Question Number 155277 by amin96 last updated on 28/Sep/21 Answered by Rasheed.Sindhi last updated on 28/Sep/21 $$\left({big}-{square}-{area}\right)−\left({non}-{shadow}-{area}\right) \\ $$$${big}-{square}-{area}=\mathrm{8}^{\mathrm{2}} =\mathrm{64}\:{sq}-{units} \\ $$$$\:\:\:\:\:\:\underset{−} {{Non}-{shadow}-{area}} \\ $$$$\mathrm{A}:{two}-{triangles}=\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}}.\mathrm{1}.\mathrm{2}\right)=\mathrm{2}\:{unit}^{\mathrm{2}}…
Question Number 89741 by cindiaulia last updated on 19/Apr/20 Commented by jagoll last updated on 19/Apr/20 $$\mathrm{area}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:\mid\begin{vmatrix}{\mathrm{0}\:\:\:\:\:\mathrm{0}}\\{\mathrm{1}\:\:\:\:\:\mathrm{1}}\end{vmatrix}+\:\begin{vmatrix}{\mathrm{1}\:\:\:\:\:\:\mathrm{1}}\\{\frac{\mathrm{2}}{\mathrm{5}}\:\:\frac{\mathrm{8}}{\mathrm{5}}}\end{vmatrix}+\:\begin{vmatrix}{\frac{\mathrm{2}}{\mathrm{5}}\:\:\:\:\frac{\mathrm{8}}{\mathrm{5}}}\\{\mathrm{0}\:\:\:\:\:\:\:\mathrm{0}}\end{vmatrix}\mid \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\:\mid\:\mathrm{0}+\frac{\mathrm{6}}{\mathrm{5}}+\mathrm{0}\:\mid\:=\:\frac{\mathrm{3}}{\mathrm{5}} \\ $$ Terms of Service Privacy…
Question Number 155272 by bekzodjumayev last updated on 28/Sep/21 Commented by bekzodjumayev last updated on 28/Sep/21 $$\boldsymbol{{P}}{rove}\:{that} \\ $$$${Please}\:{help} \\ $$ Commented by puissant last…
Question Number 89736 by cindiaulia last updated on 19/Apr/20 Commented by jagoll last updated on 19/Apr/20 $$\mathrm{vol}\:=\:\pi\:\left[\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\left(\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{2}} \:\mathrm{dx}\:+\:\underset{\mathrm{1}} {\overset{\mathrm{4}} {\int}}\:\left(\sqrt{\mathrm{x}}\right)^{\mathrm{2}} \:\mathrm{dx}\:\right] \\…
Question Number 89737 by cindiaulia last updated on 19/Apr/20 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 89735 by cindiaulia last updated on 19/Apr/20 Commented by jagoll last updated on 19/Apr/20 $$\mathrm{red}\:\mathrm{parabola}\:\mathrm{y}\:=\:\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{2x}\: \\ $$$$\mathrm{blue}\:\mathrm{parabola}\:\mathrm{y}\:=\:−\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{4} \\ $$$$\mathrm{shaded}\:\mathrm{area}\::\:\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{2x}\:=\:−\mathrm{x}^{\mathrm{2}} +\mathrm{4}…
Question Number 89732 by naka3546 last updated on 19/Apr/20 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 155265 by SANOGO last updated on 28/Sep/21 $${si}\:{E}\:{est}\:{la}\:{fonction}\:{partie}\:{entiere}\:,{et}\:{n}\:{un}\:{entier}\:{naturel} \\ $$$${alors}\:{I}=\int_{{o}} ^{{n}} {E}\left({x}\right)\:{vaut}? \\ $$ Answered by puissant last updated on 28/Sep/21 $${I}=\int_{\mathrm{0}} ^{{n}}…
Question Number 155264 by SANOGO last updated on 28/Sep/21 $${en}\:{utilisant}\:{l}'{integrale}\:{de}\:{cauchy}\:{schwarz} \\ $$$${l}'{integrale}\:\int_{{o}} ^{\mathrm{1}} \frac{{f}\left({x}\right)}{{x}+\mathrm{1}}{dx}\:\:\:\:{est}\:{majoree}\:{par}? \\ $$$$ \\ $$ Answered by puissant last updated on 28/Sep/21…
Question Number 89728 by Mr.Panoply last updated on 18/Apr/20 $${Find}\:{the}\:{area}\:{bounded}\:{by}\:\mathrm{3}{x}+\mathrm{4}{y}=\mathrm{12} \\ $$$${and}\:{the}\:{coordinate}\:{axes}? \\ $$ Answered by MJS last updated on 19/Apr/20 $$\mathrm{3}{x}+\mathrm{4}{y}=\mathrm{12}\:\mathrm{is}\:\mathrm{a}\:\mathrm{straight}\:\mathrm{line} \\ $$$${x}=\mathrm{0}\:\Rightarrow\:{y}=\mathrm{3} \\…