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Author: Tinku Tara

An-MTN-mask-is-erected-at-a-point-P-in-ilaro-town-At-a-point-B-due-west-The-angle-of-elevation-of-its-top-is-and-at-point-C-due-south-the-angle-of-elevation-is-With-the-aid-of-an-appropriat

Question Number 6723 by Tawakalitu. last updated on 16/Jul/16 $${An}\:{MTN}\:{mask}\:{is}\:{erected}\:{at}\:{a}\:{point}\:{P}\:\:{in}\:{ilaro}\:{town}.\:{At}\:{a}\:{point} \\ $$$${B}\:{due}\:{west},\:{The}\:{angle}\:{of}\:{elevation}\:{of}\:{its}\:{top}\:{is}\:\beta\:{and}\:{at}\:{point}\: \\ $$$${C}\:{due}\:{south},\:{the}\:{angle}\:{of}\:{elevation}\:{is}\:\alpha.\:{With}\:{the}\:{aid}\:{of}\:{an}\: \\ $$$${appropriate}\:{diagam}.\:{show}\:{that}\:{the}\:{angle}\:{of}\:{elevation}\:{of}\:{the}\:{top} \\ $$$${from}\:{a}\:{point}\:{due}\:{south}\:{of}\:{B}\:{and}\:{due}\:{west}\:{of}\:{C}\:{is}\: \\ $$$$ \\ $$$${cot}^{−\mathrm{1}} \left[{cot}^{\mathrm{2}} \left(\beta\right)\:+\:{cot}^{\mathrm{2}} \left(\alpha\right)\right]^{\frac{\mathrm{1}}{\mathrm{2}}}…

yz-dx-xz-dy-xy-dz-pleas-sir-help-me-

Question Number 72259 by mhmd last updated on 26/Oct/19 $$\int{yz}\:{dx}\:+\int{xz}\:{dy}\:+\int{xy}\:{dz}\:\:\:\:{pleas}\:{sir}\:{help}\:{me}\:? \\ $$ Answered by MJS last updated on 26/Oct/19 $$…={yz}\int{dx}+{xz}\int{dy}+{xy}\int{dz}=\mathrm{3}{xyz} \\ $$$$\mathrm{all}\:\mathrm{variables}\:\neq\:\mathrm{the}\:\mathrm{integral}\:\mathrm{variable}\:\mathrm{are} \\ $$$$\mathrm{considered}\:\mathrm{as}\:\mathrm{constant}\:\mathrm{factors} \\…

Prove-that-the-locus-of-a-point-which-moves-its-distance-from-the-point-b-0-is-p-times-its-distance-from-the-point-b-0-is-p-2-1-x-2-y-2-b-2-2b-p-2-1-x

Question Number 6716 by Tawakalitu. last updated on 15/Jul/16 $${Prove}\:{that}\:{the}\:{locus}\:{of}\:{a}\:{point}\:{which}\:{moves}\:{its}\:{distance}\:{from}\: \\ $$$${the}\:{point}\:\left(−{b},\:\mathrm{0}\right)\:{is}\:{p}\:{times}\:{its}\:{distance}\:{from}\:{the}\:{point}\:\left({b},\:\mathrm{0}\right)\:{is} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left({p}^{\mathrm{2}} \:−\:\mathrm{1}\right)\left({x}^{\mathrm{2}} \:+\:{y}^{\mathrm{2}} \:+\:{b}^{\mathrm{2}} \right)\:−\:\mathrm{2}{b}\left({p}^{\mathrm{2}} \:+\:\mathrm{1}\right){x}\:=\:\mathrm{0} \\ $$$${Show}\:{that}\:{this}\:{locus}\:{is}\:{a}\:{circle}\:{and}\:{find}\:{its}\:{radius}. \\ $$ Answered by…

