Question Number 102201 by Akeyz last updated on 07/Jul/20 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 167738 by HongKing last updated on 24/Mar/22 $$\mathrm{72}\:\:\:\rightarrow\:\:\:\mathrm{49} \\ $$$$\mathrm{42}\:\:\:\rightarrow\:\:\:\mathrm{16} \\ $$$$\mathrm{21}\:\:\:\rightarrow\:\:\:\mathrm{2} \\ $$$$\mathrm{16}\:\:\:\rightarrow\:\:\:? \\ $$ Answered by malwan last updated on 24/Mar/22…
Question Number 36660 by Tinkutara last updated on 03/Jun/18 Answered by tanmay.chaudhury50@gmail.com last updated on 04/Jun/18 $$\frac{\mathrm{1}}{{a}+{w}}+\frac{\mathrm{1}}{{b}+{w}}+\frac{\mathrm{1}}{{c}+{w}}+\frac{\mathrm{1}}{{d}+{w}}=\frac{\mathrm{2}{w}^{\mathrm{3}} }{{w}}=\frac{\mathrm{2}}{{w}} \\ $$$$\frac{\mathrm{1}}{{a}+{w}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}+{w}^{\mathrm{2}} }+\frac{\mathrm{1}}{{c}+{w}^{\mathrm{2}} }+\frac{\mathrm{1}}{{d}+{w}^{\mathrm{2}} }=\frac{\mathrm{2}{w}^{\mathrm{3}} }{{w}^{\mathrm{2}}…
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Question Number 167722 by SAMIRA last updated on 23/Mar/22 $$\frac{\mathrm{2}}{\boldsymbol{\mathrm{Z}}}\:=\:\frac{\mathrm{1}}{\boldsymbol{\mathrm{a}}}+\frac{\mathrm{1}}{\boldsymbol{\mathrm{b}}} \\ $$$$\boldsymbol{\mathrm{Z}}\:=\:???\:{help} \\ $$ Commented by MJS_new last updated on 23/Mar/22 $$\mathrm{law}\:\mathrm{of}\:\mathrm{adding}\:\mathrm{fractions} \\ $$$$\frac{{o}}{{p}}+\frac{{q}}{{r}}=\frac{{o}×{r}}{{p}×{r}}+\frac{{p}×{q}}{{p}×{r}}=\frac{{o}×{r}+{p}×{q}}{{p}×{r}} \\…
Question Number 167718 by mathlove last updated on 23/Mar/22 $$\left({x}−\mathrm{3}\right)^{\mathrm{2}} +\left({y}−\mathrm{5}\right)^{\mathrm{2}} +\left({z}−\mathrm{4}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\frac{{x}^{\mathrm{2}} }{\mathrm{9}}+\frac{{y}^{\mathrm{2}} }{\mathrm{25}}+\frac{{z}^{\mathrm{2}} }{\mathrm{16}}=? \\ $$$${help}\:{me} \\ $$ Commented by MJS_new…
Question Number 102178 by dw last updated on 07/Jul/20 Answered by 1549442205 last updated on 07/Jul/20 $$\mathrm{Putting}\:\mathrm{x}=\mathrm{tan}\alpha,\mathrm{y}=\mathrm{tan}\beta\:\mathrm{we}\:\mathrm{have} \\ $$$$\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }=\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \alpha}=\frac{\mathrm{1}}{\mathrm{cos}\alpha},\sqrt{\mathrm{1}+\mathrm{y}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{cos}\beta} \\ $$$$\mathrm{x}\sqrt{\mathrm{1}+\mathrm{y}^{\mathrm{2}} }+\mathrm{y}\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}}…
Question Number 102179 by dw last updated on 07/Jul/20 Answered by 1549442205 last updated on 07/Jul/20 Commented by dw last updated on 07/Jul/20 $${Thanks}\:{sir}!\:{Very}\:{good}\:{solution}. \\…
Question Number 167700 by HongKing last updated on 23/Mar/22 $$\mathrm{Evalvate}\:\mathrm{the}\:\mathrm{following}\:\mathrm{integrals}: \\ $$$$\mathrm{1}.\:\int\sqrt{\mathrm{x}^{\mathrm{3}} \:+\:\mathrm{6x}^{\mathrm{2}} \:+\:\mathrm{9x}\:\mathrm{dx}} \\ $$$$\mathrm{2}.\:\int\:\frac{\left(\mathrm{x}\:-\:\mathrm{6}\right)\:\mathrm{dx}}{\mathrm{x}^{\mathrm{4}} \:-\:\mathrm{24x}\:+\:\mathrm{3}} \\ $$ Answered by MJS_new last updated on…
Question Number 167699 by HongKing last updated on 23/Mar/22 Answered by alephzero last updated on 23/Mar/22 $$\mathrm{1}.\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}−\mathrm{cos}\:{x}}{{x}\:\mathrm{sin}\:{x}}\:\overset{\mathrm{use}\:\mathrm{L}'\mathrm{H}\hat {\mathrm{o}pital}'\mathrm{s}\:\mathrm{rule}} {=} \\ $$$$=\:\mathrm{lim}\:\frac{\mathrm{sin}\:{x}}{\mathrm{sin}\:{x}+{x}\:\mathrm{cos}\:{x}}\:\overset{\mathrm{again}} {=} \\ $$$$=\:\mathrm{lim}\:\frac{\mathrm{cos}\:{x}}{\mathrm{cos}\:{x}−\mathrm{sin}\:{x}+\mathrm{cos}\:{x}}\:=…