Question Number 147529 by mathdanisur last updated on 21/Jul/21 $${if}\:\:\:{a}_{\boldsymbol{{n}}+\mathrm{1}} \:=\:\sqrt{{a}_{\mathrm{1}} \:+\:{a}_{\boldsymbol{{n}}} } \\ $$$${find}\:\:\underset{\boldsymbol{{x}}\rightarrow\infty} {{lim}a}_{\boldsymbol{{n}}} \:=\:? \\ $$ Commented by mathdanisur last updated on…
Question Number 147500 by mathdanisur last updated on 21/Jul/21 $${x}^{\mathrm{2}} \:-\:{y}^{\mathrm{2}} \:+\:\mathrm{2}{x}\:=\:\mathrm{22} \\ $$$${what}\:{is}\:{the}\:{number}\:{of}\:{complete} \\ $$$${solutions}\:{that}\:{satisf}\:{the}\:{equation} \\ $$$$\left({x};{y}\right).? \\ $$ Commented by Mrsof last updated…
Question Number 147488 by mathdanisur last updated on 21/Jul/21 $${if}\:\:{x};{y};{z}\geqslant\mathrm{1}\:\:{then}: \\ $$$$\frac{\mathrm{1}}{\mathrm{3}{xy}−\mathrm{1}}\:+\:\frac{\mathrm{1}}{\mathrm{3}{yz}−\mathrm{1}}\:+\:\frac{\mathrm{1}}{\mathrm{3}{zx}−\mathrm{1}}\:\geqslant\:\frac{\mathrm{3}}{\mathrm{2}{xyz}} \\ $$ Answered by mindispower last updated on 21/Jul/21 $${in}\:{firt} \\ $$$$\mathrm{1}\leqslant{xy},{xy}=\frac{{xyz}}{{z}} \\…
Question Number 81944 by M±th+et£s last updated on 16/Feb/20 $${show}\:{that}\: \\ $$$${cot}\left(\mathrm{40}°\right)−{cot}\left(\mathrm{50}°\right)=\mathrm{2}{tan}\left(\mathrm{10}°\right) \\ $$$${cos}\left(\mathrm{70}°\right)\:{cos}\left(\mathrm{50}^{°} \right)\:{cos}\left(\mathrm{10}^{°} \right)=\frac{\sqrt{\mathrm{3}}}{\mathrm{8}} \\ $$ Answered by TANMAY PANACEA last updated on…
Question Number 81942 by TawaTawa last updated on 16/Feb/20 Answered by TANMAY PANACEA last updated on 16/Feb/20 $${for}\:{bank}\:{p}\:{interest}=\frac{\mathrm{3000}×\mathrm{1}×{x}}{\mathrm{100}}=\mathrm{30}{x} \\ $$$${for}\:{bank}\:{Q}\:{interst}=\frac{\mathrm{2000}×\mathrm{1}×{y}}{\mathrm{100}}=\mathrm{20}{y} \\ $$$${as}\:{per}\:{second}\:{condition} \\ $$$${interest}\:{in}\:{bank}\:{p}=\frac{\mathrm{2000}×\mathrm{1}×{x}}{\mathrm{100}}=\mathrm{20}{x} \\…
Question Number 81943 by TawaTawa last updated on 16/Feb/20 $$\mathrm{Evaluate}:\:\:\:\:\left(\frac{\sqrt{\mathrm{30}\:+\:\sqrt{\mathrm{8}}\:+\:\sqrt{\mathrm{5}}}}{\:\sqrt{\mathrm{8}}\:+\:\sqrt{\mathrm{5}}}\right)^{\mathrm{1}/\mathrm{4}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 147465 by gsk2684 last updated on 21/Jul/21 $${if}\:{n}>\mathrm{2},{n}\in{N}\:{then}\:{prove}\:{that}\: \\ $$$$\left\{\left(\mathrm{2}{n}−\mathrm{1}\right)^{{n}} +\left(\mathrm{2}{n}\right)^{{n}} \right\}<\left(\mathrm{2}{n}+\mathrm{1}\right)^{{n}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 147453 by EDWIN88 last updated on 21/Jul/21 $$\:\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}\right)\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{9}}\right)\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{16}}\right)\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{25}}\right)…=? \\ $$ Commented by gsk2684 last updated on 21/Jul/21 $$\:\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}\right)\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{9}}\right)\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{16}}\right)\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{25}}\right)…=? \\ $$$$\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\right)\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\right)\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}\right)\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}\right)\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}\right)\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}}\right)… \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\:\frac{\mathrm{3}}{\mathrm{2}}\:\frac{\mathrm{2}}{\mathrm{3}}\:\frac{\mathrm{4}}{\mathrm{3}}\:\frac{\mathrm{3}}{\mathrm{4}}\:\frac{\mathrm{5}}{\mathrm{4}}\:… \\…
Question Number 81910 by mr W last updated on 16/Feb/20 $${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 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}}} \\…