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Hallo Bertie,

\(60y^2 + 65x^2 - 128xy - 448x +448y + 768+{\color{red}60\cdot c}= 0 \qquad c > 0 \)   hyperbola between lines

Here is the answer:

I.

\(\small{ \begin{array}{lrcll} \mathbf{\text{1. line}} & 13x-10y-48=0 \\ &y &=& m_1x+b_1\\ &y &=&\frac{13}{10}x -\frac{48}{10} \\ & m_1 &=& \frac{13}{10}\\ & b_1 &=& -\frac{48}{10}\\ \text{or } & m_1x+b_1-y &=& 0 \qquad (\frac{13}{10}x -\frac{48}{10}-y=0) \\ \\ \mathbf{\text{2. line}} & 5x-6y-16=0 \\ &y &=& m_2x+b_2\\ &y &=&\frac{5}{6}x -\frac{16}{6} \\ & m_2 &=& \frac{5}{6}\\ & b_2 &=& -\frac{16}{6}\\ \text{or } & m_2x+b_2-y &=& 0 \qquad (\frac{5}{6}x -\frac{16}{6} -y=0) \\\\ \mathbf{\text{Intersection 1. line}} \\ \mathbf{\text{and 2.line}}\\ & x_c &=& \frac{b_2-b_1}{m_1-m_2} = \frac{32}{7} \\ & y_c &=& \frac{m_1b_2-b_1m_2}{m_1-m_2} = \frac{8}{7} \\ \end{array} }\)

II. The Hyperbola is the product of the two lines + c : \(\boxed{~ (m_1x+b_1-y)(m_2x+b_2-y) + c = 0 ~}\)   c > 0

\(\small{ \begin{array}{lrcll} \text{or } &\boxed{~ y^2+m_1m_2x^2+\left(m_1b_2+m_2b_1\right)x-\left(m_1+m_2\right)xy-\left(b_1+b_2\right)y+b_1b_2+c=0~} \\ \text{or } & y^2+ \frac{13}{12}x^2-\frac{112}{15}x-\frac{32}{15}xy+\frac{112}{15}y+\frac{64}{5}+c=0 \qquad \cdot 60\\ \text{or } &\boxed{~ 60y^2+ 65x^2-448x-128xy+448y+768+60\cdot c=0 ~} \end{array} }\)

III. The Hyperbola is:  \(\small{ \begin{array}{rll} Y^2 + aX^2 + bXY + c &=& 0. \qquad | \quad X = x - x_c,~ Y = y - y_c & [see\ \#6]\\ \end{array} } \)

\(\small{ \begin{array}{lclcrl} a &=& m_1m_2 &=&\frac{13}{13} & [see\ \#11]\\ b &=& -(m_1+m_2) &=& -\frac{32}{15} & [see\ \#11]\\ x_c &=& \frac{b_2-b_1}{m_1-m_2} &=& \frac{32}{7} &\\ y_c &=& \frac{m_1b_2-b_1m_2}{m_1-m_2} &=& \frac{8}{7} & \\ \end{array} }\)

let us put all things together:

\(\small{ \begin{array}{rcll} (y - \frac{m_1b_2-b_1m_2}{m_1-m_2})^2 + m_1\cdot m_2\cdot (x - \frac{b_2-b_1}{m_1-m_2})^2 - (m_1+m_2)\cdot (x - \frac{b_2-b_1}{m_1-m_2})\cdot (y - \frac{m_1b_2-b_1m_2}{m_1-m_2}) + c &=& 0\\ \cdots \\ \end{array} }\\ \)

\(\small{ \begin{array}{lcll} &=& y^2\\ &+&m_1m_2x^2\\ &+&\left[ \underbrace{ \frac{(m_1+m_2)(m_1b_2-b_1m_2)-2m_1m_2(b_2-b_1)}{m_1-m_2}}_{1.} \right] x\\ &-&(m_1+m_2)xy\\ &+&\left[ \underbrace{ \frac{(m_1+m_2)(b_2-b_1)-2(m_1b_2-b_1m_2)}{m_1-m_2}}_{2.}\right] y\\ &+&\underbrace{ \frac{(m_1b_2-b_1m_2)^2+m_1m_2(b_2-b_1)^2-(m_1+m_2)(b_2-b_1)(m_1b_2-b_1m_2)}{(m_1-m_2)^2}}_{3.}\\ &+& c = 0 \end{array} }\)

calculate 1. :

