Question Number 124041 by pticantor last updated on 30/Nov/20 $$ \\ $$$$ \\ $$$$ \\ $$$$\:\:\:\:\:\boldsymbol{{A}}_{{n}} =\left(\mathrm{1}+\sqrt{\mathrm{5}}\right)^{\boldsymbol{{n}}} −\left(\mathrm{1}−\sqrt{\mathrm{5}}\right)^{\boldsymbol{{n}}} \\ $$$$ \\ $$$$\:\:\:\boldsymbol{{A}}_{\boldsymbol{{n}}} =???? \\ $$…
Question Number 123826 by pticantor last updated on 28/Nov/20 $$ \\ $$$$\boldsymbol{{W}}_{\boldsymbol{{n}}} =\underset{\boldsymbol{{k}}=\mathrm{1}} {\overset{\mathrm{2}\boldsymbol{{n}}} {\sum}}\frac{\boldsymbol{{k}}}{\boldsymbol{{n}}^{\mathrm{2}} +\boldsymbol{{k}}^{\mathrm{2}} } \\ $$$$\boldsymbol{{montrer}}\:\boldsymbol{{que}}\:\boldsymbol{{W}}_{\boldsymbol{{n}}} \:\boldsymbol{{converge}} \\ $$$$\boldsymbol{{et}}\:\boldsymbol{{calculer}}\:\boldsymbol{{la}}\:\boldsymbol{{valeur}}\:\boldsymbol{{de}}\:\boldsymbol{{W}}_{\boldsymbol{{n}}} \\ $$ Commented…
Question Number 119150 by abdul88 last updated on 22/Oct/20 $${Given}\:{that}\:{a},\:{b}\:{and}\:{c}\:{are}\:{real}\:{numbers}\: \\ $$$${that}\:{stisfy}\:{the}\:{system}\:{of}\:{equation}\:{above} \\ $$$$ \\ $$$${a}\:−\:\sqrt{{b}^{\mathrm{2}} \:−\:\frac{\mathrm{1}}{\mathrm{16}}\:}\:=\:\sqrt{{c}^{\mathrm{2}} \:−\:\frac{\mathrm{1}}{\mathrm{16}}\:}\: \\ $$$${b}\:−\:\sqrt{{c}^{\mathrm{2}} \:−\:\frac{\mathrm{1}}{\mathrm{25}}}\:=\:\sqrt{{a}^{\mathrm{2}} \:−\:\frac{\mathrm{1}}{\mathrm{25}}\:} \\ $$$${c}\:−\:\sqrt{{a}^{\mathrm{2}} \:−\:\frac{\mathrm{1}}{\mathrm{36}}\:}\:=\:\sqrt{{b}^{\mathrm{2}}…
Question Number 182854 by manxsol last updated on 16/Dec/22 $$\left({sin}\mathrm{54}\right)^{\mathrm{10}} ={x}_{\mathrm{1}} \:\:\:\:{x}_{\mathrm{1}\:\:} {is}\:{root} \\ $$$${ax}^{\mathrm{2}} +{bx}+{c}=\mathrm{0} \\ $$$$\:\:{a},{b},{c}\:\:\:{is}\:{integer} \\ $$$${a}=? \\ $$ Commented by manxsol…
Question Number 182627 by manxsol last updated on 12/Dec/22 $$\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{\mathrm{15}} =\frac{{a}+{b}\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$${find}\:{a}+{b} \\ $$ Answered by Rasheed.Sindhi last updated on 12/Dec/22 $$\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{\mathrm{15}} =\frac{{a}+{b}\sqrt{\mathrm{5}}}{\mathrm{2}}\:;\:{a}+{b}=? \\…
Question Number 181400 by Engr_Jidda last updated on 24/Nov/22 Commented by Engr_Jidda last updated on 24/Nov/22 $${Please}\:{help}\:{me}\:{out} \\ $$ Commented by manxsol last updated on…
Question Number 179857 by Asad123 last updated on 03/Nov/22 Commented by manxsol last updated on 03/Nov/22 $$\mathrm{2}{d}\:{line}\:\:{interesante}\:{game}\:{completar}\:{a}\:\mathrm{12} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 113901 by frc2crc last updated on 16/Sep/20 $${when}\:{do}\:{I}\:{use}\:\mid{x}\mid \\ $$$${for}\:\sqrt{{x}^{\mathrm{2}} }? \\ $$ Commented by MJS_new last updated on 16/Sep/20 $$\mathrm{almost}\:\mathrm{always} \\ $$$${p}^{\mathrm{2}}…
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Question Number 178652 by doline last updated on 19/Oct/22 $${calculer}\:{la}\:{branche}\:{infinie}\:{de}\sqrt{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{4}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com