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Category: Algebra

We-are-in-C-Given-Z-0-1-Z-n-1-1-2-Z-n-1-2-i-n-N-Show-that-n-N-Z-n-lt-1-

Question Number 119159 by mathocean1 last updated on 22/Oct/20 $${We}\:{are}\:{in}\:\mathbb{C}. \\ $$$${Given}\:{Z}_{\mathrm{0}\:\:} =\mathrm{1}\:;\:\:\:\:\:{Z}_{{n}+\mathrm{1}\:} =\frac{\mathrm{1}}{\mathrm{2}}{Z}_{{n}\:} +\frac{\mathrm{1}}{\mathrm{2}}{i} \\ $$$${n}\:\in\:\mathbb{N}. \\ $$$${Show}\:{that}\:\forall\:{n}\:\in\:\mathbb{N}^{\ast\:} ,\:\mid{Z}_{{n}} \mid<\mathrm{1}. \\ $$ Answered by…

Question-119141

Question Number 119141 by Engr_Jidda last updated on 22/Oct/20 Answered by TANMAY PANACEA last updated on 23/Oct/20 $$\frac{{d}^{\mathrm{2}} {x}}{{dt}^{\mathrm{2}} }+\frac{{dy}}{{dt}}=−{sint} \\ $$$$\frac{{d}^{\mathrm{2}} {x}}{{dt}^{\mathrm{2}} }+\mathrm{4}\frac{{dx}}{{dt}}+\mathrm{3}{x}−{sint}=−{sint} \\…

Question-184669

Question Number 184669 by Shrinava last updated on 10/Jan/23 Commented by mr W last updated on 10/Jan/23 $${it}\:{is}\:{meant}: \\ $$$${before}:\:−\mathrm{2}\: \\ $$$$\left({e}.{g}.\:{your}\:{loss}\:{last}\:{year}\:{was}\:\mathrm{2}\:{mio}.\:\$\right) \\ $$$${reduced}\:{by}\:\mathrm{2}\:{times} \\…

Informatica-11110000-2-0-2-0-0-2-1-0-2-2-0-2-3-1-2-4-1-2-5-1-2-6-1-2-7-0-1-0-2-0-4-0-8-1-16-1-32-1-64-1-128-0-0-0-0-16-32-64-128-240-10-11000101-2-1-2-0-0-2-1-1-2-2-0-2-3-0-2

Question Number 119133 by Cristina last updated on 22/Oct/20 $$\mathrm{Informatica} \\ $$$$\left(\mathrm{11110000}\right)_{\mathrm{2}} =\mathrm{0}\bullet\mathrm{2}^{\mathrm{0}} +\mathrm{0}\bullet\mathrm{2}^{\mathrm{1}} +\mathrm{0}\bullet\mathrm{2}^{\mathrm{2}} +\mathrm{0}\bullet\mathrm{2}^{\mathrm{3}} +\mathrm{1}\bullet\mathrm{2}^{\mathrm{4}} +\mathrm{1}\bullet\mathrm{2}^{\mathrm{5}} +\mathrm{1}\bullet\mathrm{2}^{\mathrm{6}} +\mathrm{1}\bullet\mathrm{2}^{\mathrm{7}} = \\ $$$$=\mathrm{0}\bullet\mathrm{1}+\mathrm{0}\bullet\mathrm{2}+\mathrm{0}\bullet\mathrm{4}+\mathrm{0}\bullet\mathrm{8}+\mathrm{1}\bullet\mathrm{16}+\mathrm{1}\bullet\mathrm{32}+\mathrm{1}\bullet\mathrm{64}+\mathrm{1}\bullet\mathrm{128}= \\ $$$$\mathrm{0}+\mathrm{0}+\mathrm{0}+\mathrm{0}+\mathrm{16}+\mathrm{32}+\mathrm{64}+\mathrm{128}=\left(\mathrm{240}\right)_{\mathrm{10}}…

Question-119119

Question Number 119119 by benjo_mathlover last updated on 22/Oct/20 Answered by mindispower last updated on 22/Oct/20 $$\Rightarrow{abc}+\mathrm{1}=\frac{{a}}{{a}+{b}+{c}}…\mathrm{1} \\ $$$${abc}+\mathrm{2}=\frac{{b}}{{a}+{b}+{c}}….\mathrm{2} \\ $$$${abc}+\mathrm{7}=\frac{{c}}{{a}+{b}+{c}}…\mathrm{3} \\ $$$${abc}+\mathrm{10}=\mathrm{1} \\ $$$$\Rightarrow{abc}=−\mathrm{9}…\mathrm{2}…

Question-119107

Question Number 119107 by I want to learn more last updated on 22/Oct/20 Answered by 1549442205PVT last updated on 22/Oct/20 $$\mathrm{S}=\mathrm{1}+\mathrm{3}−\mathrm{5}+\mathrm{7}+\mathrm{9}−\mathrm{11}+\mathrm{13}+\mathrm{15}−\mathrm{17}+\mathrm{19}+\mathrm{21}−\mathrm{23}+… \\ $$$$\mathrm{S}=\left(\mathrm{1}+\mathrm{3}−\mathrm{5}\right)+\left(\mathrm{7}+\mathrm{9}−\mathrm{11}\right)+\left(\mathrm{13}+\mathrm{15}−\mathrm{17}\right)+\left(\mathrm{19}+\mathrm{21}−\mathrm{23}\right)+ \\ $$$$\left(\mathrm{25}+\mathrm{27}−\mathrm{29}\right)+…+\left[\mathrm{6n}−\mathrm{5}+\mathrm{6n}−\mathrm{3}−\left(\mathrm{6n}−\mathrm{5}\right)\right]…