Question Number 9953 by konen last updated on 18/Jan/17 $$\mathrm{x}−\mathrm{y}\:=\mathrm{8} \\ $$$$\mathrm{xy}=\mathrm{7}\:\:\Rightarrow\:\mathrm{x}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} \:=? \\ $$ Answered by mrW1 last updated on 19/Jan/17 $$\left({x}−{y}\right)^{\mathrm{2}} =\mathrm{8}^{\mathrm{2}}…
Question Number 141012 by ajfour last updated on 14/May/21 $$\:{s}^{\mathrm{2}} \left({s}+\mathrm{1}\right)^{\mathrm{2}} +\frac{{c}}{\mathrm{2}}{s}\left({s}+\mathrm{1}\right)^{\mathrm{2}} −\frac{{c}^{\mathrm{2}} }{\mathrm{2}}\left({s}+\mathrm{1}\right) \\ $$$$\:\:+\frac{{c}^{\mathrm{3}} }{\mathrm{2}}\:=\:\mathrm{0}\:\:\:\:\:{solve}\:{for}\:{s}\:{in}\:{terms} \\ $$$${of}\:\:\:\:\mathrm{0}<{c}<\frac{\mathrm{2}}{\mathrm{3}\sqrt{\mathrm{3}}}\:. \\ $$ Terms of Service Privacy…
Question Number 141004 by loveineq last updated on 14/May/21 $$\mathrm{Let}\:\mathrm{0}\:\leqslant\:{a},{b}\:<\:\mathrm{1}.\:\mathrm{Prove}\:\mathrm{that} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{4}}\centerdot\frac{\left(\mathrm{2}−{a}\right)\left(\mathrm{2}−{b}\right)}{\left(\mathrm{1}−{a}\right)\left(\mathrm{1}−{b}\right)}\:\geqslant\:\frac{\mathrm{4}+{a}+{b}}{\mathrm{4}−{a}−{b}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$ \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 141002 by EnterUsername last updated on 14/May/21 $$\mathrm{If}\:{a}>\mathrm{0}\:\mathrm{and}\:\mathrm{one}\:\mathrm{root}\:\mathrm{of}\:{a}\mathrm{x}^{\mathrm{2}} +{b}\mathrm{x}+\mathrm{c}=\mathrm{0}\:\mathrm{is}\:\mathrm{less}\:\mathrm{than}\:−\mathrm{2} \\ $$$$\mathrm{and}\:\mathrm{the}\:\mathrm{other}\:\mathrm{is}\:\mathrm{greater}\:\mathrm{than}\:\mathrm{2},\:\mathrm{then} \\ $$$$\left(\mathrm{A}\right)\:\mathrm{4}{a}+\mathrm{2}\mid{b}\mid+{c}<\mathrm{0} \\ $$$$\left(\mathrm{B}\right)\:\mathrm{4}{a}+\mathrm{2}\mid{b}\mid+{c}>\mathrm{0} \\ $$$$\left(\mathrm{C}\right)\:\mathrm{4}{a}+\mathrm{2}\mid{b}\mid+{c}=\mathrm{0} \\ $$$$\left(\mathrm{D}\right)\:{a}+{b}={c} \\ $$ Answered by…
Question Number 140997 by loveineq last updated on 14/May/21 $$\mathrm{Let}\:{a},{b},{c}\:\geqslant\:\mathrm{0}.\:\mathrm{Prove}\:\mathrm{that} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{8}}\centerdot\frac{\left(\mathrm{2}+{a}\right)\left(\mathrm{2}+{b}\right)\left(\mathrm{2}+{c}\right)}{\left(\mathrm{1}+{a}\right)\left(\mathrm{1}+{b}\right)\left(\mathrm{1}+{c}\right)}\:\geqslant\:\frac{\mathrm{4}−{a}−{b}−{c}}{\mathrm{4}+{a}+{b}+{c}}\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$ \\ $$ Commented by MJS_new last updated on 15/May/21 $$\mathrm{as}\:\mathrm{easy}\:\mathrm{as}\:\mathrm{the}\:\mathrm{last}\:\mathrm{one} \\…
