Question Number 125577 by ajfour last updated on 12/Dec/20 Commented by ajfour last updated on 12/Dec/20 $${If}\:{the}\:{cubic}\:{curve}'{s}\:{eq}.\:\:{is} \\ $$$${y}={x}^{\mathrm{3}} −{x}−{c} \\ $$$${and}\:{parabola}'{s}\:{eq}.\:{is}\: \\ $$$${y}=\left({x}−{h}\right)^{\mathrm{2}} +\left(\frac{\mathrm{2}}{\mathrm{3}\sqrt{\mathrm{3}}}−{c}\right)\:,\:{then}…
Question Number 60039 by MJS last updated on 17/May/19 $$\mathrm{find}\:\mathrm{all}\:\mathrm{solutions}\:\mathrm{for}\:{z}\in\mathbb{C} \\ $$$${z}^{\mathrm{i}} =\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{i} \\ $$$${z}^{\mathrm{1}−\mathrm{i}} =\mathrm{1}+\mathrm{i} \\ $$ Commented by maxmathsup by imad last updated…
Question Number 191111 by Shrinava last updated on 18/Apr/23 $${bb}\mathrm{55}{bb} \\ $$ Commented by mr W last updated on 18/Apr/23 $$\mathrm{12277179} \\ $$ Commented by…
Question Number 191103 by Mingma last updated on 18/Apr/23 Answered by Rasheed.Sindhi last updated on 18/Apr/23 $$\mathrm{4}^{−\frac{\mathrm{1}}{{x}}} +\mathrm{6}^{−\frac{\mathrm{1}}{{x}}} =\mathrm{9}^{−\frac{\mathrm{1}}{{x}}} \\ $$$$\frac{\mathrm{2}^{−\mathrm{2}/{x}} }{\mathrm{2}^{−\mathrm{1}/{x}} \centerdot\mathrm{3}^{−\mathrm{1}/{x}} }+\frac{\mathrm{2}^{−\mathrm{1}/{x}} \centerdot\mathrm{3}^{−\mathrm{1}/{x}}…
Question Number 191077 by Shrinava last updated on 17/Apr/23 $$\overset{\rightarrow} {\mathrm{a}}\:\centerdot\:\overset{\rightarrow} {\mathrm{b}}\:=\:\mathrm{4} \\ $$$$\overset{\rightarrow} {\mathrm{b}}\:\centerdot\:\overset{\rightarrow} {\mathrm{c}}\:=\:\mathrm{5} \\ $$$$\overset{\rightarrow} {\mathrm{7a}}\:=\:\overset{\rightarrow} {\mathrm{4b}}\:+\:\overset{\rightarrow} {\mathrm{2c}}\:\: \\ $$$$\mathrm{Find}:\:\:\:\mid\overset{\rightarrow} {\mathrm{c}}\:\mid\:=\:? \\…
Question Number 125539 by ajfour last updated on 11/Dec/20 Commented by ajfour last updated on 11/Dec/20 $${The}\:{blue}\:{cubic}\:{curve}\:{is} \\ $$$$\:\:{y}={x}^{\mathrm{3}} −\mathrm{21}{x}−\mathrm{20} \\ $$$${Find}\:{p}. \\ $$ Answered…
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Question Number 59977 by mjay last updated on 16/May/19 $$\mathrm{4}×\left(\mathrm{5}+\mathrm{5}\right) \\ $$ Answered by meme last updated on 16/May/19 $$=\mathrm{40} \\ $$ Terms of Service…
Question Number 125504 by Study last updated on 11/Dec/20 $${n}!=\mathrm{2}^{\mathrm{10}} \centerdot\mathrm{10}!\left(\mathrm{1}\centerdot\mathrm{3}\centerdot\mathrm{5}\centerdot\mathrm{7}\centerdot\centerdot\centerdot\mathrm{19}\right)\:\:\:\:\:\:\:\:{n}=??? \\ $$$${help}\:{me} \\ $$ Answered by Dwaipayan Shikari last updated on 11/Dec/20 $$\mathrm{1}.\mathrm{3}.\mathrm{5}…{k}=\frac{\mathrm{1}.\mathrm{2}.\mathrm{3}.\mathrm{4}.\mathrm{5}…\mathrm{2}{n}}{\mathrm{2}^{{k}} .{k}!}=\frac{\left(\mathrm{2}{k}\right)!}{\mathrm{2}^{{k}}…
Question Number 191024 by Rupesh123 last updated on 16/Apr/23 Answered by BaliramKumar last updated on 16/Apr/23 $$\mathrm{3600}\:=\:\mathrm{36}×\mathrm{100}\:=\:\mathrm{2}^{\mathrm{2}} ×\mathrm{3}^{\mathrm{2}} ×\mathrm{2}^{\mathrm{2}} ×\mathrm{5}^{\mathrm{2}} \: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}^{\mathrm{4}} ×\mathrm{3}^{\mathrm{2}} ×\mathrm{5}^{\mathrm{2}}…