\(\small{ \begin{array}{lcll} \frac{(m_1+m_2)(m_1b_2-b_1m_2)-2m_1m_2(b_2-b_1)}{m_1-m_2} \\ = \frac{(m_1+m_2)m_1b_2 -(m_1+m_2)b_1m_-2m_1m_2b_2+2m_1m_2b_1}{m_1-m_2} \\ \cdots \\ = \frac{-m_1m_2b_2+m_1m_2b_1+m_1^2b_2-m_2^2b_1}{m_1-m_2}\\ = \frac{b_1m_2(m_1-m_2)+b_2m_1(m_1-m_2)}{m_1-m_2}\\ \mathbf{= b_1m_2+b_2m_1}\\ \end{array} }\)

calculate 2. :

\(\small{ \begin{array}{lcll} \frac{(m_1+m_2)(b_2-b_1)-2(m_1b_2-b_1m_2)}{m_1-m_2}\\ = \frac{ m_1b_2-m_1b_1+m_2b_2-m_2b_1-2m_1b_2+2b_1m_2}{m_1-m_2}\\ = \frac{ -m_1b_2+b_1m_2-m_1b_1+m_2b_2}{m_1-m_2}\\ = \frac{-b_2(m_1-m_2)-b_1(m_1-m_2)}{m_1-m_2}\\ = -b_2-b_1\\ \mathbf{= -(b_1+b_2) }\\ \end{array} }\)

calculate 3. :

\(\small{ \begin{array}{lcll} \frac{(m_1b_2-b_1m_2)^2+m_1m_2(b_2-b_1)^2-(m_1+m_2)(b_2-b_1)(m_1b_2-b_1m_2)}{(m_1-m_2)^2} \\ \cdots \\ = \frac{-2m_1m_2b_1b_2+m_1^2b_1b_2+m_2^2b_1b_2}{(m_1-m_2)^2} \\ = \frac{b_1b_2(m_1^2-2m_1m_2+m_2^2)}{(m_1-m_2)^2} \\ = \frac{b_1b_2(m_1-m_2)^2)}{(m_1-m_2)^2} \\ \mathbf{= b_1b_2} \\ \end{array} }\)

So we have the same:

\(\small{ \begin{array}{lrcll} \boxed{~ y^2+m_1m_2x^2+\left(m_1b_2+m_2b_1\right)x-\left(m_1+m_2\right)xy-\left(b_1+b_2\right)y+b_1b_2+c=0~} \\ \end{array} }\)

Hallo Bertie,

i have finished your wonderful solution.

\(\begin{array}{rcll} \text{I set slope line 1: } ~ m_1 = \frac{13}{10} \\ \text{and slope line 2: } ~ m_2 = \frac{5}{6} \\ \text{and your equation: }~Y^2 + aX^2 + bXY + c &=& 0. \qquad | \quad X = x - \frac{32}{7},~ Y = y - \frac87\\ \text{and your equations of the two asymptotes: }~(\frac{Y}{X})^2 + b\cdot\frac{Y}{X} + a &=& 0\\ \text{and with solution of a quadratic: }~\frac{Y}{X} &=& \frac12 \cdot (-b \pm \sqrt{b^2 - 4a})\\ \text{we can find a and b: } \end{array}\)

\(\begin{array}{rcll} \frac{Y}{X} &=& \frac12 \cdot (-b \pm \sqrt{b^2 - 4a}) \\ 2\cdot \frac{Y}{X}+b &=& \pm \sqrt{b^2 - 4a} \qquad | \qquad \text{(square both sides)} \\ (2\cdot \frac{Y}{X}+b)^2 &=& b^2 - 4a \\ 4\cdot (\frac{Y}{X})^2 +4\cdot\frac{Y}{X}\cdot b +b^2 &=& b^2 - 4a \\ 4\cdot (\frac{Y}{X})^2 +4\cdot\frac{Y}{X}\cdot b &=& - 4a \\ (\frac{Y}{X})^2 + \frac{Y}{X}b &=& -a \qquad | \qquad \text{we set }~ \frac{Y}{X}=m_1 \text{ and }\frac{Y}{X}=m_2\\ m_1^2+m_1b =&-a& =m_2^2+m_2b \\ b\cdot (m_1-m_2) &=& m_2^2-m_1^2 \\ b\cdot (m_1-m_2) &=&(m_2-m_1)(m_1+m_2) \\ b\cdot (m_1-m_2) &=&-(m_1-m_2)(m_1+m_2) \\ \mathbf{b} & \mathbf{=} & \mathbf{-(m_1+m_2)} = - (\frac{13}{10}+\frac{5}{6})=-\frac{32}{15} \\\\ m_1^2+m_1b &=& -a \\ m_2^2+m_2b &=& -a \\ b&=& \frac{-a-m_1^2}{m_1} = \frac{-a-m_2^2}{m_2} \\ m_2( -a-m_1^2 ) &=& m_1(-a-m_2^2 )\\ \cdots \\ a\cdot (m_1-m_2) &=& m_1\cdot m_2\cdot (m_1-m_2) \\ \mathbf{a} & \mathbf{=} & \mathbf{m_1\cdot m_2} = \frac{13}{10}\cdot \frac{5}{6} = \frac{13}{12}\\ \end{array}\)

