Question Number 157074 by amin96 last updated on 19/Oct/21 $$\sqrt{\mathrm{4}+\sqrt{\mathrm{4}^{\mathrm{2}} +\sqrt{\mathrm{4}^{\mathrm{3}} +\sqrt{\mathrm{4}^{\mathrm{4}} +\ldots}}}}=? \\ $$ Answered by MJS_new last updated on 19/Oct/21 $$\mathrm{3} \\ $$…
Question Number 157046 by ajfour last updated on 18/Oct/21 $${q}=\left\{{a}+\left(\frac{{a}}{\mathrm{9}}−\frac{\mathrm{1}}{\mathrm{108}}−{a}^{\mathrm{3}} \right)^{\mathrm{1}/\mathrm{3}} \right\}^{\mathrm{1}/\mathrm{3}} \\ $$$$\:\:+\left\{{b}+\left(\frac{{b}}{\mathrm{9}}−\frac{\mathrm{1}}{\mathrm{108}}−{b}^{\mathrm{3}} \right)^{\mathrm{1}/\mathrm{3}} \right\}^{\mathrm{1}/\mathrm{3}} \\ $$$$\:\:\:{a}=\frac{\mathrm{9}+\sqrt{\mathrm{37}}}{\mathrm{72}}\:\:\:,\:\:{b}=\frac{\mathrm{9}−\sqrt{\mathrm{37}}}{\mathrm{72}} \\ $$$$\:{find}\:\boldsymbol{{q}}\:{correct}\:{to}\:\mathrm{5}\:{decimal} \\ $$$$\:{places}. \\ $$ Commented…
Question Number 25954 by abdo imad last updated on 16/Dec/17 $${nswer}\:{to}\:\mathrm{25929}\:\:\:{by}\:{binome}\:{formula}\:\:\: \\ $$$$\left(\mathrm{1}+{x}\right)^{{n}+{m}} \:\:=\sum_{{k}=\mathrm{0}} ^{{k}={n}+{m}} \:{C}_{{n}+{m}} ^{{k}} \:{x}^{{k}} \:\:\:\:{the}\:{coefficent}\:{of}\:{x}^{{n}} \:{is}\: \\ $$$${C}_{{n}+{m}} ^{{n}} \:{and}\:{the}\:{coefficient}\:{of}\:{x}^{{m}} \:\:{is}\:\:\:{C}_{{n}+{m}}…
Question Number 156977 by cortano last updated on 18/Oct/21 $$\:\begin{cases}{{a}_{\mathrm{1}} =\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}\\{{a}_{{n}+\mathrm{1}} =\mathrm{4}{a}_{{n}} ^{\mathrm{3}} −\mathrm{3}{a}_{{n}} \:;\:\forall{n}\geqslant\mathrm{1}}\end{cases} \\ $$$$\:{a}_{{n}} =? \\ $$ Terms of Service Privacy Policy…
Question Number 25738 by yesaditya22@gmail.com last updated on 13/Dec/17 Answered by mrW1 last updated on 13/Dec/17 $$\frac{{a}+{b}}{\:\sqrt{{ab}}}=\mathrm{6} \\ $$$$\frac{\frac{{a}}{{b}}+\mathrm{1}}{\:\sqrt{\frac{{a}}{{b}}}}=\mathrm{6} \\ $$$${let}\:{t}^{\mathrm{2}} =\frac{{a}}{{b}}\geqslant\mathrm{0} \\ $$$$\frac{{t}^{\mathrm{2}} +\mathrm{1}}{{t}}=\mathrm{6}…
Question Number 25726 by shilpa daga last updated on 13/Dec/17 $${x}−\mathrm{3}=\mathrm{11} \\ $$ Commented by tawa tawa last updated on 13/Dec/17 $$\mathrm{x}\:=\:\mathrm{11}\:+\:\mathrm{3} \\ $$$$\mathrm{x}\:=\:\mathrm{14} \\…
Question Number 156676 by mathlove last updated on 14/Oct/21 Commented by MathSh last updated on 14/Oct/21 $$\frac{\mathrm{2}}{\mathrm{a}}\:+\:\frac{\mathrm{1}}{\mathrm{b}}\:+\:\frac{\mathrm{1}}{\mathrm{c}}\:=\:\mathrm{3}\:\Rightarrow\:\frac{\mathrm{21}}{\mathrm{3}}\:=\:\mathrm{7} \\ $$ Answered by som(math1967) last updated on…
Question Number 25543 by soorajkichu last updated on 11/Dec/17 Commented by sushmitak last updated on 11/Dec/17 $$\mathrm{3}{x}+\mathrm{8}=\mathrm{17} \\ $$$$\Rightarrow\mathrm{3}{x}=\mathrm{17}−\mathrm{8}=\mathrm{9} \\ $$$$\Rightarrow{x}=\mathrm{3} \\ $$ Commented by…
Question Number 156495 by chusrang last updated on 11/Oct/21 $$\mathrm{1}.\:\mathrm{Write}\:\mathrm{the}\:\mathrm{first}\:\mathrm{three}\:\mathrm{terms}\:\mathrm{of}\:\mathrm{each}\:\mathrm{of}\:\mathrm{the}\:\mathrm{following}\:\mathrm{sequences}:\: \\ $$$$\left(\mathrm{a}\right)\:\mathrm{a}_{{n}} \:=\:\frac{\mathrm{3}^{{n}} }{\mathrm{2}^{{n}} \:+\:\mathrm{1}}\:\:\:\:\:\:\:\:\left(\mathrm{b}\right)\:\mathrm{a}_{{n}} \:=\:\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} {n}^{\mathrm{3}} \\ $$ Answered by Rasheed.Sindhi last updated on…
Question Number 156450 by cortano last updated on 11/Oct/21 Commented by john_santu last updated on 11/Oct/21 Commented by Tawa11 last updated on 12/Oct/21 $$\mathrm{Great}\:\mathrm{sir}. \\…