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Author: Tinku Tara

A-boy-ran-around-a-circular-part-of-radius-14m-in-15s-Calculate-the-average-velocity-and-the-average-speed-

Question Number 22112 by tawa tawa last updated on 11/Oct/17 $$\mathrm{A}\:\mathrm{boy}\:\mathrm{ran}\:\mathrm{around}\:\mathrm{a}\:\mathrm{circular}\:\mathrm{part}\:\mathrm{of}\:\mathrm{radius}\:\mathrm{14m}\:\mathrm{in}\:\mathrm{15s}.\:\mathrm{Calculate}\:\mathrm{the}\: \\ $$$$\mathrm{average}\:\mathrm{velocity}\:\mathrm{and}\:\mathrm{the}\:\mathrm{average}\:\mathrm{speed}. \\ $$ Answered by $@ty@m last updated on 11/Oct/17 $${Average}\:{Verlocity}=\frac{{Displacement}}{{time}} \\ $$$$=\frac{\mathrm{0}}{\mathrm{15}}=\mathrm{0}…

Given-a-random-variable-X-of-image-set-X-1-1-2-with-probabilities-P-X-1-e-a-P-X-1-e-b-and-P-X-2-e-c-where-a-b-and-c-are-in-an-Arithmetic-Progression-Assuming-the-mathematical-expec

Question Number 87643 by Ar Brandon last updated on 05/Apr/20 $${Given}\:{a}\:{random}\:{variable}\:\boldsymbol{{X}}\:{of}\:{image}\:{set} \\ $$$${X}\left(\Omega\right)=\left[\mathrm{1};−\mathrm{1};\mathrm{2}\right]\:{with}\:{probabilities}\:{P}\left({X}=\mathrm{1}\right)={e}^{{a}} , \\ $$$${P}\left({X}=−\mathrm{1}\right)={e}^{{b}} ,\:{and}\:{P}\left({X}=\mathrm{2}\right)={e}^{{c}} \:{where}\:{a},\:{b},\:{and}\:{c}\: \\ $$$${are}\:{in}\:\:{an}\:{Arithmetic}\:{Progression}. \\ $$$${Assuming}\:{the}\:{mathematical}\:{expection}\:{E}\left({X}\right)\:{of}\:{X}\: \\ $$$${is}\:{equal}\:{to}\:\mathrm{1}. \\…

use-the-first-principle-to-find-value-of-f-x-x-1-3-

Question Number 22105 by j.masanja06@gmail.com last updated on 11/Oct/17 $${use}\:{the}\:{first}\:{principle}\:{to}\:{find} \\ $$$${value}\:{of} \\ $$$${f}\left({x}\right)=\left({x}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \\ $$ Answered by $@ty@m last updated on 11/Oct/17 $$\frac{{dy}}{{dx}}=\underset{\delta{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{f}\left({x}+\delta{x}\right)−{f}\left({x}\right)}{\delta{x}}…

cosh-15-

Question Number 22102 by Parvez akhter last updated on 11/Oct/17 $$\mathrm{cosh}\:\left(\mathrm{15}\right) \\ $$ Answered by $@ty@m last updated on 11/Oct/17 $$\mathrm{cos}\:{hx}=\frac{{e}^{{x}} +{e}^{−{x}} }{\mathrm{2}} \\ $$$$\mathrm{cos}\:{h}\left(\mathrm{15}\right)=\frac{{e}^{\mathrm{15}}…