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Question Number 214457 Answers: 2 Comments: 2

If (((x + 2y + 3z)^2 )/(x^2 + y^2 + z^2 )) = 14 find: ((x + y)/z) = ?

$$\mathrm{If}\:\:\:\frac{\left(\mathrm{x}\:+\:\mathrm{2y}\:+\:\mathrm{3z}\right)^{\mathrm{2}} }{\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{y}^{\mathrm{2}} \:+\:\mathrm{z}^{\mathrm{2}} }\:=\:\mathrm{14}\:\:\:\:\:\mathrm{find}:\:\:\frac{\mathrm{x}\:+\:\mathrm{y}}{\mathrm{z}}\:=\:? \\ $$

Question Number 214456 Answers: 2 Comments: 1

If (x/(a^2 − bc)) = (y/(b^2 − ac)) = (z/(c^2 − ab)) Find: ((ax + by + cz)/(x + y + z)) = ?

$$\mathrm{If}\:\:\:\frac{\mathrm{x}}{\mathrm{a}^{\mathrm{2}} −\:\mathrm{bc}}\:=\:\frac{\mathrm{y}}{\mathrm{b}^{\mathrm{2}} −\:\mathrm{ac}}\:=\:\frac{\mathrm{z}}{\mathrm{c}^{\mathrm{2}} −\:\mathrm{ab}} \\ $$$$\mathrm{Find}:\:\:\:\frac{\mathrm{ax}\:+\:\mathrm{by}\:+\:\mathrm{cz}}{\mathrm{x}\:+\:\mathrm{y}\:+\:\mathrm{z}}\:=\:? \\ $$

Question Number 214455 Answers: 0 Comments: 4

If x+y+z=xyz Find: ((x(1−y^2 )(1−z^2 )+y(1−x^2 )(1−z^2 )+z(1−x^2 )(1−y^2 ))/(2xyz))

$$\mathrm{If}\:\:\:\mathrm{x}+\mathrm{y}+\mathrm{z}=\mathrm{xyz} \\ $$$$\mathrm{Find}: \\ $$$$\frac{\mathrm{x}\left(\mathrm{1}−\mathrm{y}^{\mathrm{2}} \right)\left(\mathrm{1}−\mathrm{z}^{\mathrm{2}} \right)+\mathrm{y}\left(\mathrm{1}−\mathrm{x}^{\mathrm{2}} \right)\left(\mathrm{1}−\mathrm{z}^{\mathrm{2}} \right)+\mathrm{z}\left(\mathrm{1}−\mathrm{x}^{\mathrm{2}} \right)\left(\mathrm{1}−\mathrm{y}^{\mathrm{2}} \right)}{\mathrm{2xyz}} \\ $$

Question Number 214454 Answers: 0 Comments: 1

a,b,c ∈ R^+ S = ((9a)/(b + c)) + ((16b)/(a + c)) + ((49c)/(a + b)) min(S) = ?

$$\mathrm{a},\mathrm{b},\mathrm{c}\:\in\:\mathbb{R}^{+} \\ $$$$\mathrm{S}\:\:=\:\:\frac{\mathrm{9a}}{\mathrm{b}\:+\:\mathrm{c}}\:\:+\:\:\frac{\mathrm{16b}}{\mathrm{a}\:+\:\mathrm{c}}\:\:+\:\:\frac{\mathrm{49c}}{\mathrm{a}\:+\:\mathrm{b}} \\ $$$$\boldsymbol{\mathrm{min}}\left(\mathrm{S}\right)\:=\:? \\ $$

Question Number 214443 Answers: 1 Comments: 0

a,b,c,d,e,f ∈ Q (1/( (√2) − (2)^(1/3) )) = 2^a + 2^b + 2^c + 2^d + 2^e + 2^f find: a,b,c,d,e,f = ?

$$\mathrm{a},\mathrm{b},\mathrm{c},\mathrm{d},\mathrm{e},\mathrm{f}\:\in\:\mathrm{Q} \\ $$$$\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}\:−\:\sqrt[{\mathrm{3}}]{\mathrm{2}}}\:=\:\mathrm{2}^{\boldsymbol{\mathrm{a}}} \:+\:\mathrm{2}^{\boldsymbol{\mathrm{b}}} \:+\:\mathrm{2}^{\boldsymbol{\mathrm{c}}} \:+\:\mathrm{2}^{\boldsymbol{\mathrm{d}}} \:+\:\mathrm{2}^{\boldsymbol{\mathrm{e}}} \:+\:\mathrm{2}^{\boldsymbol{\mathrm{f}}} \\ $$$$\mathrm{find}:\:\:\:\mathrm{a},\mathrm{b},\mathrm{c},\mathrm{d},\mathrm{e},\mathrm{f}\:=\:? \\ $$

