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Question Number 120619 Answers: 1 Comments: 0

Question Number 120614 Answers: 0 Comments: 8

Trigonometry

Question Number 120601 Answers: 1 Comments: 0

Find the range of values of x for which the expansion of the binomial (2 − 3x)^(−4) is valid. I need help with explanation please

$$\:\mathrm{Find}\:\mathrm{the}\:\mathrm{range}\:\mathrm{of}\:\mathrm{values}\:\mathrm{of}\:{x}\:\mathrm{for}\: \\ $$$$\:\mathrm{which}\:\mathrm{the}\:\mathrm{expansion}\:\mathrm{of}\:\mathrm{the}\:\mathrm{binomial} \\ $$$$\:\left(\mathrm{2}\:−\:\mathrm{3}{x}\right)^{−\mathrm{4}} \:\mathrm{is}\:\mathrm{valid}.\: \\ $$$$\:{I}\:{need}\:{help}\:{with}\:{explanation}\:{please} \\ $$

Question Number 120599 Answers: 0 Comments: 0

Let x = u^6 dx = 6u^5 du I = ∫(u^3 /((1+u^2 )^2 )) ×6u^5 du =6 ∫(u^8 /(1+2u^2 +u^4 )) du =6 ∫[((−4)/(u^2 +1))+(1/((1+u^2 )^2 ))+u^4 −2u^2 +3]du =6[−4tan^(−1) (u) + I_1 + (u^5 /5) − (2/3)u^3 +3u]+c I_1 = ∫(1/((1+u^2 )^2 )) du Put u = tan z du = sec^2 z dz I_1 = ∫(1/(sec^4 z))×sec^2 z dz = ∫cos^2 z dz = ∫(1/2)(cos 2z + 1)dz = (1/2)((1/2) sin 2z + z) = (1/2)×(u/(1+u^2 )) + (1/2) tan^(−1) u I = −4 tan^(−1) u + (u/(2(1+u^2 ))) + (1/2) tan^(−1) u +(u^5 /5) − (2/3)u^3 + 3u + c = −(7/2) tan^(−1) u + (u/(2(1+u^2 ))) + (1/5)u^5 + 3u + c = −(7/2) tan^(−1) ( ^6 (√x)) + ((x)^(1/6) /(2(1+(x^2 )^(1/6) ))) + (1/5) (x^5 )^(1/6) + 3 (x)^(1/6) + c

$${Let}\:{x}\:=\:{u}^{\mathrm{6}} \:\:\:\:\:\:\:\:\:{dx}\:=\:\mathrm{6}{u}^{\mathrm{5}} \:{du} \\ $$$${I}\:=\:\int\frac{{u}^{\mathrm{3}} }{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)^{\mathrm{2}} }\:×\mathrm{6}{u}^{\mathrm{5}} \:{du} \\ $$$$\:\:\:=\mathrm{6}\:\int\frac{{u}^{\mathrm{8}} }{\mathrm{1}+\mathrm{2}{u}^{\mathrm{2}} +{u}^{\mathrm{4}} }\:{du} \\ $$$$\:\:\:=\mathrm{6}\:\int\left[\frac{−\mathrm{4}}{{u}^{\mathrm{2}} +\mathrm{1}}+\frac{\mathrm{1}}{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)^{\mathrm{2}} }+{u}^{\mathrm{4}} −\mathrm{2}{u}^{\mathrm{2}} +\mathrm{3}\right]{du} \\ $$$$\:\:\:=\mathrm{6}\left[−\mathrm{4}{tan}^{−\mathrm{1}} \left({u}\right)\:+\:{I}_{\mathrm{1}} \:+\:\frac{{u}^{\mathrm{5}} }{\mathrm{5}}\:−\:\frac{\mathrm{2}}{\mathrm{3}}{u}^{\mathrm{3}} \:+\mathrm{3}{u}\right]+{c} \\ $$$${I}_{\mathrm{1}} \:=\:\int\frac{\mathrm{1}}{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)^{\mathrm{2}} }\:{du} \\ $$$${Put}\:{u}\:=\:{tan}\:{z}\:\:\:\:\:{du}\:=\:{sec}^{\mathrm{2}} {z}\:{dz} \\ $$$${I}_{\mathrm{1}} \:=\:\int\frac{\mathrm{1}}{{sec}^{\mathrm{4}} {z}}×{sec}^{\mathrm{2}} {z}\:{dz}\:=\:\int{cos}^{\mathrm{2}} {z}\:{dz} \\ $$$$\:\:\:\:\:=\:\int\frac{\mathrm{1}}{\mathrm{2}}\left({cos}\:\mathrm{2}{z}\:+\:\mathrm{1}\right){dz} \\ $$$$\:\:\:\:\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{2}}\:{sin}\:\mathrm{2}{z}\:+\:{z}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}}×\frac{{u}}{\mathrm{1}+{u}^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{\mathrm{2}}\:{tan}^{−\mathrm{1}} {u} \\ $$$${I}\:=\:−\mathrm{4}\:{tan}^{−\mathrm{1}} {u}\:+\:\frac{{u}}{\mathrm{2}\left(\mathrm{1}+{u}^{\mathrm{2}} \right)}\:+\:\frac{\mathrm{1}}{\mathrm{2}}\:{tan}^{−\mathrm{1}} {u}\:+\frac{{u}^{\mathrm{5}} }{\mathrm{5}}\:−\:\frac{\mathrm{2}}{\mathrm{3}}{u}^{\mathrm{3}} \:+\:\mathrm{3}{u}\:+\:{c} \\ $$$$\:\:\:=\:−\frac{\mathrm{7}}{\mathrm{2}}\:{tan}^{−\mathrm{1}} {u}\:+\:\frac{{u}}{\mathrm{2}\left(\mathrm{1}+{u}^{\mathrm{2}} \right)}\:+\:\frac{\mathrm{1}}{\mathrm{5}}{u}^{\mathrm{5}} \:+\:\mathrm{3}{u}\:+\:{c} \\ $$$$\:\:\:=\:−\frac{\mathrm{7}}{\mathrm{2}}\:{tan}^{−\mathrm{1}} \left(\overset{\mathrm{6}} {\:}\sqrt{{x}}\right)\:+\:\frac{\sqrt[{\mathrm{6}}]{{x}}}{\mathrm{2}\left(\mathrm{1}+\sqrt[{\mathrm{6}}]{{x}^{\mathrm{2}} }\right)}\:+\:\frac{\mathrm{1}}{\mathrm{5}}\:\sqrt[{\mathrm{6}}]{{x}^{\mathrm{5}} }\:+\:\mathrm{3}\:\sqrt[{\mathrm{6}}]{{x}}\:+\:{c} \\ $$

