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The-distance-S-metre-travelled-in-time-t-seconds-by-an-object-released-from-rest-and-allow-to-fall-freely-under-the-force-of-gravity-is-given-by-s-t-4-5t-2-find-a-the-average-speed-of-the-object-

Question Number 171005 by Mastermind last updated on 06/Jun/22 $${The}\:{distance}\:{S}\:{metre}\:{travelled}\:{in}\:{time} \\ $$$${t}\:{seconds}\:{by}\:{an}\:{object}\:{released}\:{from} \\ $$$${rest}\:{and}\:{allow}\:{to}\:{fall}\:{freely}\:{under}\:{the} \\ $$$${force}\:{of}\:{gravity}\:{is}\:{given}\:{by}\:{s}\left({t}\right)=\mathrm{4}.\mathrm{5}{t}^{\mathrm{2}} . \\ $$$${find}\: \\ $$$$\left.{a}\right)\:{the}\:{average}\:{speed}\:{of}\:{the}\:{object} \\ $$$${during}\:{the}\:{time}\:{interval}\:{from}\:\mathrm{2}\:{to} \\ $$$$\mathrm{2}.\mathrm{5}\:{seconds}.…

Find-the-equation-of-the-tangent-to-the-curve-x-2-y-2-4x-6y-12-0-at-the-point-2-3-Mastermind-

Question Number 170997 by Mastermind last updated on 06/Jun/22 $${Find}\:{the}\:{equation}\:{of}\:{the}\:{tangent}\:{to}\:{the} \\ $$$${curve}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{6}{y}−\mathrm{12}=\mathrm{0}\:{at}\:{the} \\ $$$${point}\:\left(\mathrm{2},\:\mathrm{3}\right). \\ $$$$ \\ $$$${Mastermind} \\ $$ Answered by depressiveshrek…

Find-the-volume-of-the-solid-obtained-by-rotating-about-x-axis-of-the-curve-y-x-on-the-interval-0-2-Mastermind-

Question Number 170996 by Mastermind last updated on 06/Jun/22 $${Find}\:{the}\:{volume}\:{of}\:{the}\:{solid}\:{obtained} \\ $$$${by}\:{rotating}\:{about}\:{x}−{axis}\:{of}\:{the}\:{curve} \\ $$$${y}=\sqrt{{x}}\:{on}\:{the}\:{interval}\:\left[\mathrm{0},\:\mathrm{2}\right]. \\ $$$$ \\ $$$${Mastermind} \\ $$ Answered by mr W last…

Question-39914

Question Number 39914 by solihin last updated on 13/Jul/18 Commented by MrW3 last updated on 13/Jul/18 $${z}=\frac{{i}}{\mathrm{2}−{i}}=\frac{{i}\:\left(\mathrm{2}+{i}\right)}{\left(\mathrm{2}−{i}\right)\left(\mathrm{2}+{i}\right)}=\frac{−\mathrm{1}+\mathrm{2}{i}}{\mathrm{5}}=−\frac{\mathrm{1}}{\mathrm{5}}+\frac{\mathrm{2}}{\mathrm{5}}{i} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\left(−\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}+\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}}{i}\right) \\ $$$$={r}\left(\mathrm{cos}\:\theta+{i}\:\mathrm{sin}\:\theta\right) \\ $$$${with}\:{r}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}} \\ $$$$\theta=\pi−\mathrm{tan}^{−\mathrm{1}}…