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Divide-10-into-two-parts-so-that-twice-the-square-of-the-first-part-plus-thrice-the-square-of-the-other-part-is-the-least-

Question Number 6179 by 314159 last updated on 17/Jun/16 $${Divide}\:\mathrm{10}\:{into}\:{two}\:{parts}\:{so}\:{that}\:{twice}\:{the} \\ $$$${square}\:{of}\:{the}\:{first}\:{part}\:{plus}\:{thrice}\:{the} \\ $$$${square}\:{of}\:{the}\:{other}\:{part}\:{is}\:{the}\:{least}. \\ $$ Commented by prakash jain last updated on 17/Jun/16 $${one}\:{part}\:{x},\:{second}\:{part}\:\left(\mathrm{10}−{x}\right)…

f-x-xf-x-f-x-

Question Number 6177 by FilupSmith last updated on 17/Jun/16 $${f}\:'\left({x}\right)={xf}\left({x}\right) \\ $$$${f}\left({x}\right)=?? \\ $$ Answered by Yozzii last updated on 17/Jun/16 $$\frac{{df}}{{dx}}={xf} \\ $$$$\Rightarrow\int\frac{{df}}{{f}}=\int{xdx} \\…

prove-the-identity-D-nds-D-2-dA-

Question Number 6174 by madscientist last updated on 17/Jun/16 $${prove}\:{the}\:{identity} \\ $$$$\int_{\partial{D}} \phi\bigtriangledown\phi\centerdot{nds}=\int\int_{{D}} \left(\phi\bigtriangledown^{\mathrm{2}\:} \phi+\bigtriangledown\phi\centerdot\bigtriangledown\phi\right){dA} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com

A-telephone-wire-hangs-from-two-points-P-Q-60m-apart-P-Q-are-on-the-same-level-the-mid-point-of-the-telephone-wire-is-3m-below-the-level-of-P-Q-Assuming-that-it-hangs-in-form-of-a-curve-

Question Number 6169 by sanusihammed last updated on 17/Jun/16 $${A}\:{telephone}\:{wire}\:{hangs}\:{from}\:{two}\:{points}\:{P},\:{Q}\:\:\mathrm{60}{m}\:{apart} \\ $$$${P},\:{Q}\:\:{are}\:{on}\:{the}\:{same}\:{level}\:.\:{the}\:{mid}\:{point}\:{of}\:{the}\:{telephone} \\ $$$${wire}\:{is}\:\:\mathrm{3}{m}\:\:{below}\:{the}\:{level}\:{of}\:{P},\:{Q}.\:{Assuming}\:{that}\:{it}\:{hangs}\:{in}\: \\ $$$${form}\:{of}\:{a}\:{curve}\:,\:\:{find}\:{it}\:{equation}. \\ $$$$ \\ $$$${please}\:{help}.\:{thanks}\:{for}\:{your}\:{time}. \\ $$ Answered by Yozzii…

If-the-sum-of-the-first-nth-terms-of-a-sequence-is-given-by-S-n-9-1-1-3-n-a-find-the-first-and-the-second-term-of-the-sequence-b-find-the-nth-term-of-the-sequence-c-show-that-th

Question Number 6166 by sanusihammed last updated on 16/Jun/16 $${If}\:{the}\:{sum}\:{of}\:{the}\:{first}\:{nth}\:{terms}\:{of}\:{a}\:{sequence}\:{is}\:{given}\:{by}\: \\ $$$${S}_{{n}\:} \:=\:\:\mathrm{9}\left(\mathrm{1}\:−\:\frac{\mathrm{1}}{\mathrm{3}^{{n}} \:}\right) \\ $$$$\left({a}\right)\:{find}\:{the}\:{first}\:{and}\:{the}\:{second}\:{term}\:{of}\:{the}\:{sequence} \\ $$$$\left({b}\right)\:{find}\:{the}\:{nth}\:{term}\:{of}\:{the}\:{sequence} \\ $$$$\left({c}\right)\:{show}\:{that}\:{the}\:{sequence}\:{is}\:{a}\:{GP}\:{and}\:{find}\:{it}\:{common}\:{ratio} \\ $$$$ \\ $$$${please}\:{help}. \\…

