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Question-190329

Question Number 190329 by yaslm last updated on 31/Mar/23 Commented by Frix last updated on 31/Mar/23 $$\mathrm{Because}\:\mathrm{everybody}\:\mathrm{here}\:\mathrm{is}\:\mathrm{clairvoyant}\:\mathrm{we} \\ $$$$\mathrm{know}\:\mathrm{Eqs}.\:\mathrm{12}.\mathrm{46},\:\mathrm{12}.\mathrm{47},\:\mathrm{12}.\mathrm{61}\:\&\:\mathrm{12}.\mathrm{62} \\ $$$$\mathrm{The}\:\mathrm{answers}\:\mathrm{are}\:\mathrm{given}\:\mathrm{in}\:{Bell},\:{Carson}\:\&\:{al}. \\ $$$$\left(\mathrm{1983}\right)\:\mathrm{on}\:\mathrm{page}\:\mathrm{436}\:\mathrm{Eq}.\:\mathrm{47}.\mathrm{38}\:\mathrm{for}\:{a}.,\:\mathrm{on}\:\mathrm{page} \\ $$$$\mathrm{214}\:\mathrm{Eq}.\:\mathrm{29}.\mathrm{4}\:\mathrm{for}\:{b}.\:\mathrm{and}\:\mathrm{in}\:{Yussuf}\:\left(\mathrm{1991}\right)\:\mathrm{on}…

Question-59226

Question Number 59226 by azizullah last updated on 06/May/19 Answered by MJS last updated on 06/May/19 $$\mathrm{trying}\:\mathrm{with}\:{x}\in\mathbb{Z}\:\mathrm{at}\:\mathrm{first} \\ $$$$\sqrt{−{x}}\:\in\mathbb{Z}\:\Rightarrow\:{x}=−{z}^{\mathrm{2}} \\ $$$${x}=−\mathrm{1}\:\Rightarrow\:\mathrm{wrong} \\ $$$${x}=−\mathrm{4}\:\Rightarrow\:\mathrm{wrong} \\ $$$${x}=−\mathrm{9}\:\Rightarrow\:\mathrm{true}…

make-x-the-subject-of-x-m-x-

Question Number 59201 by otchereabdullai@gmail.com last updated on 05/May/19 $$\mathrm{make}\:\mathrm{x}\:\mathrm{the}\:\mathrm{subject}\:\mathrm{of}\:\:\mathrm{x}=\mathrm{m}+\mathrm{x} \\ $$ Commented by Forkum Michael Choungong last updated on 05/May/19 $${x}−{x}={m} \\ $$$${x}\left(\mathrm{1}−\mathrm{1}\right)={m} \\…

Question-59193

Question Number 59193 by salahahmed last updated on 05/May/19 Answered by MJS last updated on 06/May/19 $$\mathrm{for}\:{x}\in\mathbb{N}\:\underset{\mathrm{0}} {\overset{\infty} {\int}}{t}^{{x}} \mathrm{e}^{−{t}} {dt}={x}! \\ $$$$\Rightarrow\:\mathrm{we}\:\mathrm{have}\:\mathrm{one}\:\mathrm{obvious}\:\mathrm{solution}\:\mathrm{at}\:{x}=\mathrm{5} \\ $$$${x}^{\mathrm{3}}…

Question-190270

Question Number 190270 by 073 last updated on 30/Mar/23 Answered by mr W last updated on 30/Mar/23 $${a}_{{n}} =\left({n}−\mathrm{6}\right)\left({n}+\mathrm{3}\right)<\mathrm{0} \\ $$$$\Rightarrow−\mathrm{3}<{n}<\mathrm{6} \\ $$$$\Rightarrow{n}=\mathrm{1},\mathrm{2},…,\mathrm{5}\: \\ $$$${i}.{e}.\:\mathrm{5}\:{terms}\:{are}\:{negative}.…