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I think it goes through the points (2.75,2.75) (1.75,5) and (5,1.75)

How do I find a graph that will go through these points ?? 

The formula of a hyperbola is given in general form: \(y = \frac{a}{x+c} + d \)

Set \(P_1(x_1,y_1) = (2.75,2.75)\)
Set \(P_2(x_2,y_2) = (1.75,5)\)
Set \(P_3(x_3,y_3) = (5,1.75)\)

\(\begin{array}{|lrcll|} \hline (1) & y_1 &=& \frac{a}{x_1+c} + d \\ & (y_1-d)*(x_1+c) &=& a \\ & y_1x_1+cy_1-dx_1&=& a+cd \\\\ (2) & y_2 &=& \frac{a}{x_2+c} + d \\ & (y_2-d)*(x_2+c) &=& a \\ & y_2x_2+cy_2-dx_2&=& a+cd \\\\ (3) & y_3 &=& \frac{a}{x_3+c} + d \\ & (y_3-d)*(x_3+c) &=& a \\ & y_3x_3+cy_3-dx_3&=& a+cd \\\\ \hline \end{array} \)

\(\begin{array}{|lrcll|} \hline (1) & y_1x_1+cy_1-dx_1&=& a+cd \\ (2) & y_2x_2+cy_2-dx_2&=& a+cd \\ (3) & y_3x_3+cy_3-dx_3&=& a+cd \\ \hline (1)-(2) & y_1x_1+cy_1-dx_1 -(y_2x_2+cy_2-dx_2) &=& 0 \\ & c(y_1-y_2)+d(x_2-x_1) &=& y_2x_2-y_1x_1 \\\\ (1)-(3) & y_1x_1+cy_1-dx_1 -(y_3x_3+cy_3-dx_3) &=& 0 \\ & c(y_1-y_3)+d(x_3-x_1) &=& y_3x_3-y_1x_1 \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline (1) & c(y_1-y_2)+d(x_2-x_1) &=& y_2x_2-y_1x_1 \\ (2) & c(y_1-y_3)+d(x_3-x_1) &=& y_3x_3-y_1x_1 \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline c &=& \frac{ \begin{vmatrix} y_2x_2-y_1x_1 & x_2-x_1 \\ y_3x_3-y_1x_1 & x_3-x_1 \end{vmatrix} } { \begin{vmatrix} y_1-y_2 & x_2-x_1 \\ y_1-y_3 & x_3-x_1 \end{vmatrix} } \\ c &=& \frac{ \begin{vmatrix} 5\cdot1.75-2.75\cdot2.75 & 1.75-2.75 \\ 1.75\cdot5-2.75\cdot2.75 & 5-2.75 \end{vmatrix} } { \begin{vmatrix} 2.75-5 & 1.75-2.75 \\ 2.75-1.75 & 5-2.75 \end{vmatrix} } \\ \mathbf{c} &\mathbf{=}& \mathbf{-0.95} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline d &=& \frac{ \begin{vmatrix} y_1-y_2 & y_2x_2-y_1x_1 \\ y_1-y_3 & y_3x_3-y_1x_1 \end{vmatrix} } { \begin{vmatrix} y_1-y_2 & x_2-x_1 \\ y_1-y_3 & x_3-x_1 \end{vmatrix} } \\ d &=& \frac{ \begin{vmatrix} 2.75-5 & 5\cdot 1.75-2.75\cdot 2.75 \\ 2.75-1.75 & 1.75\cdot 5-2.75\cdot 2.75 \end{vmatrix} } { \begin{vmatrix} 2.75-5 & 1.75-2.75 \\ 2.75-1.75 & 5-2.75 \end{vmatrix} } \\ \mathbf{d} &\mathbf{=}& \mathbf{0.95} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline a &=& y_1x_1+cy_1-dx_1-cd \\ a &=& 2.75\cdot 2.75-0.95\cdot 2.75-0.95\cdot 2.75 - (-0.95)\cdot 0.95 \\ \mathbf{a} &\mathbf{=}& \mathbf{3.24} \\ \hline \end{array}\)

The formula of a hyperbola is given through  \(P_1, P_2, P_3:\ y = \frac{3.24}{x-0.95} + 0.95\)

The graph is:

how to calculate 14^15 mod 15

*VERY* Hard Math Question for my profile picture

I assume two points of the hyperbola are: \(P_1(4,2) \text{ and } P_2 (2,4)\)

Then the hyperbola equation is \( xy = 8\), because the Area under the hyperbola is 4*2 = 2*4 = 8 and \(y_{\text{hyperbola}} = \frac{A}{x}\)

The other equation is a line \( y = x\)

The intersect Point:

\(\begin{array}{|rcll|} \hline xy&=&8 \quad & | \quad y=x \\ x\cdot x &=&8 \\ x^2 &=&8 \\ x^2 &=&2\cdot 4 \\ x_{\text{intersection} } &=& 2\sqrt{2} \\\\ y_{\text{intersection} } &=& x_{\text{intersection} } \\ y_{\text{intersection} } &=& 2 \sqrt{2} \\ \hline \end{array}\)

The intersect Point is: \((2\sqrt{2},\ 2 \sqrt{2} )\)

The picture is:

14^15 mod 15

\(\begin{array}{|rcll|} \hline \mathbf{ 14^{15} \pmod {15} =\ ? } \\\\ && 14^{15} \pmod {15} \quad &| \quad 14 \equiv -1 \pmod {15} \\ &\equiv& (-1)^{15} \pmod {15} \quad &| \quad (-1)^{15} = -1 \\ &\equiv& -1 \pmod {15} \\ &\equiv& -1 +15 \pmod {15} \\ &\equiv& 14 \pmod {15} \\\\ \mathbf{ 14^{15} \pmod {15} = 14 } \\ \hline \end{array} \)

10,11,12,13,14,15,16,17,18,19,20,21,?,31 

Sum of digits is strictly greater than product of digits.

