Question Number 123965 by Khalmohmmad last updated on 29/Nov/20 $${A}=\begin{bmatrix}{\mathrm{1}\:\:\:\mathrm{2}\:\:\:\:\:\mathrm{3}}\\{\mathrm{2}\:\:\:\mathrm{3}\:\:\:\:\:\mathrm{1}}\\{\mathrm{3}\:\:\:\:\mathrm{2}\:\:\:\:\mathrm{1}}\end{bmatrix} \\ $$$${det}\left({M}_{\mathrm{3}×\mathrm{3}} \right)=? \\ $$ Commented by MJS_new last updated on 29/Nov/20 $$\begin{vmatrix}{{a}}&{{b}}&{{c}}\\{{d}}&{{e}}&{{f}}\\{{g}}&{{h}}&{{i}}\end{vmatrix}={aei}+{bfg}+{cdh}−\left({afh}+{bdi}+{ceg}\right) \\ $$…
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Question Number 189495 by yaslm last updated on 17/Mar/23 Commented by mr W last updated on 18/Mar/23 $${question}\:{could}\:{be}\:{true},\:{but}\:{diagram} \\ $$$${is}\:{terrible}. \\ $$ Commented by mr…
Question Number 123950 by hatakekakashi1729gmailcom last updated on 29/Nov/20 Commented by mindispower last updated on 29/Nov/20 $${done}\:{remember}\:{this} \\ $$ Commented by Dwaipayan Shikari last updated…
Question Number 123922 by help last updated on 29/Nov/20 Answered by liberty last updated on 29/Nov/20 $${y}={p}\left({x}−\mathrm{3}\right)^{\mathrm{2}} +\mathrm{1}\:=\:{px}^{\mathrm{2}} −\mathrm{6}{px}+\mathrm{9}{p}+\mathrm{1} \\ $$$$\begin{cases}{{p}=−{a}}\\{{b}=−\mathrm{6}{p}\:}\\{{c}=\mathrm{9}{p}+\mathrm{1}}\end{cases} \\ $$ Commented by…
Question Number 58387 by azizullah last updated on 22/Apr/19 Answered by mr W last updated on 22/Apr/19 $${x}=\mathrm{1}\:{is}\:{a}\:{solution}, \\ $$$$\Rightarrow{a}+\mathrm{5}=\mathrm{3} \\ $$$$\Rightarrow{a}=−\mathrm{2} \\ $$$$−\mathrm{2}{x}^{\mathrm{2}} +\mathrm{5}{x}−\mathrm{3}=\mathrm{0}…
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Question Number 58378 by naka3546 last updated on 22/Apr/19 Commented by naka3546 last updated on 22/Apr/19 $$#\:\:\mathrm{119},\:\mathrm{120},\:{n}\:\mathrm{121}. \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 189449 by 073 last updated on 16/Mar/23 Answered by Ar Brandon last updated on 16/Mar/23 $$=\frac{\mathrm{88}×\mathrm{10}−\mathrm{62}−\mathrm{96}}{\mathrm{10}−\mathrm{2}} \\ $$ Terms of Service Privacy Policy…
Question Number 58377 by naka3546 last updated on 22/Apr/19 Answered by MJS last updated on 22/Apr/19 $$\mathrm{rectangular}\:\mathrm{triangle}\:\mathrm{with}\:\mathrm{diagonal}\:\mathrm{of}\:\mathrm{rectangle} \\ $$$$\left(\mathrm{6}−{h}\right)^{\mathrm{2}} +\mathrm{6}^{\mathrm{2}} =\left(\mathrm{2}{r}\right)^{\mathrm{2}} \\ $$$$\mathrm{short}\:\mathrm{side}\:\mathrm{of}\:\mathrm{rectangle}\:=\:\mathrm{2}{r}−\mathrm{2}{h}\:\Rightarrow \\ $$$$\mathrm{2}{r}=\mathrm{6}+{h}…