Question Number 120598 Answers: 0 Comments: 0

Let A be a subset of a Real number with dimension 2 and let x be a real number member of dimension 2. Then x is called the limit point of A if ...?

$$ \\ $$$$\mathrm{Let}\:\mathrm{A}\:\mathrm{be}\:\mathrm{a}\:\mathrm{subset}\:\mathrm{of}\:\mathrm{a}\:\mathrm{Real}\:\mathrm{number} \\ $$$$\mathrm{wit}{h}\:\mathrm{dimension}\:\mathrm{2}\:\mathrm{and}\:\mathrm{let}\:\mathrm{x}\:\mathrm{be}\:\mathrm{a}\:\mathrm{real} \\ $$$$\mathrm{num}{b}\mathrm{er}\:\mathrm{member}\:\mathrm{of}\:\mathrm{dimension}\:\mathrm{2}.\:\mathrm{Then}\: \\ $$$$\mathrm{x}\:\mathrm{is}\:\mathrm{called}\:\mathrm{the}\:\mathrm{limit}\:\mathrm{point}\:\mathrm{of}\:\mathrm{A}\:\mathrm{if}\:...? \\ $$

Question Number 120597 Answers: 0 Comments: 0

Let A ≠ −φ , A are subsets of R and A are open sets. Then each p member A applies ...?

$$ \\ $$$$\mathrm{Let}\:\mathrm{A}\:\:\neq\:−\phi\:,\:\mathrm{A}\:\mathrm{are}\:\mathrm{subsets}\:\mathrm{of}\:\mathrm{R}\:\mathrm{and}\:\mathrm{A}\:\mathrm{are} \\ $$$$\mathrm{ope}{n}\:\mathrm{sets}.\:\mathrm{Then}\:\mathrm{each}\:\mathrm{p}\:\mathrm{member}\:\mathrm{A} \\ $$$$\mathrm{applie}{s}\:...? \\ $$

Question Number 120595 Answers: 0 Comments: 0

Solution from Differential Equation System : {_(x_2 ^′ = 2x_1 − 3x_2 , x_2 (0) = −1) ^(x_1 ^′ = 2x_1 − 2x_2 , x_1 (0) = 1)

$$ \\ $$$${Solution}\:{from}\:{Differential}\:{Equation}\:{System}\:: \\ $$$$ \\ $$$$\left\{_{{x}_{\mathrm{2}} ^{'} \:=\:\:\mathrm{2}{x}_{\mathrm{1}} \:−\:\mathrm{3}{x}_{\mathrm{2}} \:\:,\:{x}_{\mathrm{2}} \:\left(\mathrm{0}\right)\:=\:−\mathrm{1}} ^{{x}_{\mathrm{1}} ^{'} \:=\:\mathrm{2}{x}_{\mathrm{1}} −\:\mathrm{2}{x}_{\mathrm{2}} \:\:\:\:,\:{x}_{\mathrm{1}} \left(\mathrm{0}\right)\:=\:\mathrm{1}} \right. \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Question Number 120593 Answers: 1 Comments: 0

lim_(z→0) ((Γ(z)+Γ(−z))/2) =^? −γ

$$ \\ $$$$\:\underset{{z}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\frac{\Gamma\left({z}\right)+\Gamma\left(−{z}\right)}{\mathrm{2}}\:\overset{?} {=}\:−\gamma \\ $$

