Question Number 221707 by Tawa11 last updated on 09/Jun/25 Commented by Tawa11 last updated on 09/Jun/25 In the ∆ABC |AB|=|BC|=|AC|=K, Area of the shaded ∆…
Question Number 221668 by mr W last updated on 09/Jun/25 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 221733 by Tawa11 last updated on 09/Jun/25 Commented by Tawa11 last updated on 09/Jun/25 $$\mathrm{Is}\:\:\mathrm{10}.\mathrm{37cm}\:\:\mathrm{correct}? \\ $$ Commented by ajfour last updated on…
Question Number 221669 by Tawa11 last updated on 09/Jun/25 Commented by Tawa11 last updated on 09/Jun/25 $$\mathrm{Solve}\:\mathrm{for}\:\:\mathrm{t}. \\ $$ Commented by AlagaIbile last updated on…
Question Number 221760 by OmoloyeMichael last updated on 09/Jun/25 Answered by shunmisaki007 last updated on 09/Jun/25 $$\left(\mathrm{For}\:{x}>\mathrm{0}\:\mathrm{and}\:{x}\neq\mathrm{1}.\right) \\ $$$$\mathrm{log}_{{x}} \left(\frac{\mathrm{log}_{\mathrm{4}} \left({x}\right)}{\mathrm{log}_{\mathrm{4}} \left({x}\right)−\mathrm{3}}\right)^{\mathrm{log}_{\mathrm{3}} \left({x}\right)} =\mathrm{2} \\…
Question Number 221697 by fantastic last updated on 09/Jun/25 $${Is}\:\sqrt{{i}}\:{an}\:{imaginary}\:{number}\:\left({i}=\sqrt{−\mathrm{1}}\right)\:{answer}\:{with}\:{logic} \\ $$ Answered by Ghisom last updated on 09/Jun/25 $$\mathrm{i}=\mathrm{e}^{\mathrm{i}\frac{\pi}{\mathrm{2}}} \\ $$$${z}={r}\mathrm{e}^{\mathrm{i}\theta} \:\Rightarrow\:\sqrt{{z}}=\sqrt{{r}}\mathrm{e}^{\mathrm{i}\frac{\theta}{\mathrm{2}}} \\ $$$$\Rightarrow\:\sqrt{\mathrm{i}}=\mathrm{e}^{\mathrm{i}\frac{\pi}{\mathrm{4}}}…
Question Number 221592 by fantastic last updated on 08/Jun/25 Answered by mr W last updated on 08/Jun/25 Commented by mr W last updated on 08/Jun/25…
Question Number 221626 by fantastic last updated on 08/Jun/25 Answered by mr W last updated on 08/Jun/25 $${side}\:{length}\:{of}\:{square}\:=\mathrm{1} \\ $$$${shaded}\:{area}\:=\frac{\pi×\mathrm{1}^{\mathrm{2}} }{\mathrm{8}}+\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} −\frac{\pi}{\mathrm{4}}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{4}} \\…
Question Number 221620 by fantastic last updated on 08/Jun/25 $${Solve}\:{for}\:{x} \\ $$$$\sqrt[{\mathrm{3}}]{\mathrm{7}{x}}=\sqrt{{x}}\left[{x}\neq\mathrm{0}\right] \\ $$ Answered by fantastic last updated on 08/Jun/25 $${or}\:\left(\mathrm{7}{x}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} ={x}^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$${or}\:\sqrt[{\mathrm{3}}]{\mathrm{7}}.{x}^{\frac{\mathrm{1}}{\mathrm{3}}}…
Question Number 221588 by Nicholas666 last updated on 08/Jun/25 $$ \\ $$$$\:\:\:\:\int\:\frac{\mathrm{8}{t}\:−\:\mathrm{8}{t}^{\:\mathrm{3}} }{{t}^{\:\mathrm{6}} \:+\:\mathrm{6}{t}^{\mathrm{5}} \:+\:\mathrm{3}{t}^{\:\mathrm{4}} \:−\:\mathrm{20}{t}^{\mathrm{3}} \:+\:\mathrm{3}{t}^{\mathrm{2}} \:+\:\mathrm{6}{t}\:+\:\mathrm{1}}\:{dt}\:\:\:\: \\ $$$$ \\ $$ Answered by Frix…