Question Number 202037 by syamilkamil1 last updated on 19/Dec/23 $$ \\ $$A board has 2, 4, and 6 written on it. One repeatedly chooses values…
Question Number 202039 by sonukgindia last updated on 19/Dec/23 Answered by mr W last updated on 19/Dec/23 Commented by mr W last updated on 19/Dec/23…
Question Number 202034 by cortano12 last updated on 19/Dec/23 Answered by MM42 last updated on 19/Dec/23 $${if}\:\frac{\mathrm{7}}{\mathrm{3}}<{a}<\frac{\mathrm{7}}{\mathrm{2}}\Rightarrow\lfloor\frac{\mathrm{7}}{{a}}\rfloor+\lfloor\frac{{a}}{\mathrm{7}}\rfloor=\mathrm{2} \\ $$$${if}\:\:\:\mathrm{14}<{a}<\mathrm{21}\Rightarrow\lfloor\frac{\mathrm{7}}{{a}}\rfloor+\lfloor\frac{{a}}{\mathrm{7}}\rfloor=\mathrm{2} \\ $$$$ \\ $$ Terms of…
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Question Number 202056 by mr W last updated on 19/Dec/23 Commented by mr W last updated on 19/Dec/23 $${the}\:{lake}\:{with}\:{center}\:{at}\:{O}\:{has}\:{a}\:{radius} \\ $$$${r}\:\left({r}=\mathrm{1}\:{km}\right).\:{the}\:{shortest}\:{distances}\: \\ $$$${from}\:{the}\:{villages}\:{A}\:{and}\:{B}\:{to}\:{the}\: \\ $$$${lake}\:{are}\:{a}\:{and}\:{b}\:{respectively}\:\left({a}=\mathrm{4}\:{km},\right.…
Question Number 202058 by BaliramKumar last updated on 19/Dec/23 $$\mathrm{Solve}\:\mathrm{by}\:\mathrm{computer}\:\mathrm{programming} \\ $$$${a},\:{b}\:\&\:{c}\:{are}\:{Prime}\:{numbers}.\:\mathrm{And}\:\mathrm{they} \\ $$$$\:\mathrm{are}\:\mathrm{in}\:\mathrm{AP}\:\mathrm{and}\:\mathrm{d}\:\mathrm{is}\:\mathrm{common}\:\mathrm{difference} \\ $$$$\mathrm{Example}\:\left(\mathrm{a},\:\mathrm{b},\:\mathrm{c},\:\mathrm{d}\right)\:=\:\left(\mathrm{3},\:\mathrm{5},\:\mathrm{7},\:\mathrm{2}\right) \\ $$$$\:\mathrm{Just}\:\mathrm{Next}\:\mathrm{Set}\left(\mathrm{a},\:\mathrm{b},\:\mathrm{c},\:\mathrm{d}\right)\:=\:? \\ $$ Commented by BaliramKumar last updated…
Question Number 202082 by Abduljalal last updated on 19/Dec/23 Answered by AST last updated on 20/Dec/23 $${p}^{\mathrm{2}} −{p}−\mathrm{2}=\mathrm{0}\left({p}={x}^{\mathrm{3}} \right)\Rightarrow{p}=\mathrm{2}\:{or}\:−\mathrm{1} \\ $$$$\Rightarrow{x}^{\mathrm{3}} =\mathrm{2}\:{or}\:{x}^{\mathrm{3}} =−\mathrm{1}\Rightarrow{x}=\sqrt[{\mathrm{3}}]{\mathrm{2}};\sqrt[{\mathrm{3}}]{\mathrm{2}}{e}^{\frac{\mathrm{2}{i}\pi}{\mathrm{3}}} ,\sqrt[{\mathrm{3}}]{\mathrm{2}}{e}^{\frac{\mathrm{4}\pi{i}}{\mathrm{3}}} \\…
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