Question Number 146776 by mathdanisur last updated on 15/Jul/21 Commented by mr W last updated on 15/Jul/21 $${AB}^{\mathrm{2}} =\mathrm{20}^{\mathrm{2}} −{x}^{\mathrm{2}} =\left(\mathrm{20}+\mathrm{5}\right)^{\mathrm{2}} +\mathrm{10}^{\mathrm{2}} −\mathrm{2}×\mathrm{25}×\mathrm{10}\:\mathrm{cos}\:{C} \\ $$$${AD}^{\mathrm{2}}…
Question Number 81242 by panky0214 last updated on 10/Feb/20 Commented by mr W last updated on 10/Feb/20 $${i}\:{got}\:{maximum}=\frac{\mathrm{120}}{\mathrm{49}}\approx\mathrm{2}.\mathrm{45} \\ $$ Commented by mr W last…
Question Number 146778 by mathdanisur last updated on 15/Jul/21 $${Find}\:{the}\:{modulus}\:{of}\:{a}\:{complex} \\ $$$${number}: \\ $$$${Z}\:=\:{cos}\:\mathrm{40}\:+\:{i}\:{sin}\:\mathrm{20}\:+\:\mathrm{1}\:=\:? \\ $$ Answered by mathmax by abdo last updated on 15/Jul/21…
Question Number 146764 by Ar Brandon last updated on 15/Jul/21 $$\mathrm{A}=\left(\frac{\left(\mathrm{x}+\sqrt[{\mathrm{3}}]{\mathrm{2ax}^{\mathrm{2}} }\right)\left(\mathrm{2a}+\sqrt[{\mathrm{3}}]{\mathrm{2a}^{\mathrm{2}} \mathrm{x}}\right)^{−\mathrm{1}} −\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\mathrm{x}}−\sqrt[{\mathrm{3}}]{\mathrm{2a}}}−\left(\mathrm{2a}\right)^{−\mathrm{1}/\mathrm{3}} \right)^{−\mathrm{6}} ,\:\left(\mathrm{a},\mathrm{b}\right)\in\mathbb{R}^{\mathrm{2}} \\ $$$$\mathrm{a}-\:\mathrm{A}=\frac{\mathrm{16a}^{\mathrm{4}} }{\mathrm{x}^{\mathrm{2}} }\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{b}-\:\mathrm{A}=\frac{\mathrm{8}}{\mathrm{ax}}\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{c}-\mathrm{A}=\frac{\sqrt[{\mathrm{3}}]{\mathrm{2a}}}{\mathrm{3x}^{\mathrm{3}} } \\ $$ Terms of…
Question Number 81223 by ubaydulla last updated on 10/Feb/20 $${x}\left(\mathrm{3}^{{x}} +\mathrm{2}\right)=\mathrm{3}\left(\mathrm{1}−\mathrm{3}^{{x}} \right)−{x}^{\mathrm{2}} \\ $$ Answered by MJS last updated on 10/Feb/20 $$\mathrm{3}^{{x}} {x}+\mathrm{2}{x}−\mathrm{3}+\mathrm{3}^{{x}} \mathrm{3}+{x}^{\mathrm{2}} =\mathrm{0}…
Question Number 146746 by mathdanisur last updated on 15/Jul/21 $$\underset{\:\mathrm{0}} {\overset{\:\mathrm{4}} {\int}}\:\sqrt{\mathrm{16}\:-\:{x}^{\mathrm{2}} }\:{dx}\:=\:? \\ $$ Answered by qaz last updated on 15/Jul/21 $$\int_{\mathrm{0}} ^{\mathrm{4}} \sqrt{\mathrm{16}−\mathrm{x}^{\mathrm{2}}…
Question Number 146744 by mathdanisur last updated on 15/Jul/21 $${if}\:\:\underset{\boldsymbol{{a}}} {\overset{\boldsymbol{{b}}} {\int}}{f}\left({x}\right){dx}\:=\:\mathrm{7}\:\:\:{and}\:\:\underset{\boldsymbol{{a}}} {\overset{\boldsymbol{{b}}} {\int}}\mathrm{4}\:{g}\left({x}\right){dx}\:=\:−\mathrm{6} \\ $$$${find}\:\:\:\underset{\boldsymbol{{a}}} {\overset{\boldsymbol{{b}}} {\int}}\left(\mathrm{3}\:{f}\left({x}\right)−\mathrm{8}\:{g}\left({x}\right)\right)\:{dx}\:=\:? \\ $$ Answered by liberty last updated…
Question Number 81206 by Power last updated on 10/Feb/20 Commented by mathmax by abdo last updated on 10/Feb/20 $${let}\:{f}\left({x}\right)=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)}\:{x}^{{n}} \:{with}\:\mid{x}\mid\leqslant\mathrm{1}\:\:{and}\:{x}\neq−\mathrm{1} \\ $$$${f}\left({x}\right)=\sum_{{n}=\mathrm{0}}…
Question Number 146730 by puissant last updated on 15/Jul/21 Commented by puissant last updated on 15/Jul/21 $$\mathrm{merci} \\ $$ Commented by Olaf_Thorendsen last updated on…
Question Number 15656 by Tinkutara last updated on 12/Jun/17 $$\mathrm{Solve}\::\:\mathrm{0}\:\leqslant\:{x}^{\mathrm{2}} \:−\:\mathrm{5}{x}\:+\:\mathrm{7}\:<\:\mathrm{1} \\ $$ Answered by ajfour last updated on 12/Jun/17 $$\:\:\:\:\:\mathrm{0}\leqslant\:\left({x}−\frac{\mathrm{5}}{\mathrm{2}}\right)^{\mathrm{2}} +\mathrm{7}−\frac{\mathrm{25}}{\mathrm{4}}\:<\:\mathrm{1} \\ $$$$\:\:\:\:\:\mathrm{0}\leqslant\:\left({x}−\frac{\mathrm{5}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{4}}\:<\:\mathrm{1}…