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Is-there-an-n-gt-11-such-that-every-digit-of-2-n-in-decimal-representation-is-even-

Question Number 227323 by CrispyXYZ last updated on 15/Jan/26 $$\mathrm{Is}\:\mathrm{there}\:\mathrm{an}\:{n}>\mathrm{11}\:\mathrm{such}\:\mathrm{that}\:\mathrm{every}\:\mathrm{digit}\:\mathrm{of}\:\mathrm{2}^{{n}} \\ $$$$\mathrm{in}\:\mathrm{decimal}\:\mathrm{representation}\:\mathrm{is}\:\mathrm{even}? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com

f-z-3-1-z-3-z-f-z-

Question Number 227319 by Lara2440 last updated on 15/Jan/26 $${f}\left({z}^{\mathrm{3}} +\frac{\mathrm{1}}{{z}^{\mathrm{3}} }\right)={z} \\ $$$${f}\left({z}\right)=? \\ $$ Answered by breniam last updated on 14/Jan/26 $$\left(\forall{z}\in\mathbb{R}\backslash\left\{\mathrm{0}\right\}\right)\left({f}\left({z}^{\mathrm{3}} +\frac{\mathrm{1}}{{z}^{\mathrm{3}}…

tg-15-ctg-5-

Question Number 227312 by hardmath last updated on 13/Jan/26 $$\mathrm{tg}\left(\mathrm{15}\right)\:+\:\mathrm{ctg}\left(\mathrm{5}\right)\:=\:? \\ $$ Answered by Kassista last updated on 13/Jan/26 $$ \\ $$$${tg}\left(\mathrm{3}\theta\right)\:=\:\frac{\mathrm{3}{tg}\left(\theta\right)−{tg}^{\mathrm{3}} \left(\theta\right)}{\mathrm{1}−\mathrm{3}{tg}^{\mathrm{2}} \left(\theta\right)}\:\therefore\:{tg}\left(\mathrm{15}\right)=\frac{\mathrm{3}{tg}\left(\mathrm{5}\right)−{tg}^{\mathrm{3}} \left(\mathrm{5}\right)}{\mathrm{1}−\mathrm{3}{tg}^{\mathrm{2}}…

Question-227276

Question Number 227276 by mr W last updated on 11/Jan/26 Commented by mr W last updated on 12/Jan/26 $${A}\:{hemispherical}\:{funnel}\:{is}\:{placed} \\ $$$${tightly}\:{against}\:{the}\:{top}\:{of}\:{a}\:{table}.\: \\ $$$${Water}\:{is}\:{poured}\:{into}\:{it}\:{slowly}\: \\ $$$${through}\:{a}\:{small}\:{hole}\:{located}\:{at}\:{the}\:…

Question-227272

Question Number 227272 by Spillover last updated on 11/Jan/26 Answered by breniam last updated on 12/Jan/26 $$\left({x}+\mathrm{1}\right){y}'\left({x}\right)−{y}\left({x}\right)={e}^{{x}} \left(\mathrm{1}+{x}\right)^{\mathrm{2}} \\ $$$${y}\left({x}\right)={a}\left({x}\right){b}\left({x}\right) \\ $$$${y}'\left({x}\right)={a}'\left({x}\right){b}\left({x}\right)+{a}\left({x}\right){b}'\left({x}\right) \\ $$$${a}\left({x}\right)\left(\left({x}+\mathrm{1}\right){b}'\left({x}\right)−{b}\left({x}\right)\right)+{a}'\left({x}\right){b}\left({x}\right)\left({x}+\mathrm{1}\right) \\…