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4-49-20-6-

Question Number 20201 by khamizan833@gmail.con last updated on 24/Aug/17 449206= Answered by Einstein Newton last updated on 24/Aug/17 $$\sqrt{\mathrm{49}\:−\:\mathrm{20}\sqrt{\mathrm{6}}}\:=\:\sqrt{\mathrm{5}^{\mathrm{2}} \:+\:\left(\mathrm{2}\sqrt{\mathrm{6}}\right)^{\mathrm{2}} \:−\:\mathrm{2}\left(\mathrm{5}\right)\left(\mathrm{2}\sqrt{\mathrm{6}}\right)} \

Question-85709

Question Number 85709 by otchereabdullai@gmail.com last updated on 24/Mar/20 Answered by MJS last updated on 24/Mar/20 wehadthismanytimesbeforethepostsare50,thecableis80,thespaceunderthecableis10$$\mathrm{the}\:\mathrm{effective}\:\mathrm{height}\:\mathrm{of}\:\mathrm{the}\:\mathrm{poles}\:\mathrm{is}\:\mathrm{40}\:\Rightarrow…