A-stone-is-thrown-vertically-upwards-with-velocity-20m-s-at-the-same-time-and-20m-vertically-above-a-second-stone-is-allowed-to-fall-After-whst-time-and-at-what-height-do-they-collide-take-

Question Number 6164 by sanusihammed last updated on 16/Jun/16 $${A}\:{stone}\:{is}\:{thrown}\:{vertically}\:{upwards}\:{with}\:{velocity}\:\mathrm{20}{m}/{s}. \\ $$$${at}\:{the}\:{same}\:{time}\:,\:{and}\:\mathrm{20}{m}\:{vertically}\:{above}\:.\:{a}\:{second}\:{stone}\:{is} \\ $$$${allowed}\:{to}\:{fall}.\:{After}\:{whst}\:{time}\:{and}\:{at}\:{what}\:{height}\:{do}\:{they}\: \\ $$$${collide}\:?. \\ $$$${take}\:\:{g}\:=\:\mathrm{10}{m}/{s}^{\mathrm{2}} \\ $$ Commented by prakash jain last…

Question-71698

Question Number 71698 by TawaTawa last updated on 18/Oct/19 Commented by MJS last updated on 19/Oct/19 $$\mathrm{we}\:\mathrm{had}\:\mathrm{a}\:\mathrm{similar}\:\mathrm{example}\:\mathrm{a}\:\mathrm{few}\:\mathrm{weeks}\:\mathrm{ago} \\ $$ Commented by TawaTawa last updated on…

The-side-of-a-square-is-measured-to-be-12cm-long-cofrect-to-the-nearest-cm-Find-the-maximum-absolute-error-and-the-maximum-percentage-error-for-a-The-length-of-the-square-Answer-0-5cm-

Question Number 71696 by TawaTawa last updated on 18/Oct/19 $$\mathrm{The}\:\mathrm{side}\:\mathrm{of}\:\mathrm{a}\:\mathrm{square}\:\mathrm{is}\:\mathrm{measured}\:\mathrm{to}\:\mathrm{be}\:\:\mathrm{12cm}\:\mathrm{long}\:\mathrm{cofrect} \\ $$$$\mathrm{to}\:\mathrm{the}\:\mathrm{nearest}\:\:\mathrm{cm}.\:\:\mathrm{Find}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{absolute}\:\mathrm{error} \\ $$$$\mathrm{and}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{percentage}\:\mathrm{error}\:\mathrm{for} \\ $$$$\left(\mathrm{a}\right)\:\:\:\:\mathrm{The}\:\mathrm{length}\:\mathrm{of}\:\mathrm{the}\:\mathrm{square}\:\:\:\:\:\left(\mathrm{Answer}:\:\:\mathrm{0}.\mathrm{5cm},\:\:\mathrm{4}.\mathrm{17\%}\right) \\ $$$$\left(\mathrm{b}\right)\:\:\:\:\mathrm{The}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{square}.\:\:\:\:\:\:\left(\mathrm{Answer}:\:\:\:\:\mathrm{12}.\mathrm{25cm},\:\:\:\mathrm{8}.\mathrm{5\%}\right) \\ $$ Answered by MJS last updated…

Question-71695

Question Number 71695 by peter frank last updated on 18/Oct/19 Answered by MJS last updated on 19/Oct/19 $$\int\frac{\sqrt{\mathrm{tan}\:{x}}}{\mathrm{1}+\sqrt{\mathrm{tan}\:{x}}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\sqrt{\mathrm{tan}\:{x}}\:\rightarrow\:{dx}=\mathrm{2}\sqrt{\mathrm{tan}\:{x}}\mathrm{cos}^{\mathrm{2}} \:{x}\:{dt}=\frac{\mathrm{2}{t}}{{t}^{\mathrm{4}} +\mathrm{1}}{dt}\right] \\ $$$$=\mathrm{2}\int\frac{{t}^{\mathrm{2}} }{\left({t}+\mathrm{1}\right)\left({t}^{\mathrm{4}}…