Question Number 120583 Answers: 0 Comments: 0

Question Number 120582 Answers: 0 Comments: 0

Let u=x^(3/5) du = (3/(5x^(2/5) )) I = (5/3)∫(u/( (√(3−2u)))) du = −(5/6) ∫((3−2u−3)/( (√(3−2u)))) du = −(5/3) ∫[(√(3−2u)) −3(3−2u)^(−(1/2)) ] du = −(5/3)[−(1/3)(3−2u)^(3/2) +3(3−2u)^(1/2) ]+c = −(5/(18))(3−2u)^(1/2) [−3+2u + 9]+c = −(5/(18))(3−2u)^(1/2) (6+2u)+c = −(5/9)(√(3−2x^(3/5) ))(3+x^(3/5) )+c

$${Let}\:{u}={x}^{\frac{\mathrm{3}}{\mathrm{5}}} \:\:\:\:\:{du}\:=\:\frac{\mathrm{3}}{\mathrm{5}{x}^{\frac{\mathrm{2}}{\mathrm{5}}} } \\ $$$${I}\:=\:\frac{\mathrm{5}}{\mathrm{3}}\int\frac{{u}}{\:\sqrt{\mathrm{3}−\mathrm{2}{u}}}\:{du}\:=\:−\frac{\mathrm{5}}{\mathrm{6}}\:\int\frac{\mathrm{3}−\mathrm{2}{u}−\mathrm{3}}{\:\sqrt{\mathrm{3}−\mathrm{2}{u}}}\:{du} \\ $$$$=\:−\frac{\mathrm{5}}{\mathrm{3}}\:\int\left[\sqrt{\mathrm{3}−\mathrm{2}{u}}\:−\mathrm{3}\left(\mathrm{3}−\mathrm{2}{u}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \right]\:{du} \\ $$$$=\:−\frac{\mathrm{5}}{\mathrm{3}}\left[−\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{3}−\mathrm{2}{u}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} +\mathrm{3}\left(\mathrm{3}−\mathrm{2}{u}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \right]+{c} \\ $$$$=\:−\frac{\mathrm{5}}{\mathrm{18}}\left(\mathrm{3}−\mathrm{2}{u}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \left[−\mathrm{3}+\mathrm{2}{u}\:+\:\mathrm{9}\right]+{c} \\ $$$$=\:−\frac{\mathrm{5}}{\mathrm{18}}\left(\mathrm{3}−\mathrm{2}{u}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \left(\mathrm{6}+\mathrm{2}{u}\right)+{c} \\ $$$$=\:−\frac{\mathrm{5}}{\mathrm{9}}\sqrt{\mathrm{3}−\mathrm{2}{x}^{\frac{\mathrm{3}}{\mathrm{5}}} }\left(\mathrm{3}+{x}^{\frac{\mathrm{3}}{\mathrm{5}}} \right)+{c} \\ $$

Question Number 120569 Answers: 1 Comments: 5

Question Number 120563 Answers: 1 Comments: 0

A man has a wife with six children and his total income is Gh$8500.He was allowed the following free tax personal:......$1200 wife:...........$300 each child:.....$250 for a maximum of 4 dependant relative:$400 insurance:........$250 Taxed: the first $2000 at 10% the next $2000 at 15% the next $2000 at 20% the next $2000 at 25% calculate a)his tax free income b)his income tax c)his monthly income d)his net monthly income

$${A}\:{man}\:{has}\:{a}\:{wife}\:{with}\:{six} \\ $$$${children}\:{and}\:{his}\:{total}\:{income}\:{is} \\ $$$${Gh\$}\mathrm{8500}.{He}\:{was}\:{allowed}\:{the} \\ $$$${following}\:{free}\:{tax} \\ $$$${personal}:......\$\mathrm{1200} \\ $$$${wife}:...........\$\mathrm{300} \\ $$$${each}\:{child}:.....\$\mathrm{250}\:{for}\:{a}\:{maximum}\:{of}\:\mathrm{4} \\ $$$${dependant}\:{relative}:\$\mathrm{400} \\ $$$${insurance}:........\$\mathrm{250} \\ $$$${Taxed}: \\ $$$${the}\:{first}\:\$\mathrm{2000}\:{at}\:\mathrm{10\%} \\ $$$${the}\:{next}\:\$\mathrm{2000}\:{at}\:\mathrm{15\%} \\ $$$${the}\:{next}\:\$\mathrm{2000}\:{at}\:\mathrm{20\%} \\ $$$${the}\:{next}\:\$\mathrm{2000}\:{at}\:\mathrm{25\%} \\ $$$${calculate} \\ $$$$\left.{a}\right){his}\:{tax}\:{free}\:{income} \\ $$$$\left.{b}\right){his}\:{income}\:{tax} \\ $$$$\left.{c}\right){his}\:{monthly}\:{income} \\ $$$$\left.{d}\right){his}\:{net}\:{monthly}\:{income} \\ $$$$ \\ $$

Question Number 120562 Answers: 1 Comments: 0