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Question Number 93847 Answers: 0 Comments: 2

log _5 (x−2)+log _8 (x−4)= log _6 (x−1)

$$\mathrm{log}\:_{\mathrm{5}} \:\left(\mathrm{x}−\mathrm{2}\right)+\mathrm{log}\:_{\mathrm{8}} \:\left(\mathrm{x}−\mathrm{4}\right)=\:\mathrm{log}\:_{\mathrm{6}} \left(\mathrm{x}−\mathrm{1}\right) \\ $$

Question Number 93844 Answers: 1 Comments: 0

let f(x)=(√x) and g(x)=(√x) find the domain of (f.g)(x) ? help me sir

$${let}\:{f}\left({x}\right)=\sqrt{{x}}\:\:{and}\:{g}\left({x}\right)=\sqrt{{x}}\:{find}\:{the}\:{domain}\:{of}\:\left({f}.{g}\right)\left({x}\right)\:? \\ $$$${help}\:{me}\:{sir} \\ $$

Question Number 93843 Answers: 1 Comments: 0

Use Horner to solve it . (a^2 −5ab−3b^2 ) : (a−b) find the remainder .

$${Use}\:\:{Horner}\:\:{to}\:\:{solve}\:{it}\:. \\ $$$$\:\:\left({a}^{\mathrm{2}} −\mathrm{5}{ab}−\mathrm{3}{b}^{\mathrm{2}} \right)\::\:\left({a}−{b}\right) \\ $$$${find}\:\:{the}\:\:{remainder}\:\:. \\ $$

Question Number 93837 Answers: 0 Comments: 5

Question Number 93836 Answers: 0 Comments: 0

If cosα+cosβ+cosγ=0 = sinα+sinβ+sinγ, prove that (i) cos3α+cos3β+cosγ = 3cos(a+β+γ) (ii) sin3α+sin3β+sinγ = 3sin(α+β+γ) (iii) cos2α+cos2β+cos2γ = 0 (iv) sin2α+sin2β+sin2γ = 0 (Hints: Take a=cis α, b=cis β, c=cis γ, a+b+c=0 ⇒ a^3 +b^3 +c^3 =3abc (1/a)+(1/b)+(1/c)=0 ⇒ a^2 +b^2 +c^2 =0) (v) cos^2 α+cos^2 β+cosγ = sin^2 α+sin^2 β+sin^2 γ = (3/2).

$${If}\:{cos}\alpha+{cos}\beta+{cos}\gamma=\mathrm{0}\:=\:{sin}\alpha+{sin}\beta+{sin}\gamma,\:{prove}\:{that} \\ $$$$\left(\mathrm{i}\right)\:{cos}\mathrm{3}\alpha+{cos}\mathrm{3}\beta+{cos}\gamma\:=\:\mathrm{3}{cos}\left({a}+\beta+\gamma\right) \\ $$$$\left(\mathrm{ii}\right)\:{sin}\mathrm{3}\alpha+{sin}\mathrm{3}\beta+{sin}\gamma\:=\:\mathrm{3}{sin}\left(\alpha+\beta+\gamma\right) \\ $$$$\left(\mathrm{iii}\right)\:{cos}\mathrm{2}\alpha+{cos}\mathrm{2}\beta+{cos}\mathrm{2}\gamma\:=\:\mathrm{0} \\ $$$$\left(\mathrm{iv}\right)\:{sin}\mathrm{2}\alpha+{sin}\mathrm{2}\beta+{sin}\mathrm{2}\gamma\:=\:\mathrm{0} \\ $$$$\left(\mathrm{Hints}:\:\mathrm{T}{ake}\:{a}=\mathrm{cis}\:\alpha,\:\mathrm{b}=\mathrm{cis}\:\beta,\:\mathrm{c}=\mathrm{cis}\:\gamma,\:\mathrm{a}+\mathrm{b}+\mathrm{c}=\mathrm{0}\:\Rightarrow\:\mathrm{a}^{\mathrm{3}} +\mathrm{b}^{\mathrm{3}} +\mathrm{c}^{\mathrm{3}} =\mathrm{3abc}\right. \\ $$$$\left.\frac{\mathrm{1}}{\mathrm{a}}+\frac{\mathrm{1}}{\mathrm{b}}+\frac{\mathrm{1}}{\mathrm{c}}=\mathrm{0}\:\Rightarrow\:\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} =\mathrm{0}\right) \\ $$$$\left(\mathrm{v}\right)\:{cos}^{\mathrm{2}} \alpha+{cos}^{\mathrm{2}} \beta+{cos}\gamma\:=\:{sin}^{\mathrm{2}} \alpha+{sin}^{\mathrm{2}} \beta+{sin}^{\mathrm{2}} \gamma\:=\:\frac{\mathrm{3}}{\mathrm{2}}. \\ $$

Question Number 93828 Answers: 1 Comments: 2

sin x ((dy/dx))−y = 2sin x

$$\mathrm{sin}\:\mathrm{x}\:\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)−\mathrm{y}\:=\:\mathrm{2sin}\:\mathrm{x}\: \\ $$

Question Number 93827 Answers: 0 Comments: 1

simplify: (((cosα+isinα)^3 )/((sinβ+icosβ)^4 ))

$${simplify}:\:\frac{\left({cos}\alpha+\boldsymbol{\mathrm{i}}{sin}\alpha\right)^{\mathrm{3}} }{\left({sin}\beta+\boldsymbol{\mathrm{i}}{cos}\beta\right)^{\mathrm{4}} } \\ $$

Question Number 93824 Answers: 1 Comments: 0