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Question Number 226622    Answers: 0   Comments: 2

Question Number 226626    Answers: 0   Comments: 0

Question Number 226624    Answers: 0   Comments: 1

Question Number 226619    Answers: 0   Comments: 1

factorise x^2 −qx−p^2 +5pq−6q^2

$${factorise} \\ $$$${x}^{\mathrm{2}} −{qx}−{p}^{\mathrm{2}} +\mathrm{5}{pq}−\mathrm{6}{q}^{\mathrm{2}} \\ $$

Question Number 226612    Answers: 0   Comments: 1

Question Number 226610    Answers: 1   Comments: 0

Question Number 226608    Answers: 0   Comments: 0

Question Number 226609    Answers: 0   Comments: 0

Question Number 226603    Answers: 1   Comments: 0

Formulate the differential equation of the solution (a)y=Ae^(bx+1) (b)y=Asin x+Bcos x

$${Formulate}\:{the}\:{differential} \\ $$$${equation}\:{of}\:{the}\:{solution} \\ $$$$\left({a}\right){y}={Ae}^{{bx}+\mathrm{1}} \\ $$$$\left({b}\right){y}={A}\mathrm{sin}\:{x}+{B}\mathrm{cos}\:{x} \\ $$$$ \\ $$

Question Number 226601    Answers: 1   Comments: 0

Question Number 226598    Answers: 1   Comments: 0

(2x+1)^5 =a_5 x^5 +a_4 x^4 +a_3 x^3 ..+a_1 x+a_0 find a_5 +a_3 +a_1 and a_4 +a_2 +a_0

$$\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{5}} ={a}_{\mathrm{5}} {x}^{\mathrm{5}} +{a}_{\mathrm{4}} {x}^{\mathrm{4}} +{a}_{\mathrm{3}} {x}^{\mathrm{3}} ..+{a}_{\mathrm{1}} {x}+{a}_{\mathrm{0}} \\ $$$${find}\:{a}_{\mathrm{5}} +{a}_{\mathrm{3}} +{a}_{\mathrm{1}} {and}\:{a}_{\mathrm{4}} +{a}_{\mathrm{2}} +{a}_{\mathrm{0}} \\ $$

Question Number 226596    Answers: 0   Comments: 0

Question Number 226595    Answers: 0   Comments: 0

Question Number 226593    Answers: 0   Comments: 2

Question Number 226588    Answers: 0   Comments: 0

Question Number 226587    Answers: 0   Comments: 0

Question Number 226586    Answers: 0   Comments: 0

Question Number 226585    Answers: 0   Comments: 0

Question Number 226581    Answers: 3   Comments: 0

find the polar of (1+i)(1+i(√3))

$$\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{polar}}\:\boldsymbol{\mathrm{of}} \\ $$$$\left(\mathrm{1}+\boldsymbol{\mathrm{i}}\right)\left(\mathrm{1}+\boldsymbol{\mathrm{i}}\sqrt{\mathrm{3}}\right) \\ $$

Question Number 226577    Answers: 2   Comments: 0

Question Number 226562    Answers: 1   Comments: 0

If x + y i = ((a + i)/(a − i)) , prove that ay − 1 = x. (x+yi)(a−i)=a+i ax − xi + ayi − yi^2 = a + i (ax + y) + (ay − x)i = a + i ay − x = 1 x = ay −1

$$\mathrm{If}\:\mathrm{x}\:+\:\mathrm{y}\:{i}\:=\:\frac{\mathrm{a}\:+\:{i}}{\mathrm{a}\:−\:{i}}\:,\:\mathrm{prove}\:\mathrm{that}\:\mathrm{ay}\:−\:\mathrm{1}\:=\:\mathrm{x}. \\ $$$$\:\left(\mathrm{x}+\mathrm{y}{i}\right)\left(\mathrm{a}−{i}\right)=\mathrm{a}+{i} \\ $$$$\:\:\:\mathrm{ax}\:−\:\mathrm{x}{i}\:+\:\mathrm{ay}{i}\:−\:\mathrm{y}{i}^{\mathrm{2}} \:=\:\mathrm{a}\:+\:{i} \\ $$$$\:\:\left(\mathrm{ax}\:+\:\mathrm{y}\right)\:+\:\left(\mathrm{ay}\:−\:\mathrm{x}\right){i}\:=\:\mathrm{a}\:+\:{i} \\ $$$$\:\:\mathrm{ay}\:−\:\mathrm{x}\:=\:\mathrm{1} \\ $$$$\:\:\mathrm{x}\:=\:\mathrm{ay}\:−\mathrm{1} \\ $$

Question Number 226561    Answers: 4   Comments: 0

Question Number 226558    Answers: 1   Comments: 0

Question Number 226554    Answers: 3   Comments: 0

Question Number 226550    Answers: 3   Comments: 0

a+b+c = x lim_(x→0) ((a^3 +b^3 +c^3 )/(abc)) =?

$$\:\:\:\: {a}+{b}+{c}\:=\:{x}\: \\ $$$$\:\:\:\: \underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} }{{abc}}\:=? \\ $$

Question Number 226542    Answers: 0   Comments: 1

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