrm{3}}\mathrm{t}^{\mathrm{3}} −\frac{\mathrm{5}}{\mathrm{2}}\mathrm{t}^{\mathrm{2}} −\mathrm{30t}+\mathrm{8x} \\ $$$$\mathrm{differentiating}\:\mathrm{both}\:\mathrm{side}\:\mathrm{with}\:\mathrm{t}, \\…
Question Number 80340 by Rio Michael last updated on 02/Feb/20 $$\mathrm{If}\:{P}\:=\:\begin{pmatrix}{{a}}&{{b}}&{{c}}&{{d}}\\{{c}}&{{d}}&{{a}}&{{b}}\end{pmatrix}\:\:,\:{Q}\:=\:\begin{pmatrix}{{a}}&{{b}}&{{c}}&{{d}}\\{{b}}&{{a}}&{{d}}&{{c}}\end{pmatrix}\:\mathrm{are} \\ $$$$\mathrm{permutations}\:\mathrm{of}\:\mathrm{the}\:\mathrm{elements}\:\left({a},{b},{c},{d}\right),\:\mathrm{then}\: \\ $$$${QP}\:\equiv \\ $$$$\: \\ $$ Answered by MJS last updated on…
Question Number 80341 by Rio Michael last updated on 02/Feb/20 $$\mathrm{A}\:\mathrm{particle}\:\mathrm{moves}\:\mathrm{round}\:\mathrm{the}\:\mathrm{polar}\:\mathrm{curve} \\ $$$${r}\:=\:{a}\left(\mathrm{1}\:+\:\mathrm{cos}\:\theta\right)\:\mathrm{with}\:\mathrm{constant}\:\mathrm{angular}\: \\ $$$$\mathrm{velocity}\:\omega\:.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{transverse}\:\mathrm{component} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{velocity}. \\ $$ Answered by mr W last updated…