Question Number 22215 by gopikrishnan005@gmail.com last updated on 13/Oct/17 $${solve}\:{the}\:{inequation}\:−{x}^{\mathrm{2}} +\mathrm{3}{x}−\mathrm{2}>\mathrm{0} \\ $$ Answered by ajfour last updated on 13/Oct/17 $${x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{2}<\mathrm{0} \\ $$$$\left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right)<\mathrm{0} \\…
Question Number 22193 by NECx last updated on 13/Oct/17 $$\mathrm{A}\:\mathrm{body}\:\mathrm{of}\:\mathrm{mass}\:\mathrm{0}.\mathrm{1kg}\:\mathrm{dropped}\: \\ $$$$\mathrm{from}\:\mathrm{a}\:\mathrm{height}\:\mathrm{of}\:\mathrm{8m}\:\mathrm{onto}\:\mathrm{a}\:\mathrm{hard} \\ $$$$\mathrm{floor}\:\mathrm{and}\:\mathrm{bounces}\:\mathrm{back}\:\mathrm{to}\:\mathrm{a}\:\mathrm{height} \\ $$$$\mathrm{of}\:\mathrm{2m}.\:\mathrm{Calculate}\:\mathrm{the}\:\mathrm{chaange}\:\mathrm{in} \\ $$$$\mathrm{momentum}.\mathrm{If}\:\mathrm{the}\:\mathrm{body}\:\mathrm{is}\:\mathrm{in} \\ $$$$\mathrm{contact}\:\mathrm{with}\:\mathrm{the}\:\mathrm{floor}\:\mathrm{for}\:\mathrm{0}.\mathrm{1s}, \\ $$$$\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{force}\:\mathrm{exerted}\:\mathrm{on}\:\mathrm{the} \\ $$$$\mathrm{body}?\left(\mathrm{g}=\mathrm{10m}/\mathrm{s}^{\mathrm{2}} \right)…
Question Number 153257 by naka3546 last updated on 06/Sep/21 $${Find}\:\:{set}\:\:{of}\:\:{k}\:\:{value}\:\:{so}\:\:{that} \\ $$$$\:\:\:\:\:\:\:\mid{x}\mid\:+\:\mid{x}−\mathrm{1}\mid\:+\:\mid{x}−\mathrm{4}\mid\:=\:{k} \\ $$$${a}.\:{has}\:\:{one}\:\:{solution} \\ $$$${b}.\:{has}\:\:{two}\:\:{solutions} \\ $$$${c}.\:{has}\:\:{many}\:\:{solutions} \\ $$$${d}.\:{has}\:\:{no}\:\:{solution} \\ $$ Answered by ajfour…
Question Number 153249 by SANOGO last updated on 06/Sep/21 Answered by qaz last updated on 06/Sep/21 $$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left(\mathrm{l}−\mathrm{t}^{\mathrm{2}} \right)}{\mathrm{t}^{\mathrm{2}} }\mathrm{dt} \\ $$$$=−\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{n}}\int_{\mathrm{0}}…
Question Number 22174 by Adoy last updated on 12/Oct/17 Answered by $@ty@m last updated on 13/Oct/17 $$ \\ $$$$=\frac{\mathrm{4}−\mathrm{5}\sqrt{\mathrm{3}}}{\mathrm{2}+\sqrt{\mathrm{6}}} \\ $$$$=\frac{\mathrm{4}−\mathrm{5}\sqrt{\mathrm{3}}}{\mathrm{2}+\sqrt{\mathrm{6}}}×\frac{\mathrm{2}−\sqrt{\mathrm{6}}}{\mathrm{2}−\sqrt{\mathrm{6}}} \\ $$$$=\frac{\left(\mathrm{4}−\mathrm{5}\sqrt{}\mathrm{3}\right)\left(\mathrm{2}−\sqrt{\mathrm{6}}\right)}{\left(\mathrm{2}+\sqrt{\mathrm{6}}\right)\left(\mathrm{2}−\sqrt{\left.\mathrm{6}\right)}\right.} \\ $$$$=\frac{\mathrm{8}−\mathrm{4}\sqrt{}\mathrm{6}−\mathrm{10}\sqrt{\mathrm{3}}+\mathrm{5}\sqrt{\mathrm{18}}}{\mathrm{2}^{\mathrm{2}}…
Question Number 153239 by rexford last updated on 06/Sep/21 Answered by MJS_new last updated on 06/Sep/21 $${z}={a}+{b}\mathrm{i} \\ $$$${w}=−{b}+{a}\mathrm{i} \\ $$$${z}+{w}\mathrm{i}={a}+{b}\mathrm{i}+\left(−{b}\mathrm{i}−{a}\right)=\mathrm{0} \\ $$$${zw}=\left({a}+{b}\mathrm{i}\right)\left(−{b}+{a}\mathrm{i}\right)=−\mathrm{2}{ab}+\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)\mathrm{i}…
Question Number 153214 by pticantor last updated on 05/Sep/21 $$\boldsymbol{{let}}\:\boldsymbol{{a}},\boldsymbol{{b}}\in\mathbb{N}^{\ast} \: \\ $$$$\:\boldsymbol{{a}}\ast\boldsymbol{{b}}=\boldsymbol{{a}}+\boldsymbol{{b}}+\boldsymbol{{ab}} \\ $$$$\boldsymbol{{a}}^{\left(\boldsymbol{{n}}\right)} =\boldsymbol{{a}}^{\left(\boldsymbol{{n}}−\mathrm{1}\right)} \ast\boldsymbol{{a}} \\ $$$$\boldsymbol{{explicite}}\:\boldsymbol{{a}}^{\left(\boldsymbol{{n}}\right)} \:\boldsymbol{{en}}\:\boldsymbol{{fonction}}\:\boldsymbol{{de}}\:\boldsymbol{{a}} \\ $$$$ \\ $$$$ \\…
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Question Number 22120 by Sirius last updated on 11/Oct/17 $$\mathrm{hi},\:{i}'{m}\:{new}\:{here}. \\ $$ Commented by Rasheed.Sindhi last updated on 11/Oct/17 $$\mathrm{You}\:\mathrm{are}\:\mathrm{welcome}! \\ $$ Commented by Sirius…
Question Number 153172 by SANOGO last updated on 05/Sep/21 $$\underset{{x}\rightarrow+{oo}} {\mathrm{li}\underset{} {{m}}}\underset{{k}={o}} {\overset{{n}^{\mathrm{2}} } {\sum}}\frac{{n}}{\:{n}^{\mathrm{2}} +{k}^{\mathrm{2}} } \\ $$ Answered by Ar Brandon last updated…