Question Number 75456 by peter frank last updated on 11/Dec/19 $${Find}\:{the}\:{domain}\:{and}\: \\ $$$${range}\:{of}\:{relation} \\ $$$$\left({a}\right){R}=\left\{\left({x},{y}\right):{y}=\sqrt{{x}^{\mathrm{2}} −\mathrm{6}}\:\right\} \\ $$$$\left({b}\right){R}=\left\{\left({x},{y}\right):{x}^{\mathrm{2}} −{y}^{\mathrm{2}} =\mathrm{1}\:\right\} \\ $$ Answered by Kunal12588…
Question Number 9923 by konen last updated on 16/Jan/17 $$\mathrm{i}=\sqrt{−\mathrm{1}} \\ $$$$\mathrm{z}=\:\left(\mathrm{1}−\mathrm{i}\right)^{\mathrm{17}} −\left(\mathrm{1}+\mathrm{i}\right)^{\mathrm{17}} \:\Rightarrow\:\mathrm{z}=?\: \\ $$ Commented by RasheedSoomro last updated on 16/Jan/17 $$\mathrm{z}=\:\left(\mathrm{1}−\mathrm{i}\right)^{\mathrm{17}} −\left(\mathrm{1}+\mathrm{i}\right)^{\mathrm{17}}…
Question Number 75457 by Master last updated on 11/Dec/19 Commented by MJS last updated on 11/Dec/19 $$\mathrm{no}\:\mathrm{exact}\:\mathrm{solution}\:\mathrm{possible} \\ $$$$\mathrm{try}\:\mathrm{to}\:\mathrm{approximate},\:\mathrm{there}\:\mathrm{are}\:\mathrm{3}\:\mathrm{solutions}, \\ $$$$\mathrm{one}\:\mathrm{real}\:\mathrm{and}\:\mathrm{two}\:\mathrm{conjugated}\:\mathrm{complex} \\ $$ Commented by…
Question Number 140981 by EnterUsername last updated on 14/May/21 $$\mathrm{Let}\:\alpha\neq\mathrm{1}\:\mathrm{and}\:\alpha^{\mathrm{13}} =\mathrm{1}.\:\mathrm{If}\:{a}=\alpha+\alpha^{\mathrm{3}} +\alpha^{\mathrm{4}} +\alpha^{−\mathrm{4}} +\alpha^{−\mathrm{3}} + \\ $$$$\alpha^{−\mathrm{1}} \:\mathrm{and}\:{b}=\alpha^{\mathrm{2}} +\alpha^{\mathrm{5}} +\alpha^{\mathrm{6}} +\alpha^{−\mathrm{6}} +\alpha^{−\mathrm{5}} +\alpha^{−\mathrm{2}} \:\mathrm{then}\:\mathrm{the} \\…
Question Number 140973 by EnterUsername last updated on 14/May/21 $$\mathrm{If}\:\alpha\:\mathrm{and}\:\beta\:\mathrm{are}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{2x}^{\mathrm{2}} +{a}\mathrm{x}+{b}=\mathrm{0}, \\ $$$$\mathrm{then}\:\mathrm{one}\:\mathrm{of}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{2}\left(\alpha\mathrm{x}+\beta\right)^{\mathrm{2}} + \\ $$$${a}\left(\alpha\mathrm{x}+\beta\right)+{b}=\mathrm{0}\:\mathrm{is} \\ $$$$\left(\mathrm{A}\right)\:\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{B}\right)\:\frac{\alpha+\mathrm{2}{b}}{\alpha^{\mathrm{2}} } \\ $$$$\left(\mathrm{C}\right)\:\frac{{a}\alpha+{b}}{\mathrm{2}\alpha^{\mathrm{2}} }\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{D}\right)\:\frac{{a}\alpha−\mathrm{2}{b}}{\mathrm{2}\alpha^{\mathrm{2}} } \\ $$…