Question Number 168948 by cortano1 last updated on 22/Apr/22 Answered by mahdipoor last updated on 22/Apr/22 $${S}\left({ABCDEF}\right)=\mathrm{28}.\mathrm{75}=\frac{\mathrm{115}}{\mathrm{4}} \\ $$$$\left\{{consider}\:{a}\:{rectangle}\:{around}\:{it}\:{and}\:\right. \\ $$$$\left.{calculate}\:{S}\right\} \\ $$$${a}={coefficient}\:{of}\:{similar}=\frac{{AE}}{{AE}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${S}\left({ABCDEF}\right)={a}^{\mathrm{2}}…
Question Number 37862 by ajfour last updated on 18/Jun/18 Commented by MrW3 last updated on 18/Jun/18 $${yes},\:\sqrt{{r}^{\mathrm{2}} +{q}^{\mathrm{2}} }<\mathrm{2}{a}. \\ $$ Commented by ajfour last…
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Question Number 103228 by ajfour last updated on 13/Jul/20 Commented by ajfour last updated on 13/Jul/20 $${Calculate}\:\boldsymbol{{h}},\:\boldsymbol{{k}}\:\:{in}\:{terms}\:{of}\:{ellipse} \\ $$$${parameters}\:\boldsymbol{{a}},\boldsymbol{{b}}. \\ $$ Answered by mr W…
Question Number 103159 by 281981 last updated on 13/Jul/20 $$\boldsymbol{\mathrm{how}}\:\boldsymbol{\mathrm{do}}\:\boldsymbol{\mathrm{you}}\:\boldsymbol{\mathrm{represent}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{distance}}\:\boldsymbol{\mathrm{between}} \\ $$$$\boldsymbol{\mathrm{M}}\:{and}\boldsymbol{{N}}\:\mathrm{is}\:\mathrm{7} \\ $$ Commented by 9027201563 last updated on 13/Jul/20 $$\mathrm{from}\:\mathrm{the}\:\mathrm{distance}\:\mathrm{formuar} \\ $$$$\mathrm{distance}=\sqrt{\left(\mathrm{x}_{\mathrm{2}} −\mathrm{x}_{\mathrm{1}}…
Question Number 168609 by infinityaction last updated on 14/Apr/22 Commented by infinityaction last updated on 14/Apr/22 $${coordinate}\:{of}\:{point}\:\:\left({a},{b}\right)\: \\ $$ Answered by mr W last updated…
Question Number 103062 by ajfour last updated on 12/Jul/20 Commented by ajfour last updated on 12/Jul/20 $${If}\:{both}\:{coloured}\:{regions}\:{have} \\ $$$${equal}\:{areas},\:{find}\:{radius}\:{of}\:{the} \\ $$$${semicircle}. \\ $$ Answered by…
Question Number 103043 by bemath last updated on 12/Jul/20 $$\begin{cases}{{x}^{\mathrm{2}} +\mathrm{2}{xy}+\mathrm{21}=\mathrm{6}{x}}\\{{y}^{\mathrm{2}} +\mathrm{2}{yz}−\mathrm{8}=\mathrm{6}{y}}\\{{z}^{\mathrm{2}} +\mathrm{2}{zx}−\mathrm{4}=\mathrm{6}{z}}\end{cases} \\ $$$${find}\:{x}+{y}+{z}\: \\ $$ Answered by maths mind last updated on 12/Jul/20…
Question Number 37485 by ajfour last updated on 13/Jun/18 Commented by ajfour last updated on 13/Jun/18 $${Relate}\:\boldsymbol{{r}}\:{and}\:\boldsymbol{{R}}\:{if}\:\:\boldsymbol{\alpha},\:\boldsymbol{{d}}\:{are}\:{given}. \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on…
Question Number 168427 by cortano1 last updated on 10/Apr/22 Commented by som(math1967) last updated on 10/Apr/22 $$\measuredangle{ABC}={x} \\ $$$$\boldsymbol{{cosx}}=\frac{\mathrm{24}}{\mathrm{64}}=\frac{\mathrm{3}}{\mathrm{8}} \\ $$$$\:\frac{{BE}}{\mathrm{2}}={AB}×{S}\boldsymbol{{in}}\left(\mathrm{270}−\mathrm{3}\boldsymbol{{x}}\right) \\ $$$$\boldsymbol{{B}}{E}=\mathrm{2}{AB}×−{Cos}\mathrm{3}{x} \\ $$$$\:=\mathrm{2}×\mathrm{64}×\left(\mathrm{3}\boldsymbol{{cosx}}−\mathrm{4}\boldsymbol{{cos}}^{\mathrm{3}}…