Question Number 214449 Answers: 2 Comments: 1

Question Number 214448 Answers: 0 Comments: 0

x^− = ((Σf_i x_i )/(Σf_i )) , u^− = ((Σf_i u_i )/(Σf_i )) , u = ((x−a)/h) , proved that x^− = a + hu^−

$$ \\ $$$$\overset{−} {{x}}\:=\:\frac{\Sigma{f}_{{i}} {x}_{{i}} }{\Sigma{f}_{{i}} \:}\:,\:\:\:\overset{−} {{u}}\:=\:\frac{\Sigma{f}_{{i}} {u}_{{i}} }{\Sigma{f}_{{i}} }\:,\:{u}\:=\:\frac{{x}−{a}}{{h}}\:, \\ $$$${proved}\:{that}\:\overset{−} {{x}}\:=\:{a}\:+\:{h}\overset{−} {{u}} \\ $$

Question Number 214447 Answers: 1 Comments: 1

Question Number 214427 Answers: 3 Comments: 1

Question Number 214421 Answers: 1 Comments: 0

Question Number 214419 Answers: 1 Comments: 0

Question Number 214414 Answers: 1 Comments: 0

Given that the roots of the equation ax^2 +bx+c=0 are α and β, show that; λμb^2 =ac(λ+μ)^2 where (α/β)=(λ/μ) Mr Hans

$$\mathrm{Given}\:\mathrm{that}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\mathrm{ax}^{\mathrm{2}} +\mathrm{bx}+\mathrm{c}=\mathrm{0}\:\mathrm{are}\:\alpha\:\mathrm{and}\:\beta, \\ $$$$\:\mathrm{show}\:\mathrm{that}; \\ $$$$\lambda\mu\mathrm{b}^{\mathrm{2}} =\mathrm{ac}\left(\lambda+\mu\right)^{\mathrm{2}} \:\mathrm{where}\:\frac{\alpha}{\beta}=\frac{\lambda}{\mu} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{Mr}\:{Hans} \\ $$

Question Number 214408 Answers: 2 Comments: 0

Question Number 214409 Answers: 3 Comments: 0

Question Number 214402 Answers: 3 Comments: 0

If ((a+b)/c) = ((b+c)/a) = ((a+c)/b) then find ((a+b)/c) .

$$\:\:\:\mathrm{If}\:\frac{\mathrm{a}+\mathrm{b}}{\mathrm{c}}\:=\:\frac{\mathrm{b}+\mathrm{c}}{\mathrm{a}}\:=\:\frac{\mathrm{a}+\mathrm{c}}{\mathrm{b}}\:\mathrm{then}\:\mathrm{find}\: \\ $$$$\:\:\:\:\frac{\mathrm{a}+\mathrm{b}}{\mathrm{c}}\:. \\ $$

Question Number 214398 Answers: 1 Comments: 1

Question Number 214395 Answers: 1 Comments: 1

Question Number 214372 Answers: 2 Comments: 1

Question Number 214369 Answers: 0 Comments: 0

Question Number 214366 Answers: 0 Comments: 0

f(α)=∫_(−∞) ^( ∞) e^(−αx^2 ) dx ∫ e^(−αt^2 ) dt=(1/2)(√(π/α))∙erf((√α)t)+Const ∴∫_(−∞) ^( ∞) e^(−αt^2 ) dt=(√(π/α)) , α∈(0,∞)

$${f}\left(\alpha\right)=\int_{−\infty} ^{\:\infty} \:{e}^{−\alpha{x}^{\mathrm{2}} } \mathrm{d}{x} \\ $$$$\int\:\:{e}^{−\alpha{t}^{\mathrm{2}} } \mathrm{d}{t}=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\frac{\pi}{\alpha}}\centerdot\mathrm{erf}\left(\sqrt{\alpha}{t}\right)+\mathrm{Const} \\ $$$$\therefore\int_{−\infty} ^{\:\infty} \:{e}^{−\alpha{t}^{\mathrm{2}} } \mathrm{d}{t}=\sqrt{\frac{\pi}{\alpha}}\:,\:\alpha\in\left(\mathrm{0},\infty\right) \\ $$

