Question Number 74978 by aliesam last updated on 05/Dec/19 Commented by aliesam last updated on 05/Dec/19 $${find}\:{the}\:{area}\:{of}\:{the}\:{red}\:{part} \\ $$ Answered by mind is power last…
Question Number 74904 by necxxx last updated on 03/Dec/19 Commented by mr W last updated on 03/Dec/19 $${no}\:{unique}\:{solution} \\ $$ Terms of Service Privacy Policy…
Question Number 74882 by ajfour last updated on 03/Dec/19 Commented by ajfour last updated on 03/Dec/19 $${If}\:{each}\:{small}\:{blue}\:{triangles} \\ $$$${have}\:{the}\:{same}\:{area},\:{and}\:{each} \\ $$$${small}\:{quadrilaterals}\:{have}\:{the} \\ $$$${same}\:{area},\:{find}\:{ratio}\:{of}\:{length} \\ $$$${to}\:{breadth}\:{of}\:{the}\:{outer}\:{rectangle}.…
Question Number 9298 by tawakalitu last updated on 29/Nov/16 Answered by mrW last updated on 29/Nov/16 $$\mathrm{radius}\:\mathrm{r}=\left(\mathrm{7}+\mathrm{3}\right)/\mathrm{2}=\mathrm{5}\:\mathrm{cm} \\ $$$$\mathrm{height}\:\mathrm{of}\:\mathrm{segment}\:\mathrm{h}=\mathrm{3}\:\mathrm{cm} \\ $$$$\mathrm{area}\:\mathrm{of}\:\mathrm{segment}\:=\:\mathrm{r}^{\mathrm{2}} \mathrm{cos}\:^{−\mathrm{1}} \left(\frac{\mathrm{r}−\mathrm{h}}{\mathrm{r}}\right)−\left(\mathrm{r}−\mathrm{h}\right)\sqrt{\mathrm{2rh}−\mathrm{h}^{\mathrm{2}} } \\…
Question Number 9203 by tawakalitu last updated on 23/Nov/16 $$\mathrm{The}\:\mathrm{position}\:\mathrm{of}\:\mathrm{a}\:\mathrm{point}\:\mathrm{A}\:\mathrm{is}\:\left(\mathrm{41}°\mathrm{N},\:\mathrm{18}°\mathrm{E}\right) \\ $$$$\mathrm{point}\:\mathrm{B}\:\mathrm{is}\:\mathrm{8000}\:\mathrm{km}\:\mathrm{due}\:\mathrm{west}\:\mathrm{of}\:\mathrm{A}.\:\mathrm{Find}\:\mathrm{the} \\ $$$$\mathrm{longitude}\:\mathrm{of}\:\mathrm{B}\:\mathrm{to}\:\mathrm{the}\:\mathrm{nearest}\:\mathrm{degree}. \\ $$ Answered by mrW last updated on 23/Nov/16 $$\mathrm{The}\:\mathrm{radius}\:\mathrm{of}\:\mathrm{the}\:\mathrm{Earth}\:\mathrm{is}\:\mathrm{roughly} \\…
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Question Number 8972 by tawakalitu last updated on 09/Nov/16 $$\mathrm{A}\:\mathrm{cylindrical}\:\mathrm{can}\:\mathrm{of}\:\mathrm{internal}\:\mathrm{diameter}\:\mathrm{8cm} \\ $$$$\mathrm{contains}\:\mathrm{water}\:\mathrm{to}\:\mathrm{a}\:\mathrm{deep}\:\mathrm{of}\:\mathrm{6cm}.\:\mathrm{24000}\:\mathrm{heavy} \\ $$$$\mathrm{spherical}\:\mathrm{pullet}\:\mathrm{of}\:\mathrm{diameter}\:\mathrm{2mm}\:\mathrm{are}\:\mathrm{dropped} \\ $$$$\mathrm{into}\:\mathrm{the}\:\mathrm{can}.\:\mathrm{how}\:\mathrm{far}\:\mathrm{does}\:\mathrm{the}\:\mathrm{water}\:\mathrm{level}\:\mathrm{rise}\:? \\ $$ Answered by Rasheed Soomro last updated on…
