Question Number 36691 by Tinkutara last updated on 04/Jun/18 Commented by Rasheed.Sindhi last updated on 05/Jun/18 $${ax}+{cy}+{bz}={X},{cx}+{by}+{az}={Y},{bx}+{ay}+{cz}={Z} \\ $$$$\mathrm{Adding}\:\mathrm{the}\:\mathrm{three}: \\ $$$$\left({a}+{b}+{c}\right)\left({x}+{y}+{z}\right)={X}+{Y}+{Z} \\ $$$$…… \\ $$$$….…
Question Number 36692 by math2018 last updated on 04/Jun/18 $${x}^{\mathrm{3}} +{y}^{\mathrm{3}} =\mathrm{5} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{3} \\ $$ Commented by behi83417@gmail.com last updated on 04/Jun/18…
Question Number 167750 by mathlove last updated on 24/Mar/22 Answered by Rasheed.Sindhi last updated on 24/Mar/22 $${f}\left({x}−{y}\right)={f}\left({x}\right)−{f}\left({y}\right);\:{f}\left(\mathrm{2}\right)=\mathrm{5};\:{f}\left(\mathrm{16}\right)=? \\ $$$$\bullet{x}=\mathrm{2}{y} \\ $$$${f}\left(\mathrm{2}{y}−{y}\right)={f}\left(\mathrm{2}{y}\right)−{f}\left({y}\right) \\ $$$${f}\left({y}\right)={f}\left(\mathrm{2}{y}\right)−{f}\left({y}\right) \\ $$$$\mathrm{2}{f}\left({y}\right)={f}\left(\mathrm{2}{y}\right)…
Question Number 167746 by mathlove last updated on 24/Mar/22 $${log}_{\left({x}^{\mathrm{2}} +\mathrm{2}\right)} \left({x}^{\mathrm{2}} +\mathrm{4}{x}\right)=? \\ $$ Commented by MJS_new last updated on 24/Mar/22 $$\mathrm{log}_{{x}^{\mathrm{2}} +\mathrm{2}} \:\left({x}^{\mathrm{2}}…
Question Number 36676 by tawa tawa last updated on 04/Jun/18 $$\mathrm{if}\:\:\mathrm{z}\:=\:−\:\mathrm{27},\:\:\mathrm{find}\:\mathrm{all}\:\mathrm{the}\:\mathrm{root}\:\mathrm{of}\:\mathrm{z}\:\mathrm{in}\:\mathrm{complex}\:\mathrm{plain} \\ $$ Commented by abdo mathsup 649 cc last updated on 04/Jun/18 $${the}\:{roots}\:{are}?{the}\:{comlex}\:{z}\:/\:{z}^{\mathrm{2}} \:=−\mathrm{27}\:=\left({i}\sqrt{\mathrm{27}}\right)^{\mathrm{2}}…
Question Number 167737 by HongKing last updated on 24/Mar/22 $$\mathrm{3}\:\:\:\Box\:\:\:\mathrm{4}\:\:\:\rightarrow\:\:\:\mathrm{27} \\ $$$$\mathrm{4}\:\:\:\Box\:\:\:\mathrm{2}\:\:\:\rightarrow\:\:\:\mathrm{36} \\ $$$$\mathrm{2}\:\:\:\Box\:\:\:\mathrm{7}\:\:\:\rightarrow\:\:\:\mathrm{18} \\ $$$$\mathrm{1}\:\:\:\Box\:\:\:\mathrm{9}\:\:\:\rightarrow\:\:\:? \\ $$ Answered by Rasheed.Sindhi last updated on 24/Mar/22…
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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}}…
Question Number 102186 by bemath last updated on 07/Jul/20 Terms of Service Privacy Policy Contact: info@tinkutara.com