Question Number 81913 by ahmadshahhimat775@gmail.com last updated on 16/Feb/20 Commented by john santu last updated on 16/Feb/20 $$\Rightarrow\sqrt{{x}}+\sqrt[{\mathrm{6}\:}]{{y}}\:+\frac{{y}\:\sqrt[{\mathrm{6}\:}]{{y}}+{x}\:\sqrt[{\mathrm{6}\:}]{{x}}}{\:\sqrt[{\mathrm{3}\:}]{{x}}\:\sqrt{{y}}}\:= \\ $$$$\frac{\sqrt[{\mathrm{6}\:}]{{x}^{\mathrm{5}} \:}\sqrt{{y}}\:+\:\sqrt[{\mathrm{3}\:}]{{x}}\:\sqrt[{\mathrm{6}\:}]{{y}^{\mathrm{4}} }+\sqrt[{\mathrm{6}\:}]{{y}^{\mathrm{7}} }+\sqrt[{\mathrm{6}\:}]{{x}^{\mathrm{7}} }}{\:\sqrt[{\mathrm{3}\:}]{{x}}\:\sqrt{{y}}} \\…
Question Number 81910 by mr W last updated on 16/Feb/20 $${a}_{\mathrm{1}} =\mathrm{1} \\ $$$${a}_{\mathrm{2}} =\mathrm{2} \\ $$$${a}_{{n}+\mathrm{1}} =\left({n}+\mathrm{1}\right){a}_{{n}} −\mathrm{2}{a}_{{n}−\mathrm{1}} \\ $$$${find}\:{a}_{{n}} =? \\ $$ Commented…
Question Number 147444 by aliibrahim1 last updated on 20/Jul/21 Answered by SEKRET last updated on 21/Jul/21 $$\int\:\boldsymbol{\mathrm{x}}^{\frac{\mathrm{1}}{\mathrm{2}}} \centerdot\boldsymbol{\mathrm{x}}^{\frac{\mathrm{1}}{\mathrm{2}\centerdot\mathrm{3}}} \centerdot\boldsymbol{\mathrm{x}}^{\frac{\mathrm{1}}{\mathrm{2}\centerdot\mathrm{3}\centerdot\mathrm{4}}} \centerdot….\boldsymbol{\mathrm{dx}}= \\ $$$$\int\:\boldsymbol{\mathrm{x}}^{\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}\centerdot\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{2}\centerdot\mathrm{3}\centerdot\mathrm{4}}+} …\:\mathrm{d}\boldsymbol{\mathrm{x}}=\:\int\:\boldsymbol{\mathrm{x}}^{\frac{\mathrm{1}}{\mathrm{2}!}+\frac{\mathrm{1}}{\mathrm{3}!}+\frac{\mathrm{1}}{\mathrm{4}!}+..} \boldsymbol{\mathrm{dx}} \\…
Question Number 81911 by jagoll last updated on 16/Feb/20 Commented by jagoll last updated on 16/Feb/20 $${a}_{{n}+\mathrm{1}} \:=\:\frac{\mathrm{3}{a}_{{n}} +\mathrm{1}}{\mathrm{2}}\: \\ $$$${a}_{{n}+\mathrm{1}} \:+{A}\:=\:{A}+\frac{\mathrm{3}{a}_{{n}} +\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{3}{a}_{{n}} +\mathrm{2}{A}+\mathrm{1}}{\mathrm{2}} \\…
Question Number 16374 by gourav~ last updated on 21/Jun/17 $${The}\:{Value}\:{of}\:{the}\:{sum}..\:\underset{{n}=\mathrm{1}} {\overset{\mathrm{13}} {\sum}}\left({i}^{{n}} +{i}^{{n}+\mathrm{1}} \right),\:{Where}\:{i}=\sqrt{−\mathrm{1}\:\:} \\ $$$${is}.. \\ $$$$\left({a}.\right)\:{i} \\ $$$$\left({b}.\right)\:{i}−\mathrm{1} \\ $$$$\left({c}.\right)\:−{i} \\ $$$$\left({d}.\right)\:\mathrm{0} \\…
Question Number 16373 by gourav~ last updated on 21/Jun/17 $${if}\:\underset{{k}=\mathrm{0}} {\overset{\mathrm{200}} {\sum}}{i}^{{k}} +\underset{{p}=\mathrm{1}} {\overset{\mathrm{50}} {\prod}}{i}^{{p}} ={x}+{iy}\:{then}..\left({x},{y}\right){is}… \\ $$$${a}.\:\left(\mathrm{0},\mathrm{1}\right) \\ $$$${b}.\:\left(\mathrm{1},−\mathrm{1}\right) \\ $$$${c}.\:\left(\mathrm{2},\mathrm{3}\right) \\ $$$${d}.\:\left(\mathrm{4},\mathrm{8}\right) \\…
Question Number 81906 by ahmadshahhimat775@gmail.com last updated on 16/Feb/20 Commented by mr W last updated on 16/Feb/20 $$\left.{w}\left.{hat}'{s}\:{the}\:{difference}\:{between}\:\mathrm{2}\right)\:{and}\:\mathrm{4}\right)? \\ $$ Terms of Service Privacy Policy…
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Question Number 16364 by Tinkutara last updated on 21/Jun/17 Answered by ajfour last updated on 21/Jun/17 Commented by ajfour last updated on 21/Jun/17 $${In}\:\bigtriangleup{BCG}\:,\:{applying}\:{the}\:{sine}\:{rule}\:: \\…