Question Number 8465 by tawakalitu last updated on 12/Oct/16 Answered by sandy_suhendra last updated on 12/Oct/16 $$\left.\mathrm{a}\right)\:\mathrm{let}\:\mathrm{r}=\mathrm{7}\:\mathrm{cm}\:\mathrm{and}\:\mathrm{h}=\mathrm{5}\:\mathrm{cm} \\ $$$$\mathrm{the}\:\mathrm{surface}\:\mathrm{area}=\mathrm{2}×\frac{\mathrm{30}}{\mathrm{360}}\:\pi\mathrm{r}^{\mathrm{2}} +\mathrm{2rh}+\frac{\mathrm{30}}{\mathrm{360}}×\mathrm{2}\pi\mathrm{rh} \\ $$$$=\frac{\mathrm{1}}{\mathrm{6}}×\frac{\mathrm{22}}{\mathrm{7}}×\mathrm{7}^{\mathrm{2}} +\mathrm{2}×\mathrm{7}×\mathrm{5}+\frac{\mathrm{1}}{\mathrm{6}}×\frac{\mathrm{22}}{\mathrm{7}}×\mathrm{7}×\mathrm{5} \\ $$$$=\mathrm{25}.\mathrm{67}+\mathrm{70}+\mathrm{18}.\mathrm{33}…
Question Number 8016 by Nayon last updated on 28/Sep/16 $$ \\ $$$$ \\ $$$$\:{if}\:{x}\neq\mathrm{0}\:{then} \\ $$$${prove}\rightarrow \\ $$$$\:\:\:\:\frac{{x}^{\mathrm{4}} +{x}^{−\mathrm{4}} +\mathrm{1}}{{x}^{\mathrm{3}} +{x}^{−\mathrm{3}} }=\frac{{x}^{\mathrm{2}} +\mathrm{1}}{{x}}−\frac{{x}}{{x}^{\mathrm{2}} +\mathrm{1}} \\…
Question Number 73221 by aliesam last updated on 08/Nov/19 Answered by MJS last updated on 08/Nov/19 $${A}=\begin{pmatrix}{\mathrm{0}}\\{\mathrm{0}}\end{pmatrix}\:\:{B}=\begin{pmatrix}{{a}}\\{\mathrm{0}}\end{pmatrix}\:\:{C}=\begin{pmatrix}{{a}}\\{{a}}\end{pmatrix}\:\:{D}=\begin{pmatrix}{\mathrm{0}}\\{{a}}\end{pmatrix} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{E}=\begin{pmatrix}{{b}}\\{\mathrm{0}}\end{pmatrix}\:\:{F}=\begin{pmatrix}{{b}}\\{{b}}\end{pmatrix}\:\:{G}=\begin{pmatrix}{\mathrm{0}}\\{{b}}\end{pmatrix} \\ $$$${T}=\begin{pmatrix}{\frac{{a}+{b}}{\mathrm{2}}}\\{\mathrm{0}}\end{pmatrix}\:\:{I}=\begin{pmatrix}{\frac{{a}+{b}}{\mathrm{2}}}\\{\frac{{a}−{b}}{\mathrm{2}}}\end{pmatrix} \\ $$$${M}=\begin{pmatrix}{{b}}\\{\frac{{a}−{b}}{\mathrm{2}}}\end{pmatrix}\:\:{N}=\begin{pmatrix}{\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}\left({a}+{b}\right)+{b}}\\{\frac{\mathrm{3}{a}−{b}}{\mathrm{4}}}\end{pmatrix}\:\:{L}=\begin{pmatrix}{{b}}\\{{a}}\end{pmatrix} \\ $$$$\mathrm{radius}\:\mathrm{of}\:\mathrm{incircle}\:\mathrm{of}\:{ETIM}=\frac{{r}}{\mathrm{2}}=\frac{{a}−{b}}{\mathrm{4}}…