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Author: Tinku Tara

L-lim-i-1-i-1-i-i-1-L-

Question Number 4587 by FilupSmith last updated on 09/Feb/16 $${L}=\underset{{i}\rightarrow\infty} {\mathrm{lim}}\:\frac{\left(−\mathrm{1}\right)^{{i}+\mathrm{1}} {i}}{{i}+\mathrm{1}} \\ $$$${L}=? \\ $$ Commented by Yozzii last updated on 14/Feb/16 $${L}\:{does}\:{not}\:{exist}.\:{While}\:\mid{L}\mid=\underset{{i}\rightarrow\infty} {\mathrm{lim}}\frac{{i}}{{i}+\mathrm{1}}…

500-108-54-and-15-are-are-not-perfect-cubes-how-do-we-factorize-1-500X-3-108Y-3-2-54X-3-15Y-3-please-help-me-with-well-explained-steps-or-shortcut-thanks-

Question Number 135656 by otchereabdullai@gmail.com last updated on 14/Mar/21 $$\mathrm{500},\:\mathrm{108},\mathrm{54}\:\mathrm{and}\:\mathrm{15}\:\mathrm{are}\:\mathrm{are}\:\mathrm{not}\: \\ $$$$\mathrm{perfect}\:\mathrm{cubes}\:\mathrm{how}\:\mathrm{do}\:\mathrm{we}\:\mathrm{factorize} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{500X}^{\mathrm{3}} −\mathrm{108Y}^{\mathrm{3}} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{54X}^{\mathrm{3}} +\mathrm{15Y}^{\mathrm{3}} \\ $$$$\mathrm{please}\:\mathrm{help}\:\mathrm{me}\:\mathrm{with}\:\mathrm{well}\:\mathrm{explained}\: \\ $$$$\mathrm{steps}\:\mathrm{or}\:\mathrm{shortcut}\:\mathrm{thanks} \\ $$ Commented…

Question-70121

Question Number 70121 by ahmadshahhimat775@gmail.com last updated on 01/Oct/19 Answered by mind is power last updated on 01/Oct/19 $${let}\:{a}\:{angle}\:{left}\:−,{c}\:{side}\:{of}\:{squar} \\ $$$$\Rightarrow{c}=\mathrm{6}{sin}\left({a}\right) \\ $$$$\frac{\mathrm{6}{sin}\left({a}\right)+\mathrm{6}{cos}\left({a}\right)}{{sin}\left(\mathrm{45}\right)}=\frac{\mathrm{10}}{{sin}\left(\mathrm{45}+{a}\right)} \\ $$$$\Rightarrow\mathrm{6}.\left({sin}\left({a}\right)+{cos}\left({a}\right)\right).{sin}\left(\mathrm{45}+{a}\right)=\mathrm{10}{sin}\left(\mathrm{45}\right)…

Question-4584

Question Number 4584 by MrGreg last updated on 08/Feb/16 $$ \\ $$ Answered by FilupSmith last updated on 09/Feb/16 $$\:\:\:\:{S}=\mathrm{1}−\mathrm{2}+\mathrm{4}−\mathrm{8}+\mathrm{16}−\mathrm{32}+\mathrm{64}−… \\ $$$$+{S}=\mathrm{0}+\mathrm{1}−\mathrm{2}+\mathrm{4}−\mathrm{8}+\mathrm{16}−\mathrm{32}+\mathrm{64}−… \\ $$$$\mathrm{2}{S}=\mathrm{1}+\left(−\mathrm{1}+\mathrm{2}−\mathrm{4}+\mathrm{8}−\mathrm{16}+…\right) \\…

In-a-right-triangle-the-mid-point-of-the-hypotenuse-is-equidistant-from-all-the-three-vertices-of-the-triangle-

Question Number 4579 by Rasheed Soomro last updated on 08/Feb/16 $${In}\:{a}\:{right}\:{triangle},\:{the}\:{mid}-{point}\:{of}\:{the} \\ $$$${hypotenuse}\:{is}\:{equidistant}\:{from}\:{all}\:{the} \\ $$$${three}\:{vertices}\:{of}\:{the}\:{triangle}. \\ $$ Commented by Yozzii last updated on 08/Feb/16 $${Prove}\:{this}\:{theorem}?…