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Question Number 8667 by tawakalitu last updated on 20/Oct/16 Answered by sandy_suhendra last updated on 20/Oct/16 $$\mathrm{at}\:\mathrm{t}=\mathrm{5}\:\mathrm{sec}\:\mathrm{the}\:\mathrm{body}\:\mathrm{reaches}\:\mathrm{the}\:\mathrm{ground}\:\Rightarrow\:\mathrm{y}=\mathrm{0} \\ $$$$\mathrm{y}=\mathrm{V}_{\mathrm{o}} .\mathrm{t}\:−\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{g}.\mathrm{t}^{\mathrm{2}} \\ $$$$\mathrm{0}=\mathrm{V}_{\mathrm{o}} .\mathrm{5}\:−\:\frac{\mathrm{1}}{\mathrm{2}}.\mathrm{10}.\mathrm{5}^{\mathrm{2}} \\ $$$$\mathrm{5V}_{\mathrm{o}}…
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Question Number 74181 by MASANJAJ last updated on 19/Nov/19 Answered by Rio Michael last updated on 19/Nov/19 $$\mathrm{1}{cm}\:=\:\mathrm{500000}{cm} \\ $$$$\mathrm{1}{cm}\:=\:\mathrm{5}{km} \\ $$$$\:{x}\:=\:\mathrm{30}{km} \\ $$$$\mathrm{30}\:=\:\mathrm{5}{x} \\…
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Question Number 74163 by MASANJAJ last updated on 19/Nov/19 $${factorize} \\ $$$$\left({x}−\mathrm{1}\right)^{\mathrm{2}} \:−\mathrm{4}{y}^{\mathrm{2}} \\ $$ Answered by MJS last updated on 19/Nov/19 $${a}^{\mathrm{2}} −{b}^{\mathrm{2}} =\left({a}−{b}\right)\left({a}+{b}\right)…