Question Number 27094 by abdo imad last updated on 02/Jan/18 $${if}\:\mathrm{1}+{x}+{x}^{\mathrm{2}} =\mathrm{0}\:{find}\:{the}\:{value}\:{of}\: \\ $$$${A}=\:\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{6}} \:+\left(\:{x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)^{\mathrm{6}} \:\:+…\:\left(\:\:{x}^{\mathrm{100}} +\frac{\mathrm{1}}{{x}^{\mathrm{100}} }\right)^{\mathrm{6}} \:. \\ $$ Commented by…
Question Number 92608 by jagoll last updated on 08/May/20 Commented by john santu last updated on 08/May/20 $$\mathrm{g}\left(\mathrm{t}\right)\:=\:\begin{cases}{−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{t}\:,−\mathrm{4}\leqslant\mathrm{t}\leqslant\mathrm{0}}\\{−\sqrt{\mathrm{9}−\left(\mathrm{t}−\mathrm{3}\right)^{\mathrm{2}} },\:\mathrm{0}\leqslant\mathrm{t}\leqslant\mathrm{6}}\\{\mathrm{3}\:,\:\mathrm{6}<\mathrm{t}\leqslant\mathrm{8}}\end{cases} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{3x}+\underset{\mathrm{0}} {\overset{\mathrm{x}} {\int}}\:\mathrm{g}\left(\mathrm{t}\right)\:\mathrm{dt}\: \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{3x}+\begin{cases}{−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{x}^{\mathrm{2}}…
Question Number 27061 by math solver last updated on 01/Jan/18 Commented by math solver last updated on 02/Jan/18 $${pLz}\:{help}? \\ $$ Answered by prakash jain…
Question Number 27060 by math solver last updated on 01/Jan/18 Commented by math solver last updated on 02/Jan/18 $${plz}\:{help}? \\ $$$$ \\ $$ Answered by…
Question Number 27059 by math solver last updated on 01/Jan/18 Commented by prakash jain last updated on 01/Jan/18 $$\mathrm{1}+{z}+{z}^{\mathrm{2}} ={x}−{xz}+{xz}^{\mathrm{2}} \\ $$$${z}^{\mathrm{2}} \left(\mathrm{1}−{x}\right)+{z}\left(\mathrm{1}+{x}\right)+\left(\mathrm{1}−{x}\right)=\mathrm{0} \\ $$$$\mathrm{Given}\:{z}\:\mathrm{is}\:\mathrm{not}\:\mathrm{real},\:\mathrm{root}\:\mathrm{are}…
Question Number 158124 by HongKing last updated on 31/Oct/21 $$\mathrm{let}\:\boldsymbol{\omega}\:\mathrm{be}\:\mathrm{a}\:\mathrm{root}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\mathrm{x}^{\mathrm{4}} \:+\:\left(\mathrm{x}\:-\:\mathrm{1}\right)^{\mathrm{4}} \:+\:\mathrm{1}\:=\:\mathrm{0} \\ $$$$\mathrm{find}\:\:\boldsymbol{\Omega}\:=\:\boldsymbol{\omega}^{\mathrm{300}} \:+\:\boldsymbol{\omega}^{\mathrm{301}} \\ $$ Answered by Rasheed.Sindhi last updated on…
Question Number 158118 by HongKing last updated on 31/Oct/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 27046 by ajfour last updated on 01/Jan/18 $${Considering}\:\boldsymbol{{y}}=\boldsymbol{{x}}^{\mathrm{3}} +\boldsymbol{{px}}+\boldsymbol{{q}} \\ $$$${If}\:\:\:\:\:\frac{{dy}}{{dx}}\mid_{{x}=\alpha} =\mathrm{0}\:\:\Rightarrow\:\:\alpha^{\mathrm{2}} =−\frac{{p}}{\mathrm{3}} \\ $$$${if}\:\:\:\frac{{d}\left({y}/{x}\right)}{{dx}}\mid_{{x}=\beta} =\mathrm{0}\:\:\:\Rightarrow\:\beta^{\:\mathrm{3}} =\frac{{q}}{\mathrm{2}} \\ $$$${roots}\:{of}\:{the}\:{cubic}\:\:{eq}^{{n}} \:{are}: \\ $$$$\:\:\:\:{x}=\left[−\beta^{\:\mathrm{3}} \pm\sqrt{\beta^{\:\mathrm{6}}…
Question Number 158114 by cortano last updated on 31/Oct/21 $$\:\begin{cases}{{x}^{\mathrm{2}} −\mathrm{3}{xy}+\mathrm{2}{y}^{\mathrm{2}} =\mathrm{35}}\\{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\:\mathrm{13}}\end{cases} \\ $$$$\:\Rightarrow{x}=?\:\wedge\:{y}=?\: \\ $$ Commented by bobhans last updated on 31/Oct/21…
Question Number 158104 by zainaltanjung last updated on 31/Oct/21 $$ \\ $$The comparison between Rahman and Aditya's books is 2: 3. If the number of…