Question Number 184706 by Ml last updated on 10/Jan/23 $$\mathrm{deg}\left[\mathrm{3p}\left(\mathrm{x}\right)+\mathrm{Q}\left(\mathrm{x}\right)\right]=\mathrm{6} \\ $$$$\mathrm{deg}\left[\mathrm{p}\left(\mathrm{x}\right)+\mathrm{x}^{\mathrm{4}} \right]=\mathrm{5} \\ $$$$\mathrm{deg}\left[\frac{\left(\mathrm{x}^{\mathrm{4}} +\mathrm{1}\right)\mathrm{p}\left(\mathrm{x}^{\mathrm{2}} \right)}{\mathrm{x}^{\mathrm{3}} \centerdot\mathrm{Q}\left(\mathrm{x}\right)}\right]=? \\ $$$$\mathrm{deg}=\mathrm{degree}\: \\ $$ Answered by Rasheed.Sindhi…
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Question Number 119158 by mathdave last updated on 22/Oct/20 Answered by MJS_new last updated on 22/Oct/20 $${p}=\mathrm{tan}\:\alpha \\ $$$${q}=\mathrm{tan}\:\beta \\ $$$${p}+{q}=\mathrm{2} \\ $$$$\sqrt{{p}^{\mathrm{2}} +\mathrm{1}}+\sqrt{{q}^{\mathrm{2}} +\mathrm{1}}=\mathrm{3}…
Question Number 119157 by mathdave last updated on 22/Oct/20 Answered by mindispower last updated on 23/Oct/20 $${x}=\frac{\mathrm{1}}{{t}}\Rightarrow \\ $$$$=\int_{\mathrm{0}} ^{\infty} \frac{{tan}^{−} \left(\frac{\mathrm{1}}{{t}}\right)}{\mathrm{1}+\frac{\mathrm{1}}{{t}}}.\frac{{dt}}{{t}^{\mathrm{2}} .\frac{\mathrm{1}}{\:\sqrt{{t}}}}=\int_{\mathrm{0}} ^{\infty} \frac{{tan}^{−}…
Question Number 119155 by zakirullah last updated on 22/Oct/20 Commented by zakirullah last updated on 22/Oct/20 $${if}\:{a}\:=\:{i}+{j}+{k},\:{b}\:=\:\mathrm{4}{i}−\mathrm{2}{n}+\mathrm{3}{k},\:{c}\:=\:{i}−\mathrm{2}{j}+{k}\:{then} \\ $$$${find}\:{a}\:{vecter}\:{of}\:{magnitude}\:\mathrm{6}\:{unit} \\ $$$${which}\:{is}\:{parellel}\:{to}\:{the}\:{vecter}\:\mathrm{2}{a}−{b}+\mathrm{3}{c}. \\ $$ Commented by…
Question Number 119138 by 675480065 last updated on 22/Oct/20 $$\sqrt{\mathrm{2}\sqrt{\mathrm{4}\sqrt{\mathrm{8}\sqrt{\mathrm{16}\sqrt{\mathrm{32}…}}}}}=? \\ $$$$\mathrm{pls}\:\mathrm{help} \\ $$ Commented by Dwaipayan Shikari last updated on 22/Oct/20 $$\mathrm{4}=\sqrt{\mathrm{2}.\mathrm{8}} \\ $$$$\mathrm{4}=\sqrt{\mathrm{2}\sqrt{\mathrm{64}}}…
Question Number 119136 by Hassen_Timol last updated on 22/Oct/20 Commented by MJS_new last updated on 22/Oct/20 $$\mathrm{to}\:\mathrm{understand}\:\mathrm{what}\:\mathrm{happens}\:\mathrm{draw}\:\mathrm{it}! \\ $$$$\mathrm{use}\:{p}=\mathrm{2}\:\Rightarrow\:{n}=\mathrm{5} \\ $$$${w}^{{n}} =\mathrm{1}\:\Rightarrow\:\mathrm{draw}\:{x}−\mathrm{axis}\:\left(\mathrm{real}\right)\:\mathrm{and}\:{y}−\mathrm{axis} \\ $$$$\left(\mathrm{imaginary}\right)\:\mathrm{and}\:\mathrm{a}\:\mathrm{unit}\:\mathrm{circle}.\:\mathrm{then}\:\mathrm{scetch} \\…
Question Number 119114 by Hassen_Timol last updated on 22/Oct/20 Commented by Hassen_Timol last updated on 22/Oct/20 Could you please help me... I don't understand at all Commented by MJS_new last updated on 22/Oct/20 $${T}\:\mathrm{is}\:\mathrm{not}\:\mathrm{defined}\:\mathrm{because}\:{k}\:\mathrm{is}\:\mathrm{an}\:\mathrm{index}\:\mathrm{with}…
Question Number 184645 by yaslm last updated on 09/Jan/23 Answered by TheSupreme last updated on 10/Jan/23 $${y}>\mathrm{0} \\ $$$${x}>{ln}\left({y}\right) \\ $$$${y}<{e}^{{x}} \\ $$$${D}\:{is}\:{the}\:{area}\:{between}\:{y}=\mathrm{0}\:{and}\:{y}={e}^{{x}} \\ $$…