Question Number 214360 Answers: 1 Comments: 0

Find ∫_(−∞) ^∞ e^(−ax^2 ) dx when a is constant without changing the coordinate.

$$\mathrm{Find}\:\underset{−\infty} {\overset{\infty} {\int}}{e}^{−{ax}^{\mathrm{2}} } {dx}\:\mathrm{when}\:{a}\:\mathrm{is}\:\mathrm{constant}\:\mathrm{without}\:\mathrm{changing}\:\mathrm{the}\:\mathrm{coordinate}. \\ $$

Question Number 214351 Answers: 1 Comments: 3

Question Number 214350 Answers: 2 Comments: 0

Question Number 214340 Answers: 1 Comments: 0

evaluate ((∮_C (z/(2∙sin(z)−2z∙cos(z)−π))dz)/(∮_( C) (2/(2∙sin(z)−2z∙cos(z)−π))dz)) where C is the circle∣z−((3π)/4)∣=(π/4).

$$\mathrm{evaluate} \\ $$$$\frac{\oint_{{C}} \:\frac{{z}}{\mathrm{2}\centerdot\mathrm{sin}\left({z}\right)−\mathrm{2}{z}\centerdot\mathrm{cos}\left({z}\right)−\pi}\mathrm{d}{z}}{\oint_{\:{C}} \:\frac{\mathrm{2}}{\mathrm{2}\centerdot\mathrm{sin}\left({z}\right)−\mathrm{2}{z}\centerdot\mathrm{cos}\left({z}\right)−\pi}\mathrm{d}{z}} \\ $$$$\mathrm{where}\:{C}\:\mathrm{is}\:\mathrm{the}\:\mathrm{circle}\mid{z}−\frac{\mathrm{3}\pi}{\mathrm{4}}\mid=\frac{\pi}{\mathrm{4}}. \\ $$

Question Number 214322 Answers: 0 Comments: 2

find the errors 7x+4=10x−6 7x+4−10x=10x−6−10x −3x+4=−6 −3x+4+6=−6+6 −3x+10=0 −3x+10−10=0−10 −3x=−10 x=((−3)/(−10)) x=(3/(10)) if this error show the real one

$$\mathrm{find}\:\mathrm{the}\:\mathrm{errors} \\ $$$$\mathrm{7}{x}+\mathrm{4}=\mathrm{10}{x}−\mathrm{6} \\ $$$$\mathrm{7}{x}+\mathrm{4}−\mathrm{10}{x}=\mathrm{10}{x}−\mathrm{6}−\mathrm{10}{x} \\ $$$$−\mathrm{3}{x}+\mathrm{4}=−\mathrm{6} \\ $$$$−\mathrm{3}{x}+\mathrm{4}+\mathrm{6}=−\mathrm{6}+\mathrm{6} \\ $$$$−\mathrm{3}{x}+\mathrm{10}=\mathrm{0} \\ $$$$−\mathrm{3}{x}+\mathrm{10}−\mathrm{10}=\mathrm{0}−\mathrm{10} \\ $$$$−\mathrm{3}{x}=−\mathrm{10} \\ $$$${x}=\frac{−\mathrm{3}}{−\mathrm{10}} \\ $$$${x}=\frac{\mathrm{3}}{\mathrm{10}} \\ $$$$\mathrm{if}\:\mathrm{this}\:\mathrm{error}\:\mathrm{show}\:\mathrm{the}\:\mathrm{real}\:\mathrm{one} \\ $$

Question Number 214321 Answers: 1 Comments: 0

15x(((6x)/(10))+((12x)/(20))+((15x)/(30))+...+((60x)/(100)))=160 solve for x

$$\mathrm{15}{x}\left(\frac{\mathrm{6}{x}}{\mathrm{10}}+\frac{\mathrm{12}{x}}{\mathrm{20}}+\frac{\mathrm{15}{x}}{\mathrm{30}}+...+\frac{\mathrm{60}{x}}{\mathrm{100}}\right)=\mathrm{160} \\ $$$$\mathrm{solve}\:\mathrm{for}\:{x} \\ $$

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