Solve-the-following-equation-for-0-lt-lt-360-o-cosx-cos3x-cos5x-cos7x-0-

Question Number 6715 by Tawakalitu. last updated on 15/Jul/16 $${Solve}\:{the}\:{following}\:{equation}\:{for}\:\mathrm{0}\:<\:\Theta\:<\:\mathrm{360}^{{o}} \\ $$$${cosx}\:+\:{cos}\mathrm{3}{x}\:+\:{cos}\mathrm{5}{x}\:+\:{cos}\mathrm{7}{x}\:=\:\mathrm{0} \\ $$ Answered by Yozzii last updated on 16/Jul/16 $${We}\:{know}\:{the}\:{identity}\:{cosa}+{cosb}=\mathrm{2}{cos}\frac{{a}+{b}}{\mathrm{2}}{cos}\frac{{a}−{b}}{\mathrm{2}}\:{for}\:{a},{b}\in\mathbb{R}….\left(\mathrm{1}\right) \\ $$$${Let}\:{u}={cosx}+{cos}\mathrm{3}{x}+{cos}\mathrm{5}{x}+{cos}\mathrm{7}{x}. \\…

Question-72251

Question Number 72251 by aliesam last updated on 26/Oct/19 Commented by mathmax by abdo last updated on 26/Oct/19 $${let}\:{f}\left({x}\right)=\frac{{ln}\left(\mathrm{1}+{ax}+{bx}^{\mathrm{2}} \right)−{ln}\left(\mathrm{1}−{ax}+{bx}^{\mathrm{2}} \right)}{\mathrm{1}−{cosx}} \\ $$$${we}\:{have}\:{ln}^{'} \left(\mathrm{1}+{u}\right)=\frac{\mathrm{1}}{\mathrm{1}+{u}}\:=\mathrm{1}−{u}\:+{o}\left({u}\right)\:\left({u}\rightarrow\mathrm{0}\right)\:\Rightarrow \\…

A-string-AB-of-lenght-2l-has-a-particle-attached-to-its-midpoint-C-The-ends-A-and-B-of-the-string-are-fastened-to-two-fixed-points-with-A-distance-l-vertically-above-B-With-both-parts-of-the-string-

Question Number 137787 by physicstutes last updated on 06/Apr/21 $$\mathrm{A}\:\mathrm{string}\:{AB}\:\mathrm{of}\:\mathrm{lenght}\:\mathrm{2}{l}\:\mathrm{has}\:\mathrm{a}\:\mathrm{particle}\:\mathrm{attached}\:\mathrm{to}\:\mathrm{its}\:\mathrm{midpoint}\:\mathrm{C}.\: \\ $$$$\mathrm{The}\:\mathrm{ends}\:{A}\:\mathrm{and}\:{B}\:\mathrm{of}\:\mathrm{the}\:\mathrm{string}\:\mathrm{are}\:\mathrm{fastened}\:\mathrm{to}\:\mathrm{two}\:\mathrm{fixed}\:\mathrm{points} \\ $$$$\mathrm{with}\:{A}\:\mathrm{distance}\:{l}\:\mathrm{vertically}\:\mathrm{above}\:\mathrm{B}.\mathrm{With}\:\mathrm{both}\:\mathrm{parts}\:\mathrm{of}\:\mathrm{the}\:\mathrm{string} \\ $$$$\mathrm{taunt},\mathrm{the}\:\mathrm{particle}\:\mathrm{describes}\:\mathrm{a}\:\mathrm{horizontal}\:\mathrm{circle}\:\mathrm{about}\:\mathrm{the}\:\mathrm{line}\:{AB}\:\mathrm{with} \\ $$$$\mathrm{constant}\:\mathrm{angular}\:\mathrm{speed}\:\omega.\:\mathrm{If}\:\mathrm{the}\:\mathrm{tension}\:\mathrm{in}\:{CA}\:\mathrm{is}\:\mathrm{three}\:\mathrm{times} \\ $$$$\mathrm{that}\:\mathrm{in}\:{CB},\:\mathrm{prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{angular}\:\mathrm{velocity}\:\mathrm{is}\:\mathrm{2}\sqrt{\frac{\mathrm{g}}{{l}}}\:. \\ $$ Answered by mr…