Question Number 158220 by ajfour last updated on 01/Nov/21 $$\:\:\:{source}:\:{myself} \\ $$$$\:{x}^{\mathrm{5}} +\mathrm{3}{cx}^{\mathrm{2}} −{x}−\mathrm{5}{c}=\mathrm{0} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 158185 by HongKing last updated on 31/Oct/21 Commented by MJS_new last updated on 31/Oct/21 $$\mathrm{obviously}\:{x}=\mathrm{1}\:\mathrm{is}\:\mathrm{a}\:\mathrm{solution} \\ $$ Commented by HongKing last updated on…
Question Number 158187 by HongKing last updated on 31/Oct/21 $$\mathrm{Solve}\:\mathrm{for}\:\mathrm{real}\:\mathrm{numbers}: \\ $$$$\begin{cases}{\sqrt{\mathrm{x}}\:-\:\mathrm{y}^{\mathrm{5}} \:=\:\mathrm{3}}\\{\sqrt[{\mathrm{5}}]{\sqrt{\mathrm{x}}\:-\:\mathrm{3}}\:-\:\sqrt[{\mathrm{5}}]{\mathrm{y}^{\mathrm{5}} \:+\:\mathrm{6}}\:=\:-\:\mathrm{1}}\end{cases}\: \\ $$$$ \\ $$ Answered by MJS_new last updated on 31/Oct/21…
Question Number 27112 by NECx last updated on 02/Jan/18 Commented by NECx last updated on 02/Jan/18 $${please}\:{help}\:{with}\:{the}\:{both}…\:{Thanks} \\ $$ Commented by prakash jain last updated…
Question Number 27103 by ktomboy1992 last updated on 02/Jan/18 $$\mathrm{the}\:\mathrm{intrest}\:\mathrm{on}\:\mathrm{a}\:\mathrm{certain}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{money}\:\mathrm{at}\:\mathrm{the} \\ $$$$\mathrm{end}\:\mathrm{of}\:\mathrm{6}.\mathrm{25}\:\mathrm{year}\:\mathrm{was}\:\frac{\mathrm{5}}{\mathrm{16}}\:\mathrm{of}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{itself}.\mathrm{what} \\ $$$$\mathrm{is}\:\mathrm{the}\:\mathrm{rate}\:\mathrm{percent}? \\ $$ Commented by mrW1 last updated on 02/Jan/18 $$\Rightarrow\:{Q}\mathrm{27016} \\…
Question Number 27094 by abdo imad last updated on 02/Jan/18 $${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}…