heureka

(3x+2)/x=x/(-2x+1)

\(\begin{array}{rcll} \dfrac{3x+2} {x} &=& \dfrac{x} {-2x+1} \\\\ \dfrac{3x+2} {x} &=& \dfrac{x} {1-2x} \\\\ 3 + \dfrac{2} {x} &=& \dfrac{1} { \frac{1-2x}{x} } \\\\ 3 + \dfrac{2} {x} &=& \dfrac{1} { \frac{1}{x} - 2 } \qquad & | \cdot \dfrac{1}{x} - 2\\\\ \left(\dfrac{1}{x} - 2 \right)\cdot \left(3 + \dfrac{2} {x} \right) &=& 1 \\\\ \left(\dfrac{1}{x} - 2 \right)\cdot \left(3 + \dfrac{2} {x} \right) &=& 1 \\\\ \left(\dfrac{1}{x} - 2 \right)\cdot \left(3 + 2\cdot \dfrac{1} {x} \right) &=& 1 \\\\ \end{array}\)

We substitute:  \(u=\dfrac{1}{x}\)

\(\begin{array}{rcll} \left(u - 2 \right)\cdot \left(3 + 2\cdot u \right) &=& 1 \\ 3u+2u^2-6-4u &=& 1 \\ 2u^2 -u-7 &=& 0 \\ \end{array}\)

\(\boxed{~ \begin{array}{rcll} ax^2+bx+c &=& 0 \\ x &=& \dfrac{-b \pm \sqrt{b^2-4ac} }{2a} \end{array} ~}\)

\(\begin{array}{rcll} 2u^2 -u-7 &=& 0 \qquad & a = 2 \qquad b = -1 \qquad c = -7\\ u_{1,2} &=& \dfrac{-(-1) \pm \sqrt{(-1)^2-4\cdot 2 \cdot (-7)} }{2\cdot 2}\\ u_{1,2} &=& \dfrac{1 \pm \sqrt{1+56} }{4}\\ u_{1,2} &=& \dfrac{1 \pm \sqrt{57} }{4}\\\\ u_1 &=& \dfrac{1 + \sqrt{57} }{4}\\ x_1 &=& \frac{1}{u_1} \\ \mathbf{x_1} &\mathbf{=}& \mathbf{\dfrac{4}{1 + \sqrt{57} } }\\\\ u_2 &=& \dfrac{1 - \sqrt{57} }{4}\\ x_2 &=& \frac{1}{u_2} \\ \mathbf{x_2} &\mathbf{=}& \mathbf{\dfrac{4}{1 - \sqrt{57} } } \end{array}\)

Good morning, was wondering if anyone can figure this out, i'll post the answer later 

Verify that \(y=5e^{x^2}\) is a solution of \(y'=2xy\)

\(\begin{array}{rcll} y &=& 5e^{x^2} \qquad & \ln{()} \\ \ln{(y)} &=& \ln{( 5e^{x^2} ) } \\ \ln{(y)} &=& \ln{( 5 ) } + \ln{( e^{x^2} ) } \\ \ln{(y)} &=& \ln{( 5 ) } + x^2\cdot \ln{( e ) } \qquad & \ln{(e)} = 1 \\ \ln{(y)} &=& \ln{( 5 ) } + x^2 \qquad & \frac{d()}{dx} \\ \frac{y'}{y} &=& 0 + 2\cdot x \\ \frac{y'}{y} &=& 2\cdot x \\ y' &=& 2\cdot x \cdot y \\ \end{array}\)

Change a term or sign so that the equation has the solution x=2 

2(x-3) = 1-(2x-5)

This is really important i'd appreciate the help very much 

Change a term or sign so that the equation has the solution x=2 

2(x-3) = 1-(2x-5)

This is really important i'd appreciate the help very much 

if \(x = 2\) we have:

\(\begin{array}{rcll} 2(x-3) &\ne& 1-(2x-5) \\ 2(2-3) &\ne& 1-(2\cdot 2-5)\\ 2(-1) &\ne& 1-(4-5)\\ 2(-1) &\ne& 1-(-1)\\ -2 &\ne& 1+1\\ -2 &\ne& 2\\ \end{array}\)

We must change the sign:

\(\begin{array}{rcll} {\color{red}-}2(x-3) &=& 1-(2x-5) \\ -2(2-3) &=& 1-(2\cdot 2-5)\\ -2(-1) &=& 1-(4-5)\\ -2(-1) &=& 1-(-1)\\ 2 &=& 1+1\\ 2 &=& 2\\ \end{array}\)

What method did you use for your solution?

What are the coords of the endpoints of the other diagonal \(G(x_G,y_G) \)and \(H(x_H,y_H)\).