\(\begin{array}{lcll} \text{your equation is now: } \end{array}\\ \begin{array}{rcll} Y^2 + m_1\cdot m_2\cdot X^2 - (m_1+m_2)\cdot XY + c &=& 0. \qquad | \quad X = x - \frac{32}{7},~ Y = y - \frac87\\ \end{array}\\ \begin{array}{lcll} \text{or: } \end{array}\\ \begin{array}{rcll} \mathbf{(y-\frac87)^2 + \frac{13}{10}\cdot \frac{5}{6} \cdot (x-\frac{32}{7})^2 - (\frac{13}{10}+\frac{5}{6})\cdot (x-\frac{32}{7}) \cdot (y-\frac87)+c }&\mathbf{=}& \mathbf{0} \end{array}\)

All of the water from a full cylinder tank is drained into a rectangular aquarium. The cylinder has a height of 50 cm and a diameter of 30 cm. How deep is the water in the aquarium, if it has a rectangular base measuring 40 cm by 20 cm?

A cylinder with radius r units and height h units

has a volume of V cubic units given by 

\(\begin{array}{lcll} \boxed{~ \begin{array}{lcll} V_{\text{cylinder tank}} &=& \pi \cdot r^2 \cdot h \\ &=& \pi \cdot 30^2 \cdot 50 \ cm^3 \\ &=& 45000 \cdot \pi \ cm^3 \\ &=& 141371.669412\ cm^3 \\ \end{array} ~}\\\\ \end{array}\\\)

\(\begin{array}{lcll} \boxed{~ \begin{array}{lcll} V_{\text{aquarium}} &=& a\cdot b \cdot h \\ &=& 40\ cm \cdot 20\ cm \cdot h \\ &=& 800\ cm^2 \cdot h \\ \end{array} ~}\\\\ \end{array}\\\)

\(\begin{array}{lcll} \boxed{~ \begin{array}{lcll} V_{\text{cylinder tank}} &=& V_{\text{aquarium}} \\ 141371.669412\ cm^3 &=& 800\ cm^2 \cdot h_{\text{aquarium}} \\ h_{\text{aquarium}} &=& \frac{141371.669412\ cm^3}{800\ cm^2} \\ h_{\text{aquarium}} &=& 176.714586764\ cm \end{array} ~} \end{array}\)

The water in the aquarium is 176.7 cm deep.

Or in meters: 1.767 m

Find the first term, the common difference, and tn for a series having Sn = 4n^2+n.

Find the first term, the common difference, and tn for a series having Sn = 4n^2+n.

\(\begin{array}{lcll} \boxed{~ \begin{array}{lcll} S_n &=& 4n^2+n \end{array} ~}\\\\ \end{array}\\ \begin{array}{lcll} t_n &=& S_n - S_{n-1} \\ t_n &=&4n^2+n - [ 4(n-1)^2+(n-1) ] \\ t_n &=&4n^2+n - 4(n-1)^2 - (n-1) ] \\ t_n &=&4n^2+n - 4(n-1)^2 - n +1 ] \\ t_n &=&4n^2 - 4(n-1)^2 +1 ] \\ t_n &=&4n^2 - 4(n^2-2n+1) +1 ] \\ t_n &=&4n^2 - 4n^2 +8n-4 +1 ] \\ t_n &=&8n-4 +1 ] \\ \mathbf{t_n} & \mathbf{=} & \mathbf{8n-3} \\\\ t_1 &=& 8\cdot 1 - 3 \\ t_1 &=& 8-3\\ \mathbf{t_1} &\mathbf{=}& \mathbf{5}\\\\ d &=& t_n - t_{n-1} \\ d &=& 8n-3 -[ 8\cdot (n-1)-3 ]\\ d &=& 8n-3 - 8\cdot (n-1) +3 \\ d &=& 8n - 8\cdot (n-1) \\ d &=& 8n - 8n+8 \\ \mathbf{d} &\mathbf{=}& \mathbf{8} \\ \end{array}\)

Find the first term, the common difference, and tn for a series having Sn= 4n−3n^2.

