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AlgebraQuestion and Answers: Page 307

Question Number 63175 Answers: 0 Comments: 2

solve for x x^x^x = 16 x = 2, but how to use Lambert W function

$$\mathrm{solve}\:\mathrm{for}\:\mathrm{x}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{x}^{\mathrm{x}^{\mathrm{x}} } \:=\:\:\mathrm{16} \\ $$$$\mathrm{x}\:=\:\mathrm{2},\:\:\:\:\:\mathrm{but}\:\mathrm{how}\:\mathrm{to}\:\mathrm{use}\:\mathrm{Lambert}\:\mathrm{W}\:\mathrm{function} \\ $$

Question Number 63162 Answers: 1 Comments: 3

Find the set of values of x which satisfy the inequalities (2/(x−1))≤(1/x) and x^2 −∣3x∣+2<0

$${Find}\:{the}\:{set}\:{of}\:{values}\:{of}\:{x}\:{which}\:{satisfy}\:{the}\:{inequalities}\: \\ $$$$\frac{\mathrm{2}}{{x}−\mathrm{1}}\leqslant\frac{\mathrm{1}}{{x}}\:\:{and}\:\:{x}^{\mathrm{2}} −\mid\mathrm{3}{x}\mid+\mathrm{2}<\mathrm{0} \\ $$

Question Number 63095 Answers: 1 Comments: 0

Question Number 63090 Answers: 0 Comments: 0

s=(√(a^2 +(a^2 −d)^2 ))+(√((b−a)^2 +(b^2 −a^2 )^2 )) +(√(b^2 +(c−b^2 )^2 ))+c−d p= a(a^2 −d)+(a+b)(b^2 −a^2 ) +b(c−b^2 ) Find a,b,c, or d in terms of s if p is maximum. Assume a,b,c,d ≥0 .

$${s}=\sqrt{{a}^{\mathrm{2}} +\left({a}^{\mathrm{2}} −{d}\right)^{\mathrm{2}} }+\sqrt{\left({b}−{a}\right)^{\mathrm{2}} +\left({b}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\sqrt{{b}^{\mathrm{2}} +\left({c}−{b}^{\mathrm{2}} \right)^{\mathrm{2}} }+{c}−{d} \\ $$$$\:{p}=\:{a}\left({a}^{\mathrm{2}} −{d}\right)+\left({a}+{b}\right)\left({b}^{\mathrm{2}} −{a}^{\mathrm{2}} \right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+{b}\left({c}−{b}^{\mathrm{2}} \right) \\ $$$${Find}\:{a},{b},{c},\:{or}\:{d}\:\:{in}\:{terms}\:{of}\:{s} \\ $$$${if}\:\:{p}\:{is}\:{maximum}.\: \\ $$$${Assume}\:\:\:\:{a},{b},{c},{d}\:\geqslant\mathrm{0}\:. \\ $$

Question Number 63076 Answers: 0 Comments: 0

show that f:A→B is bijection then f(A_1 ^c )=[f(A_1 )]^c

$${show}\:{that}\:{f}:{A}\rightarrow{B}\:{is}\:{bijection}\:{then}\:{f}\left({A}_{\mathrm{1}} ^{{c}} \right)=\left[{f}\left({A}_{\mathrm{1}} \right)\right]^{{c}} \\ $$

Question Number 63021 Answers: 2 Comments: 3

solve this equation x^y =y^x x,y∈R.

$${solve}\:{this}\:{equation} \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}^{{y}} ={y}^{{x}} \\ $$$$ \\ $$$$ \\ $$$${x},{y}\in\mathbb{R}. \\ $$

Question Number 62998 Answers: 0 Comments: 12

Solve for x: 5^x +6x=7

$${Solve}\:{for}\:{x}:\:\:\mathrm{5}^{\boldsymbol{{x}}} +\mathrm{6}\boldsymbol{{x}}=\mathrm{7} \\ $$

Question Number 62945 Answers: 1 Comments: 0

Find the greatest coefficient in the expansion of (6 − 4x)^(−3)

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{greatest}\:\mathrm{coefficient}\:\mathrm{in}\:\mathrm{the}\:\mathrm{expansion}\:\mathrm{of}\:\:\:\left(\mathrm{6}\:−\:\mathrm{4x}\right)^{−\mathrm{3}} \\ $$

Question Number 62942 Answers: 1 Comments: 10

Make r the subject of the formular: S = ((a(r^n − 1))/(r − 1))

$$\mathrm{Make}\:\:\mathrm{r}\:\:\mathrm{the}\:\mathrm{subject}\:\mathrm{of}\:\mathrm{the}\:\mathrm{formular}: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{S}\:\:=\:\:\frac{\mathrm{a}\left(\mathrm{r}^{\mathrm{n}} \:−\:\mathrm{1}\right)}{\mathrm{r}\:−\:\mathrm{1}} \\ $$

Question Number 62895 Answers: 1 Comments: 2

Question Number 62844 Answers: 1 Comments: 0

Let p(x) = ax^2 + bx + c be such that p(x) takes real values for real values of x and non−real values for non−real values of x . Prove that a = 0 and find all possible values of c.

$${Let}\:{p}\left({x}\right)\:=\:{ax}^{\mathrm{2}} \:+\:{bx}\:+\:{c}\:\:{be}\:{such}\:{that}\:{p}\left({x}\right)\:{takes}\:{real}\:{values} \\ $$$${for}\:{real}\:{values}\:{of}\:{x}\:{and}\:{non}−{real}\:{values}\:{for}\:{non}−{real} \\ $$$${values}\:{of}\:{x}\:.\:{Prove}\:{that}\:{a}\:=\:\mathrm{0}\:{and}\:{find}\:{all} \\ $$$${possible}\:{values}\:{of}\:{c}. \\ $$

Question Number 62798 Answers: 0 Comments: 0

Calculate tg(20°)+4sin(20°)+1

$$\boldsymbol{\mathrm{Calculate}} \\ $$$$\boldsymbol{\mathrm{tg}}\left(\mathrm{20}°\right)+\mathrm{4}\boldsymbol{\mathrm{sin}}\left(\mathrm{20}°\right)+\mathrm{1} \\ $$

Question Number 62676 Answers: 1 Comments: 1