\(\begin{array}{rcll} \vec{E} &=& \binom{2}{3} \\ \vec{F} &=& \binom{0}{-3} \\\\ \vec{G} &=& \vec{F} + \frac12 \cdot ( \vec{E}-\vec{F} ) + \frac12 \cdot ( \vec{E}-\vec{F} )_\perp\\ \vec{H} &=& \vec{F} + \frac12 \cdot ( \vec{E}-\vec{F} ) - \frac12 \cdot ( \vec{E}-\vec{F} )_\perp\\\\ \end{array}\)

\(\begin{array}{lcll} \text{please note, if }~ \vec{a} = \binom{x_a}{y_a} ~\text{ then }~ \vec{a}_\perp = \binom{-y_a}{x_a}~\text{ or }~ \vec{a}_\perp = \binom{y_a}{-x_a} \\ \text{because }~ \vec{a} \cdot \vec{a}_\perp ~\text{ must be null, if be perpendicular to one another.}\\ \binom{x_a}{y_a} \cdot \binom{-y_a}{x_a} = -x_a\cdot y_a + x_a\cdot y_a = 0\\ \binom{x_a}{y_a} \cdot \binom{y_a}{-x_a} = x_a\cdot y_a - x_a\cdot y_a = 0\\ \end{array}\)

1/2 log base 3 of p + 2 log base 9 of p

\(\begin{array}{rcll} p &=& p \\ 3^{\log_3{(p)}} &=& 9^{\log_9{(p)}} \qquad & | \qquad \log_3{()} \\ \log_3{ ( 3^{\log_3{(p)}} ) }&=& \log_3{ ( 9^{\log_9{(p)}} ) } \\ \log_3{(p)}\cdot \log_3{( 3 )}&=& \log_9{(p)}\cdot \log_3{( 9 )} \qquad & | \qquad \log_3{(3)} = 1\\ \end{array}\\ \boxed{~ \begin{array}{rcll} \log_3{(p)}&=& \log_9{(p)} \cdot \log_3{( 9 )} \qquad & | \qquad \log_3{( 9 )} = \log_3{( 3^2 )} = 2\cdot \log_3{(3)} = 2 \\ \log_3{(p)}&=& \log_9{(p)} \cdot 2 \\ \log_9{(p)} &=& \frac{ \log_3{(p)} }{2} \end{array} ~}\)

\(\begin{array}{rcll} \frac12 \cdot \log_3{(p)} + 2\cdot \log_9{(p)} &=& \frac12 \cdot \log_3{(p)} + 2\cdot \frac{ \log_3{(p)} }{2} \\ &=& \frac12 \cdot \log_3{(p)} + \log_3{(p)} \\ &=& \frac32 \cdot \log_3{(p)} \\ \end{array}\)

A square graphed on the coordinate plane has a diagonal with endpoints E(2,3) and F(0,-3). What are the coords of the endpoints of the other diagonal?

1. Diagonal: \(E (x_E=2,y_E=3)\) and \(F(x_F=0,y_F=-3)\).  

What are the coords of the endpoints of the other diagonal \(G(x_G,y_G)\) and \(H(x_H,y_H)\).

\(\begin{array}{rcll} \binom{x_G}{y_G} &=& \binom{x_F}{y_F} +\frac12 \cdot \binom{x_E-x_F}{y_E-y_F} + \frac12 \cdot \binom{-(y_E-y_F)}{x_E-x_F} \\ \binom{x_G}{y_G} &=& \binom{0}{-3} +\frac12 \cdot \binom{2-0}{3-(-3)} + \frac12 \cdot \binom{-(3-(-3))}{2-0} \\ \binom{x_G}{y_G} &=& \binom{0}{-3} +\frac12 \cdot \binom{2}{6} + \frac12 \cdot \binom{-6}{2} \\ \binom{x_G}{y_G} &=& \binom{0}{-3} +\binom{\frac12\cdot 2}{\frac12\cdot 6} + \binom{\frac12\cdot (-6)}{\frac12\cdot 2} \\ \binom{x_G}{y_G} &=& \binom{0}{-3} +\binom{1}{3} + \binom{-3}{1} \\ \binom{x_G}{y_G} &=& \binom{0+1-3}{-3+3+1} \\ \binom{x_G}{y_G} &=& \binom{-2}{1} \\ \end{array}\)

\(\begin{array}{rcll} \binom{x_H}{y_H} &=& \binom{x_F}{y_F} +\frac12 \cdot \binom{x_E-x_F}{y_E-y_F} - \frac12 \cdot \binom{-(y_E-y_F)}{x_E-x_F} \\ \binom{x_H}{y_H} &=& \binom{0}{-3} +\frac12 \cdot \binom{2-0}{3-(-3)} - \frac12 \cdot \binom{-(3-(-3))}{2-0} \\ \binom{x_H}{y_H} &=& \binom{0}{-3} +\frac12 \cdot \binom{2}{6} - \frac12 \cdot \binom{-6}{2} \\ \binom{x_H}{y_H} &=& \binom{0}{-3} +\binom{\frac12\cdot 2}{\frac12\cdot 6} - \binom{\frac12\cdot (-6)}{\frac12\cdot 2} \\ \binom{x_H}{y_H} &=& \binom{0}{-3} +\binom{1}{3} - \binom{-3}{1} \\ \binom{x_H}{y_H} &=& \binom{0+1-(-3)}{-3+3-1} \\ \binom{x_H}{y_H} &=& \binom{0+1+3}{-3+3-1} \\ \binom{x_H}{y_H} &=& \binom{4}{-1} \\ \end{array}\)

Imagine you have a large supply of 3kg and 8kg weights. Two 3kg weights and three 8kg weights have a mean of 6kg, because (3+3+8+8+8)/5=6 Can you find other combinations of 3kg and 8kg weights whose mean weight is is whole number of kg? What's the smallest? Whats the largest? Can you make all the whole numbers inbetween?

