Question Number 30119 by rahul 19 last updated on 16/Feb/18 $$\mathrm{Let}\:\:\mathrm{N}=\:\mathrm{2}^{\mathrm{1224}} \:−\mathrm{1}. \\ $$$$\mathrm{S}=\:\mathrm{2}^{\mathrm{153}} +\mathrm{2}^{\mathrm{77}} +\mathrm{1}. \\ $$$$\mathrm{T}=\:\mathrm{2}^{\mathrm{408}} −\mathrm{2}^{\mathrm{204}} +\mathrm{1}. \\ $$$$\mathrm{then}\:\mathrm{which}\:\mathrm{of}\:\mathrm{the}\:\mathrm{following}\:\mathrm{statment}\:\mathrm{is} \\ $$$$\:\mathrm{correct}? \\…
Question Number 95648 by Abdulrahman last updated on 26/May/20 $$\mathrm{if}\:\mathrm{6}^{\mathrm{x}} =\mathrm{18}\:\mathrm{and}\:\:\mathrm{12}^{\mathrm{y}} =\mathrm{3} \\ $$$$\mathrm{then}\:\:\mathrm{x}=? \\ $$$$\frac{\mathrm{3y}−\mathrm{4}}{\mathrm{y}+\mathrm{4}}\:\:\:\:\:\:\:\:\frac{\mathrm{3y}+\mathrm{2}}{\mathrm{y}+\mathrm{2}}\:\:\:\:\:\:\:\:\frac{\mathrm{3y}−\mathrm{3}}{\mathrm{3y}−\mathrm{3}}\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{3y}+\mathrm{1}}{\mathrm{y}+\mathrm{1}} \\ $$ Commented by john santu last updated on…
Question Number 161177 by azadsir last updated on 13/Dec/21 Answered by Rasheed.Sindhi last updated on 13/Dec/21 $${Let}\:\theta\:{is}\:{the}\:{angle}\:{of}\:\boldsymbol{{Regular}}\:\:\boldsymbol{{n}}-\boldsymbol{{gon}} \\ $$$$\left(\mathrm{180}−\theta\right)×{n}=\mathrm{360} \\ $$$$\theta=\mathrm{180}−\frac{\mathrm{360}}{{n}}=\frac{\mathrm{180}\left({n}−\mathrm{2}\right)}{{n}};\:\:\:\:\:{n}\geqslant\mathrm{3} \\ $$ Terms of…
Question Number 30092 by math1967 last updated on 16/Feb/18 $${If}\:{x}+\frac{\mathrm{1}}{{x}}=\mathrm{3}\:{find}\:{x}^{\mathrm{5}} +\frac{\mathrm{1}}{{x}^{\mathrm{5}} } \\ $$ Answered by math1967 last updated on 16/Feb/18 $${x}+\frac{\mathrm{1}}{{x}}=\mathrm{3}\:\therefore{x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }=\mathrm{7}\:{and}\:{x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}}…
Question Number 161162 by metamorfose last updated on 13/Dec/21 $$\left.{f}\::\right]\mathrm{0},+\infty\left[\rightarrow\right]\mathrm{0},+\infty\left[\:{is}\:{convex}\:{function}\right. \\ $$$${for}\:{n}\geqslant\mathrm{2}\:{an}\:{integer}\:,\:{prove}\:: \\ $$$$\left({f}\left(\mathrm{1}\right)^{{f}\left(\mathrm{1}\right)} {f}\left(\mathrm{2}\right)^{{f}\left(\mathrm{2}\right)} …{f}\left({n}\right)^{{f}\left({n}\right)} \right)^{\frac{\mathrm{1}}{{f}\left(\mathrm{1}\right)+{f}\left(\mathrm{2}\right)+…+{f}\left({n}\right)}} +\left({f}\left(\mathrm{1}\right){f}\left(\mathrm{2}\right)…{f}\left({n}\right)\right)^{\frac{\mathrm{1}}{{n}}} \leqslant{f}\left(\mathrm{1}\right)+{f}\left({n}\right) \\ $$ Terms of Service Privacy…
Question Number 161159 by daus last updated on 13/Dec/21 Answered by Rasheed.Sindhi last updated on 13/Dec/21 $${x}=\frac{\mathrm{2}+{y}^{\mathrm{2}} }{{y}}\:;{y}\neq\mathrm{0} \\ $$$$\left(\frac{\mathrm{2}+{y}^{\mathrm{2}} }{{y}}\right)^{\mathrm{2}} −\mathrm{5}\left(\frac{\mathrm{2}+{y}^{\mathrm{2}} }{{y}}\right){y}+\mathrm{6}{y}^{\mathrm{2}} =\mathrm{0} \\…
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Question Number 161150 by mathlove last updated on 13/Dec/21 Commented by MJS_new last updated on 13/Dec/21 $${y}=\sqrt{\mathrm{2}{x}^{\mathrm{2}} }\geqslant\mathrm{0} \\ $$$${y}=\sqrt{\mathrm{2}}\mid{x}\mid\:\Rightarrow\:{xy}=\sqrt{\mathrm{2}}\mid{x}\mid{x}=\begin{cases}{{x}<\mathrm{0};\:−\sqrt{\mathrm{2}}{x}^{\mathrm{2}} }\\{{x}\geqslant\mathrm{0};\:\sqrt{\mathrm{2}}{x}^{\mathrm{2}} }\end{cases} \\ $$ Terms…
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Question Number 95608 by bobhans last updated on 26/May/20 Commented by john santu last updated on 26/May/20 $$\mathrm{another}\:\mathrm{way} \\ $$$$\mathrm{set}\:{x}+{y}\:=\:{h}\:{and}\:{xy}\:=\:{g} \\ $$$$\Rightarrow{xy}\:\left({x}+{y}\right)\:=\:\mathrm{30}\:\&\:{g}\:+\:{h}\:=\:\mathrm{11} \\ $$$$\Rightarrow{gh}\:=\:\mathrm{30}\:;\:{g}\:=\:\mathrm{11}−{h} \\…