Question Number 126130 by Algoritm last updated on 17/Dec/20
Commented by talminator2856791 last updated on 17/Dec/20
$$\:\mathrm{this}\:\mathrm{looks}\:\mathrm{like}\:\mathrm{physics}\:\mathrm{question} \\ $$
Answered by mr W last updated on 17/Dec/20
Commented by mr W last updated on 17/Dec/20
$$\sqrt{\mathrm{10}^{\mathrm{2}} +\mathrm{10}^{\mathrm{2}} }+\sqrt{\mathrm{10}^{\mathrm{2}} +\mathrm{20}^{\mathrm{2}} }+\sqrt{\mathrm{10}^{\mathrm{2}} +\mathrm{30}^{\mathrm{2}} } \\ $$$$=\mathrm{10}\left(\sqrt{\mathrm{2}}+\sqrt{\mathrm{5}}+\sqrt{\mathrm{10}}\right) \\ $$$$=\mathrm{68}.\mathrm{1}\:{cm} \\ $$
Commented by talminator2856791 last updated on 18/Dec/20
$$\mathrm{why}\:\mathrm{can}\:\mathrm{you}\:\mathrm{assume}\:\mathrm{it}\:\mathrm{forms}\:\mathrm{a}\:\mathrm{straight}\:\mathrm{line}? \\ $$
Commented by mr W last updated on 18/Dec/20
$${the}\:{centers}\:{of}\:{mass}\:{of}\:{all}\:{three}\:{objects} \\ $$$${and}\:{the}\:{support}\:{point}\:\left({nose}\right)\:{must}\:{be} \\ $$$${collinear}\:{and}\:{perpendicular}. \\ $$