Imagine you have a large supply of 3kg and 8kg weights. Two 3kg weights and three 8kg weights have a mean of 6kg, because (3+3+8+8+8)/5=6 Can you find other combinations of 3kg and 8kg weights whose mean weight is is whole number of kg? What's the smallest? Whats the largest? Can you make all the whole numbers inbetween?

Part 2:

\(\text{proof}:\\ \begin{array}{rcll} \text{mean } &=& \frac{ u\cdot n \times 3\ kg + v\cdot n \times 8\ kg } { u\cdot n + v\cdot n } \\ &=& \frac{ n\cdot ( u \times 3\ kg + v \times 8\ kg ) } { n\cdot ( u + v ) } \\ &=& \frac{ u \times 3\ kg + v \times 8\ kg } { u + v } \qquad & | \qquad 3\ kg = 8\ kg - 5\ kg \\ &=& \frac{u \times (8\ kg - 5\ kg) + v \times 8\ kg }{ u + v } \\ &=& \frac{u \times (8 - 5 ) + v \times 8} { u + v} \\ &=& \frac{ 8u - 5u + 8v }{ u + v} \\ &=& \frac{ 8(u+v) - 5u}{ u + v} \\ &=& \frac{ 8(u+v) }{ u + v} - \frac{ 5u }{ u + v} \\ &=& 8 - \frac{ 5u }{ u + v} \\ \end{array}\)

mean is a whole number, if \(\frac{ 5u }{ u + v}\) is a whole number.

The divisors of 5 is 5 and 1.

So \(\frac{ 5u }{ u + v}\) is a whole number, if \(u+v= 1\) or \(u+v = 5\)

1.)  u+v= 1

\(\begin{array}{|rcrcc|ccc|} \hline u && v && u+v & 8 - \frac{ 5u }{ u + v}&& mean \\ \hline 1 &+& 0 & = & 1 & 8 - \frac{ 5\cdot 1 }{ 1 + 0} &=& 3 \\ 0 &+& 1 & = & 1 & 8 - \frac{ 5\cdot 0 }{ 0 + 1} &=& 8 \\ \hline \end{array}\)

2.)  u+v= 5

\(\begin{array}{|rcrcc|ccc|} \hline u && v && u+v & 8 - \frac{ 5u }{ u + v}&& mean \\ \hline 1 &+& 4 & = & 5 & 8 - \frac{ 5\cdot 1 }{ 1 + 4} &=& 7 \\ 2 &+& 3 & = & 5 & 8 - \frac{ 5\cdot 2 }{ 2 + 3} &=& 6 \\ 3 &+& 2 & = & 5 & 8 - \frac{ 5\cdot 3 }{ 3 + 2} &=& 5 \\ 4 &+& 1 & = & 5 & 8 - \frac{ 5\cdot 4 }{ 4 + 1} &=& 4 \\ \hline \end{array}\)

Imagine you have a large supply of 3kg and 8kg weights.

Two 3kg weights and three 8kg weights have a mean of 6kg, because (3+3+8+8+8)/5=6 .

Can you find other combinations of 3kg and 8kg weights whose mean weight is whole number of kg?

What's the smallest?

Whats the largest?

Can you make all the whole numbers inbetween?

\(\begin{array}{|r|r|r|r|} \hline \text{mean} & \text{all combinations} & \text{example combination }~ (n= 1) & \text{example mean }~ (n= 1) \\ \hline 3 & 1\times 3\ kg & \\ 4 & 4\cdot n\times 3\ kg + n\times 8\ kg & 4\cdot 1\times 3\ kg + 1\times 8\ kg & \frac{12+8}{4+1} = 4 \\ 5 & 3\cdot n\times 3\ kg + 2\cdot n\times 8\ kg & 3\cdot 1\times 3\ kg + 2\cdot 1\times 8\ kg & \frac{9+16}{3+2} = 5 \\ 6 & 2\cdot n\times 3\ kg + 3\cdot n\times 8\ kg & 2\cdot 1\times 3\ kg + 3\cdot 1\times 8\ kg & \frac{6+24}{2+3} = 6 \\ 7 & n\times 3\ kg + 4\cdot n\times 8\ kg & 1\times 3\ kg + 4\cdot 1\times 8\ kg & \frac{3+32}{1+4} = 7 \\ 8 & 1\times 8\ kg & \\ \hline \end{array}\)