Question Number 166112 by mathls last updated on 13/Feb/22 $${prove}\:{that}\:\mathrm{0}!=\mathrm{1} \\ $$ Commented by mathls last updated on 13/Feb/22 $$? \\ $$ Answered by satyam123…
Question Number 166113 by mathls last updated on 13/Feb/22 $${prove}\:{that}\:\mathrm{1}!=\mathrm{1} \\ $$ Commented by mathls last updated on 13/Feb/22 $$? \\ $$ Commented by MJS_new…
Question Number 166102 by Tawa11 last updated on 13/Feb/22 Answered by mr W last updated on 13/Feb/22 $${we}\:{have}\:{totally}\:\mathrm{12}\:{rolls}\:{for}\:{the}\:{numbers} \\ $$$$\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{5},\mathrm{6},\:{the}\:{sum}\:{of}\:{them}\:{is}\:\mathrm{47}.\: \\ $$$${only}\:{one}\:{number}\:{occurs}\:\mathrm{3}\:{times}.\:{it}'{s} \\ $$$${easy}\:{to}\:{get}\:{that}\:{an}\:{other}\:{number}\:{must} \\…
Question Number 100567 by bramlex last updated on 27/Jun/20 $$\begin{cases}{{x}−\sqrt{{yz}}\:=\:\mathrm{42}}\\{{y}−\sqrt{{xz}}\:=\:\mathrm{6}}\\{{z}−\sqrt{{xy}}\:=\:−\mathrm{30}}\end{cases} \\ $$$${find}\:{x}+{y}+{z}\:= \\ $$ Commented by Dwaipayan Shikari last updated on 27/Jun/20 $$\mathrm{x}=\mathrm{54} \\ $$$$\mathrm{y}=\mathrm{24}\:\:\:\:\:\mathrm{z}=\mathrm{6}…
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Question Number 166087 by mathlove last updated on 13/Feb/22 Commented by cortano1 last updated on 13/Feb/22 $$\mathrm{1} \\ $$ Answered by cortano1 last updated on…
Question Number 100540 by ajfour last updated on 27/Jun/20 Commented by ajfour last updated on 27/Jun/20 $${p},\:{q},\:{r}\:{are}\:{roots}\:{of}\:{equation} \\ $$$${x}^{\mathrm{3}} −{bx}−{c}=\mathrm{0}\:\:,\:\:{b}>\mathrm{0},\:{c}>\mathrm{0} \\ $$$${Find}\:{equation}\:{of}\:{parabola}\:{in} \\ $$$${terms}\:{of}\:{b}\:{and}\:{c}. \\…
Question Number 166070 by mathlove last updated on 12/Feb/22 Commented by behi834171 last updated on 13/Feb/22 $$. \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 166058 by cortano1 last updated on 12/Feb/22 Answered by som(math1967) last updated on 12/Feb/22 $$\bigtriangleup{ABC}\sim\bigtriangleup{ALB}\sim\bigtriangleup{CLB} \\ $$$${let}\:\measuredangle{BAC}=\measuredangle{DCL}=\alpha \\ $$$${AL}×{AC}={a}^{\mathrm{2}} \\ $$$${AL}=\frac{{a}^{\mathrm{2}} }{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}}…
Question Number 166053 by neinhaltsieger last updated on 12/Feb/22 $$\:\boldsymbol{\mathrm{Find}}\:\:\boldsymbol{\mathrm{x}}\:\:\boldsymbol{\mathrm{in}}\:\:\mathbb{R}\overset{\:} {:} \\ $$$$\: \\ $$$$\:\boldsymbol{\mathrm{x}}^{\sqrt{\mathrm{2}}} \:\:+\:\:\boldsymbol{\mathrm{x}}\:\:=\:\:\mathrm{6}\:\:\:\:\:\left(\boldsymbol{\mathrm{How}}\:\:\boldsymbol{\mathrm{to}}\:\:\boldsymbol{\mathrm{solve}}?\right) \\ $$ Commented by MJS_new last updated on 12/Feb/22…