Wie sieht die 2. Ableitung zu folgender Funktion aus?

f(x)= ln(x+2)-ln(x+1)

\(\begin{array}{lcll} \boxed{~ \begin{array}{lcll} y&=& \ln{(~f(x)~)}\\ y' &=& \frac{f'(x)}{f(x)} \end{array} ~} \end{array}\)

\(\begin{array}{lcll} y&=& \ln{(~x+2~)}\\ y' &=& \frac{1}{(x+2)}\\\\ y&=& \ln{(~x+1~)}\\ y' &=& \frac{1}{(x+1)} \end{array}\)

\(\begin{array}{lcll} y&=& \ln{(~x+2~)} - \ln{(~x+1~)}\\ y' &=& \frac{1}{(x+2)}-\frac{1}{(x+1)} = -\frac{1}{x^2+3x+2}\\\\ \end{array}\)

\(\begin{array}{lcll} \boxed{~ \begin{array}{lcll} y&=& \frac{1}{f(x)} \\ &=& f(x)^{-1} \qquad \text{(nach der Kettenregel abgeleitet)}\\ y' &=& (-1) \cdot f(x)^{-1-1} \cdot f'(x) \\ &=& - f(x)^{-2} \cdot f'(x) \\ &=& - \frac{f'(x)}{f(x)^{2}} \end{array} ~} \end{array}\)

\(\begin{array}{lcll} y&=& \frac{1}{x+2}\\ y' &=& - \frac{1}{(x+2)^{2}}\\\\ y&=& \frac{1}{x+1}\\ y' &=& - \frac{1}{(x+1)^{2}}\\\\ \end{array}\)

\(\begin{array}{lcll} y'&=& \frac{1}{x+2} - \frac{1}{x+1}\\ y'' &=& - \frac{1}{(x+2)^{2}} - \left( - \frac{1}{(x+1)^{2}} \right)\\ &=& - \frac{1}{(x+2)^{2}} + \frac{1}{(x+1)^{2}} \\\\ \mathbf{y''} &\mathbf{=}& \mathbf{\frac{2x+3}{(x^2+3x+2)^2} } \end{array}\)

y^3 = 4x^2

If x is doubled, by what factor does y increase?

\(\begin{array}{lcll} \boxed{~ \begin{array}{lcll} y^3 &=& 4\cdot x^2 \end{array} ~} \end{array}\\ \begin{array}{rcll} \\ (y\cdot f)^3 &=& 4\cdot (2x)^2 \\ y^3\cdot f^3 &=& 4\cdot 4 \cdot x^2 \qquad & | \qquad 4 \cdot x^2 = y^3\\ y^3\cdot f^3 &=& 4\cdot y^3\\ f^3 &=& 4\\ \mathbf{f} &\mathbf{=}& \mathbf{\sqrt[3]{4}}\\ \end{array}\\\)

sqrt(sqrt(x-5) + x) = 5 

Find all real solutions 

The only real solution is 21

sqrt(sqrt(x-5) + x) = 5 

Find all real solutions 

\(\begin{array}{lcll} \sqrt{ \sqrt{x-5} + x } &=& 5 \qquad &|\qquad \text{(square both sides)} \\ \sqrt{x-5} + x &=& 5^2 \\ \sqrt{x-5} + x &=& 25 \qquad &|\qquad - x \\ \sqrt{x-5} &=& 25-x \\ \sqrt{x-5} &=& 25-x \qquad &|\qquad \text{(square both sides)} \\ x-5 &=& (25-x)^2 \\ x-5 &=& 25^2-2\cdot 25x+x^2 \\ x-5 &=&625-50x+x^2 \\ 625-50x+x^2 &=& x-5 \qquad &|\qquad - x+5 \\ 625-50x+x^2 - x+5 &=&0 \\ x^2 -51x +630 &=&0 \\ x_{1,2} &=& \frac12 \cdot ( 51\pm \sqrt{51^2-4\cdot 630} ) \\ x_{1,2} &=& \frac12 \cdot( 51\pm \sqrt{2601-2530} )\\ x_{1,2} &=& \frac12 \cdot( 51\pm \sqrt{81} )\\ x_{1,2} &=& \frac12 \cdot ( 51\pm 9 )\\\\ x_{1} &=& \frac12 \cdot ( 51 + 9 )\\ \mathbf{ x_{1} } &\mathbf{=}& \mathbf{30}\\\\ x_{2} &=& \frac12 \cdot ( 51 - 9 )\\ \mathbf{ x_{2} } &\mathbf{=}& \mathbf{21} \end{array}\)