Question Number 112624 by Aina Samuel Temidayo last updated on 09/Sep/20 $$\mathrm{Minimum}\:\mathrm{value}\:\mathrm{of}\:\mathrm{x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}−\mathrm{3}\:\mathrm{is} \\ $$ Answered by MJS_new last updated on 09/Sep/20 $${x}^{\mathrm{2}} +\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}}…
Question Number 112626 by Aina Samuel Temidayo last updated on 09/Sep/20 Answered by som(math1967) last updated on 09/Sep/20 $$\left.\mathrm{a}\right)\frac{\mathrm{x}−\mathrm{y}}{\mathrm{b}−\mathrm{a}}=\frac{\mathrm{y}−\mathrm{z}}{\mathrm{c}−\mathrm{b}}=\frac{\mathrm{z}−\mathrm{x}}{\mathrm{a}−\mathrm{c}} \\ $$ Commented by Aina Samuel…
Question Number 112625 by Aina Samuel Temidayo last updated on 09/Sep/20 $$\mathrm{If}\:\mathrm{a}+\mathrm{b}=\mathrm{5},\:\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} =\mathrm{13},\:\mathrm{the}\:\mathrm{value}\:\mathrm{of} \\ $$$$\mathrm{a}−\mathrm{b}\:\left(\mathrm{where}\:\mathrm{a}>\mathrm{b}\right)\:\mathrm{is} \\ $$ Commented by MJS_new last updated on 09/Sep/20…
Question Number 47068 by 143jesus last updated on 04/Nov/18 $$\left(\mathrm{a}−\mathrm{b}\right)^{\mathrm{2}} \\ $$ Answered by ajfour last updated on 04/Nov/18 $$=\:\left({a}−{b}\right)\left({a}−{b}\right) \\ $$$$=\:{a}\left({a}−{b}\right)−{b}\left({a}−{b}\right) \\ $$$$=\:{a}^{\mathrm{2}} −{ab}−{ba}+{b}^{\mathrm{2}}…
Question Number 178136 by Acem last updated on 13/Oct/22 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{Journey}}\:\boldsymbol{{inside}}\:\boldsymbol{{a}}\:\boldsymbol{{regular}}\:\boldsymbol{{hexagon}} \\ $$$$\:{The}\:{operation}\:{is}\:{to}\:{connect}\:{three}\:{dots}\:{of} \\ $$$$\:{a}\:{regular}\:{hexagon}'{s}\:{heads}. \\ $$$$\mathrm{1}\bullet\:{How}\:{many}\:{types}\:{of}\:{geometric}\:{shapes}\:{will}\:{we}\:{get}? \\ $$$$\mathrm{2}\bullet\:{How}\:{many}\:{each}\:{type}? \\ $$$$ \\ $$ Terms of Service…
Question Number 178137 by Acem last updated on 13/Oct/22 $${Let}\:{f}\left({x}\right)=\:\left({ax}+\mathrm{1}\right)^{\mathrm{5}} .\left(\mathrm{1}+{bx}\right)^{\mathrm{4}} \:;\:{a},{b}\:\in\:\mathbb{N} \\ $$$$\:{if}\:{times}\:{of}\:{x}\:{equal}\:\mathrm{62}\:{so}\:{what}\:{are}\:{possible}\:{values} \\ $$$$\:{of}\:{the}\:{sum}\:{a},\:{b}? \\ $$$$ \\ $$ Commented by Rasheed.Sindhi last updated…
Question Number 178134 by Acem last updated on 13/Oct/22 $${Am}\:{not}\:{a}\:{friend}\:{with}\:{this}\:{isssue}: \\ $$$$\mathrm{6}\:{red},\:\mathrm{1}\:{black}\:{and}\:\mathrm{3}\:{white}\:{balls} \\ $$$${We}\:{draw}\:\mathrm{3}\:{balls}\:{in}\:{a}\:{raw},\:{returning}\:{the} \\ $$$$\:{drawn}\:{ball}\:{each}\:{time}. \\ $$$${How}\:{many}\:{different}\:{results}\:{which}\:{include} \\ $$$$\:\boldsymbol{{at}}\:\boldsymbol{{least}}\:{one}\:{black}\:{ball}. \\ $$$$ \\ $$$$\:{Way}_{\mathrm{1}} :\:\mathrm{1000}−\mathrm{9}^{\mathrm{3}}…
Question Number 178113 by Acem last updated on 12/Oct/22 $$\:\mathrm{8}\:{yellow},\:\mathrm{8}\:{red},\:\mathrm{8}\:{green}\:{and}\:\mathrm{8}\:{blue}\:{cards},\: \\ $$$${each}\:{grouo}\:{is}\:{numbered}\:{from}\:\mathrm{1}\:{to}\:\mathrm{8}\:. \\ $$$$\:{We}\:{want}\:{to}\:{know}\:{how}\:{many}\:{pulls}\:{contain} \\ $$$$\:{at}\:{least}\:{one}\:{card}\:{with}\:{number}\:\mathrm{1}\:{through} \\ $$$$\:{pull}\:{processes}\:\mathrm{5}\:{cards}\:{each}\:{time} \\ $$$$ \\ $$$$ \\ $$ Answered…
Question Number 47019 by ajfour last updated on 04/Nov/18 Commented by ajfour last updated on 04/Nov/18 $${If}\:{at}\:{its}\:{roots}\:\:{x}=−\mathrm{3}\:{and}\:{x}=\mathrm{2} \\ $$$${the}\:{biquadratic}\:{also}\:{has}\:{its} \\ $$$${points}\:{of}\:{inflexion},\:{find}\:{its} \\ $$$${other}\:{two}\:{roots}\:\alpha\:{and}\:\beta. \\ $$…
Question Number 178073 by infinityaction last updated on 12/Oct/22 Commented by Rasheed.Sindhi last updated on 12/Oct/22 $$\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{1}} −\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} −\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} −\mathrm{1}}+…+\frac{\mathrm{1}}{\mathrm{2}^{{n}} −\mathrm{1}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}−\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{4}−\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{8}−\mathrm{1}}+…+\frac{\mathrm{1}}{\mathrm{2}^{{n}} −\mathrm{1}} \\…