Question Number 67749 by TawaTawa last updated on 31/Aug/19 Answered by JDamian last updated on 31/Aug/19 $${Assuming}\:{A},\:{B}\:{and}\:{C}\:{points}\:{are}\:{the}\:{centers} \\ $$$${of}\:{their}\:{circles}: \\ $$$$ \\ $$$${AC}={R}_{{A}} −{R}_{{C}} =\mathrm{11}−\mathrm{5}=\mathrm{6}…
Question Number 2196 by Rasheed Soomro last updated on 08/Nov/15 $$\left({a}+{b}\right)^{\mathrm{2}} ={a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{2}{ab} \\ $$$$\left({a}+{b}+{c}\right)^{\mathrm{2}} ={a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +\mathrm{2}{ab}+\mathrm{2}{bc}+\mathrm{2}{ca} \\ $$$$\left({a}+{b}+{c}+{d}\right)^{\mathrm{2}} =? \\ $$$$\left({a}_{\mathrm{1}}…
Question Number 67728 by TawaTawa last updated on 30/Aug/19 Commented by mr W last updated on 30/Aug/19 $${question}\:{is}\:{not}\:{clear}. \\ $$$${is}\:\angle{BAD}=\mathrm{90}°\:{as}\:{marked}\:{or}\:{is} \\ $$$$\angle{BAC}=\mathrm{90}°\:{as}\:{shown}? \\ $$ Commented…
Question Number 67716 by TawaTawa last updated on 30/Aug/19 Answered by mind is power last updated on 30/Aug/19 $${let}\:{c}_{\mathrm{1}} ={demi}\:{circle}\:{of}\:{r}=\mathrm{2} \\ $$$${c}\mathrm{2}={circle}\:{r}=\mathrm{1} \\ $$$${c}_{\mathrm{1}} ..{x}^{\mathrm{2}}…
Question Number 67692 by mr W last updated on 30/Aug/19 Answered by mr W last updated on 30/Aug/19 Commented by mr W last updated on…
Question Number 2138 by Filup last updated on 06/Nov/15 $$\mathrm{Is}\:\mathrm{the}\:\mathrm{following}\:\mathrm{proof}\:\mathrm{correct}? \\ $$$$ \\ $$$$\Delta=\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{i}} \mathrm{2}^{{i}} =\mathrm{1}−\mathrm{2}+\mathrm{4}−\mathrm{8}+\mathrm{16}−\mathrm{32}+… \\ $$$$ \\ $$$$\mathrm{Let}: \\ $$$$\Delta_{\mathrm{1}} =\mathrm{1}−\mathrm{2}+\mathrm{4}−\mathrm{8}+\mathrm{16}−\mathrm{32}+……
Question Number 67623 by mr W last updated on 29/Aug/19 Commented by mr W last updated on 29/Aug/19 $${find}\:{the}\:{shaded}\:{area}\:{A}=? \\ $$ Commented by Prithwish sen…
Question Number 67615 by TawaTawa last updated on 29/Aug/19 Commented by TawaTawa last updated on 29/Aug/19 $$\mathrm{i}\:\mathrm{did}.\:\:\:\:\:\:\mathrm{A}\:\mathrm{of}\:\mathrm{Big}\:\mathrm{semi}\:\mathrm{circle}\:=\:\frac{\pi\mathrm{r}^{\mathrm{2}} }{\mathrm{2}}\:\:=\:\:\frac{\pi.\mathrm{2}^{\mathrm{2}} }{\mathrm{2}}\:\:=\:\:\mathrm{2}\pi \\ $$$$\mathrm{A}\:\mathrm{of}\:\mathrm{second}\:\mathrm{big}\:\mathrm{semicircle}\:\:=\:\:\frac{\pi\mathrm{r}^{\mathrm{2}} }{\mathrm{2}}\:\:=\:\:\frac{\mathrm{1}}{\mathrm{2}}\pi \\ $$$$\mathrm{But}\:\mathrm{i}\:\mathrm{don}'\mathrm{t}\:\mathrm{know}\:\mathrm{what}\:\mathrm{to}\:\mathrm{do}\:\mathrm{again} \\…
Question Number 2001 by Fitrah last updated on 29/Oct/15 $${Prove}\:{that}\:: \\ $$$$\frac{\mathrm{1}}{\mathrm{15}}\:<\:\frac{\mathrm{1}}{\mathrm{2}}\:\centerdot\:\frac{\mathrm{3}}{\mathrm{4}}\:\centerdot\:\frac{\mathrm{5}}{\mathrm{6}}\:\centerdot\:\centerdot\:\centerdot\:\centerdot\:\centerdot\:\frac{\mathrm{99}}{\mathrm{100}}\:<\:\frac{\mathrm{1}}{\mathrm{10}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 1999 by Fitrah last updated on 29/Oct/15 $${if}\: \\ $$$$\left(\mathrm{1}\:+\:{tan}\:\mathrm{1}°\right)\left(\mathrm{1}\:+\:{tan}\:\mathrm{2}°\right)\left(\mathrm{1}\:+\:{tan}\:\mathrm{3}°\right)…… \\ $$$$\left…..\right)\left(\mathrm{1}\:+\:{tan}\:\mathrm{44}°\right)\left(\mathrm{1}\:+\:{tan}\:\mathrm{45}°\right)\overset{} {\:}=\:\:\left\{\:\:\:\sqrt[{\mathrm{3}}]{\left(\sqrt{\mathrm{50}}\:+\:\mathrm{7}\right)}\:−\:\sqrt[{\mathrm{3}}]{\left(\sqrt{\mathrm{50}}−\mathrm{7}\right)}\:\overset{\left({x}\:−\:\mathrm{7}\right)} {\right\}} \\ $$$$\: \\ $$$${find}\:{x}\:=\:…? \\ $$ Answered by Yozzi…