Question Number 186442 by Mingma last updated on 04/Feb/23 Answered by HeferH last updated on 04/Feb/23 Commented by HeferH last updated on 04/Feb/23 $$\left(\frac{\mathrm{11}}{\mathrm{2}}\right)^{\mathrm{2}} −{r}^{\mathrm{2}}…
Question Number 186401 by Rupesh123 last updated on 04/Feb/23 Answered by ajfour last updated on 04/Feb/23 Commented by ajfour last updated on 04/Feb/23 $${Red}_{{area}} =…
Question Number 120790 by peter frank last updated on 02/Nov/20 Commented by TANMAY PANACEA last updated on 02/Nov/20 $${eqn}\:{big}\:{circle}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} ={R}^{\mathrm{2}} \:\:{and}\:{area}\:\pi{R}^{\mathrm{2}} \\ $$$${small}\:{circle}\:{centre}\left(−{a},\mathrm{0}\right)\:{radius}\:{r} \\…
Question Number 186302 by ajfour last updated on 03/Feb/23 Commented by ajfour last updated on 03/Feb/23 $${Can}\:{it}\:{be}\:=\:\frac{\mathrm{1}}{\mathrm{6}}\:? \\ $$ Commented by Frix last updated on…
Question Number 55058 by gunawan last updated on 16/Feb/19 $$\mathrm{Prove}\:\mathrm{that}: \\ $$$$\mathrm{P}_{{n}} \left({z}\right)={a}_{\mathrm{0}} {z}^{{n}} +{a}_{\mathrm{1}} {z}^{{n}−\mathrm{1}} +…+{a}_{{n}−\mathrm{1}} {z}+{a}_{{n}} \\ $$$$\mathrm{at}\:\mathrm{least}\:\mathrm{have}\:\mathrm{one}\:\:\mathrm{value}\:\mathrm{of}\:\mathrm{zero} \\ $$ Terms of Service…
Question Number 55055 by peter frank last updated on 16/Feb/19 Answered by mr W last updated on 16/Feb/19 $${T}_{{n}} =\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)\left(\mathrm{1}−\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\right)=\frac{\left({n}−\mathrm{1}\right)}{{n}}×\frac{\left({n}+\mathrm{1}\right)}{{n}}×\frac{\left({n}+\mathrm{1}\right)}{{n}} \\ $$$${T}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{3}}{\mathrm{2}}×\frac{\mathrm{3}}{\mathrm{2}} \\…
Question Number 55054 by peter frank last updated on 16/Feb/19 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 186052 by Rupesh123 last updated on 31/Jan/23 Answered by HeferH last updated on 31/Jan/23 Commented by HeferH last updated on 31/Jan/23 $${S}\::\:{area}\:{of}\:{regular}\:{octagon} \\…
Question Number 186029 by Rupesh123 last updated on 31/Jan/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 186030 by Rupesh123 last updated on 31/Jan/23 Answered by ajfour last updated on 31/Jan/23 $${x}^{\mathrm{2}} +{y}^{\mathrm{2}} ={d}^{\mathrm{2}} \\ $$$$\left({a}−{x}\right){y}={d}^{\mathrm{2}} \\ $$$$\left({b}−{y}\right){x}={d}^{\mathrm{2}} \\ $$$${d}\left(\frac{{x}}{{y}}+\mathrm{1}+\frac{{y}}{{x}}\right)={c}…