Question Number 144922 by mathdanisur last updated on 30/Jun/21 $${if}\:\:{z}^{\mathrm{2}} \:-\:\mathrm{16}\sqrt{{z}}\:=\:\mathrm{12} \\ $$$${find}\:\:{z}\:-\:\mathrm{2}\sqrt{{z}}\:=\:? \\ $$ Answered by liberty last updated on 30/Jun/21 $$\:\mathrm{x}^{\mathrm{4}} −\mathrm{16x}−\mathrm{12}=\mathrm{0} \\…
Question Number 144917 by mathdanisur last updated on 30/Jun/21 $${if}\:\:{x};{y}>\mathrm{0}\:\:{then}: \\ $$$$\mathrm{10}\:\centerdot\:\sqrt{\frac{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }{\mathrm{2}}}\:+\:\frac{\mathrm{8}{xy}}{{x}+{y}}\:\geqslant\:\mathrm{7}{x}+\mathrm{7}{y} \\ $$ Answered by justtry last updated on 30/Jun/21 $${remember}\:: \\…
Question Number 144901 by loveineq last updated on 30/Jun/21 $$\mathrm{Let}\:{a},{b}\:>\:\mathrm{0}\:\mathrm{and}\:{a}+{b}+\mathrm{1}\:=\:\mathrm{3}{ab}.\:\mathrm{Prove}\:\mathrm{that} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\frac{{a}}{{a}^{\mathrm{2}} +\mathrm{1}}+\frac{{b}}{{b}^{\mathrm{2}} +\mathrm{1}}\:\leqslant\:\mathrm{1}\:\leqslant\:\frac{{a}^{\mathrm{3}} }{{a}^{\mathrm{2}} +\mathrm{1}}+\frac{{b}^{\mathrm{3}} }{{b}^{\mathrm{2}} +\mathrm{1}} \\ $$$$ \\ $$$$\mathrm{Let}\:{a},{b}\:>\:\mathrm{0},\:{n}\:\in\:\mathbb{Z}^{+} \:\mathrm{and}\:{a}+{b}+\mathrm{1}\:=\:\mathrm{3}{ab}.\:\mathrm{Prove}\:\mathrm{or}\:\mathrm{disprove} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\frac{{a}^{{n}−\mathrm{1}}…
Question Number 144897 by physicstutes last updated on 30/Jun/21 $$\underset{\mathrm{0}} {\overset{\frac{\left[{x}\right]}{\mathrm{3}}} {\int}}\frac{\mathrm{8}^{{x}} }{\mathrm{2}^{\left[\mathrm{3}{x}\right]} }\:{dx}=\:???\:\mathrm{where}\:\left[.\right]\:\mathrm{is}\:\mathrm{the}\:\mathrm{greatest}\:\mathrm{integer}\:\mathrm{function}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 144888 by imjagoll last updated on 30/Jun/21 Commented by imjagoll last updated on 30/Jun/21 $$\mathrm{find}\:\mathrm{the}\:\mathrm{shaded}\:\mathrm{area}. \\ $$ Answered by nimnim last updated on…
Question Number 144887 by imjagoll last updated on 30/Jun/21 Answered by nimnim last updated on 30/Jun/21 $$=\sqrt{\left(\sqrt{\mathrm{125}+\mathrm{44}}\right)\left(\mathrm{9}\right)+\mathrm{4}}=\sqrt{\left(\sqrt{\mathrm{169}}\right)\left(\mathrm{9}\right)+\mathrm{4}} \\ $$$$=\sqrt{\mathrm{13}\left(\mathrm{9}\right)+\mathrm{4}}=\sqrt{\mathrm{117}+\mathrm{4}}=\sqrt{\mathrm{121}}=\mathrm{11} \\ $$$$\: \\ $$ Terms of…
Question Number 144877 by imjagoll last updated on 30/Jun/21 $$\:\:\mathrm{u}+\sqrt{\mathrm{u}}+\sqrt[{\mathrm{3}}]{\mathrm{u}}+\sqrt[{\mathrm{4}}]{\mathrm{u}}+\sqrt[{\mathrm{5}}]{\mathrm{u}}+…\:+\infty=? \\ $$$$ \\ $$ Answered by MJS_new last updated on 30/Jun/21 $$\forall{u}>\mathrm{0}:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{u}^{\mathrm{1}/{n}} \:=+\infty…
Question Number 79340 by TawaTawa last updated on 24/Jan/20 Commented by john santu last updated on 24/Jan/20 $${let}\:{this}\:{number}\::\:{a},{ar},{ar}^{\mathrm{2}} \\ $$$$\left({i}\right)\:{a}\left(\frac{{r}^{\mathrm{3}} −\mathrm{1}}{{r}−\mathrm{1}}\right)={p}\Rightarrow\frac{{r}^{\mathrm{3}} −\mathrm{1}}{{r}−\mathrm{1}}=\frac{{p}}{{a}} \\ $$$${r}^{\mathrm{2}} +{r}+\mathrm{1}=\frac{{p}}{{a}}\:…
Question Number 144860 by mathdanisur last updated on 29/Jun/21 $$\mid\frac{{x}}{{x}-\mathrm{1}}\mid\:+\:\mid{x}\mid\:=\:\frac{{x}^{\mathrm{2}} }{\mid{x}-\mathrm{1}\mid}\:\:\:{find}\:\:{x}=? \\ $$ Commented by hknkrc46 last updated on 29/Jun/21 $$\bigstar\:\mid\boldsymbol{{x}}\mid\:=\:\frac{\boldsymbol{{x}}^{\mathrm{2}} \:−\:\mid\boldsymbol{{x}}\mid}{\mid\boldsymbol{{x}}\:−\:\mathrm{1}\mid} \\ $$$$\bigstar\:\mid\boldsymbol{{x}}^{\mathrm{2}} \:−\:\boldsymbol{{x}}\mid\:=\:\boldsymbol{{x}}^{\mathrm{2}}…
Question Number 144869 by mathdanisur last updated on 29/Jun/21 $${if}\:\:\mathrm{3}^{\boldsymbol{{z}}} \:=\:\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{5}\sqrt{\mathrm{3}}} \:\centerdot\:\mathrm{3}^{\mathrm{2}} }\:\:{find}\:\:\boldsymbol{{z}}=? \\ $$ Answered by liberty last updated on 30/Jun/21 $$\:\mathrm{3}^{\mathrm{z}} \:=\:\mathrm{3}^{−\left(\mathrm{2}+\mathrm{5}\sqrt{\mathrm{3}}\right)} \:\Rightarrow\mathrm{z}=−\mathrm{2}−\mathrm{5}\sqrt{\mathrm{3}}…