\(10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, \mathbf{30}, 31, \\ 40, 41, 50, 51, 60, 61, 70, 71, 80, 81, 90, 91, 100, 101, \\ 102, 103, 104, 105, 106, 107, 108, 109, 110, 111, 112, \\ 113, 114, 115, 116, 117, 118, 119, 120, 121, 122, 130, 131, \\ 140, 141, 150, 151, 160, 161, 170, \ldots\)

Example:

\(\begin{array}{|r|r|c|r|} \hline & \text{ Sum of digits } && \text{product of digits} \\ \hline 10 & 1+0 = 1 && 1\cdot 0 = 0 \\ & & 1 > 0 & \\\\ 11 & 1+1 = 2 && 1\cdot 1 = 1 \\ & & 2 > 1 & \\\\ 12 & 1+2 = 3 && 1\cdot 2 = 2 \\ & & 3 > 2 & \\ \ldots & && \\ \\ 20 & 2+0 = 2 && 2\cdot 0 = 0 \\ & & 2 > 0 & \\\\ 21 & 2+1 = 3 && 2\cdot 1 = 2 \\ & & 3 > 2 & \\ \ldots & && \\ \\ 30 & 3+0 = 3 && 3\cdot 0 = 0 \\ & & 3 > 0 & \\\\ 31 & 3+1 = 4 && 3\cdot 1 = 3 \\ & & 4 > 3 & \\ \hline \end{array}\)

use 1, 2, 3, 4, 5, 6 in the equation ?/? + ?/? - ?/? = 1/30

\(\dfrac16 + \dfrac23-\dfrac45 = \dfrac{1}{30}\)

Suppose that for some a,b,c we have

a+b+c = 6,

ab + ac + bc = 5, and

abc = -12.

What is a^3 + b^3 + c^3?

1. 

\(\small{ \begin{array}{|rcll|} \hline (a+b+c)\times (ab + ac + bc) &=& 6\cdot 5 \\ (a+b+c)\times (ab + ac + bc) &=& 30 \\ a^2b+a^2c+abc+ab^2+abc+b^2c+abc+ac^2+bc^2 &=& 30 \\ a^2(b+c) +b^2(a+c) +c^2(a+b)+ 3abc &=& 30 \\ a^2(b+c) +b^2(a+c) +c^2(a+b) &=& 30 - 3abc \quad & | \quad abc = -12 \\ a^2(b+c) +b^2(a+c) +c^2(a+b) &=& 30 - 3(-12) \\ a^2(b+c) +b^2(a+c) +c^2(a+b) &=& 30 +36 \\ \mathbf{a^2(b+c) +b^2(a+c) +c^2(a+b)} & \mathbf{=} &\mathbf{66} \\ \hline \end{array} }\)

2.

\(\small{ \begin{array}{|rcll|} \hline (a+b+c)^3 &=& (a+b+c)^2\times (a+b+c) \\ 6^3 &=& \Big(a^2+b^2+c^2+2(ab + ac + bc ) \Big)\times (a+b+c) \quad | \quad ab + ac + bc = 5 \\ 216 &=& (a^2+b^2+c^2+2\cdot 5 )\times (a+b+c) \\ 216 &=&(a^2+b^2+c^2+10 )\times (a+b+c) \\ 216 &=&a^3+a^2(b+c)+b^3+b^2(a+c)+c^3+c^2(a+b)+10(a+b+c) \\ 216 &=&a^3+b^3+c^3+a^2(b+c)+b^2(a+c)+c^2(a+b)+10(a+b+c) \quad | \quad a+b+c = 6\\ 216 &=&a^3+b^3+c^3+a^2(b+c)+b^2(a+c)+c^2(a+b)+10\cdot 6\\ 216 &=&a^3+b^3+c^3+a^2(b+c)+b^2(a+c)+c^2(a+b)+60 \\ 216-60 &=&a^3+b^3+c^3+a^2(b+c)+b^2(a+c)+c^2(a+b)\\ 156 &=&a^3+b^3+c^3+\underbrace{a^2(b+c)+b^2(a+c)+c^2(a+b)}_{=66} \\ 156 &=&a^3+b^3+c^3+66 \\ 156-66 &=&a^3+b^3+c^3 \\ 90 &=&a^3+b^3+c^3 \\ \mathbf{a^3+b^3+c^3} & \mathbf{=} &\mathbf{90} \\ \hline \end{array} }\)

\frac{4}{.2\cdot \frac{4}{6\cdot \frac{3}{3\cdot \frac{3.06}{\frac{6}{\frac{3}{955\cdot 6}}}}}}

\(\begin{array}{|l|ll|} \hline && \text{repeating decimal}:\\ \dfrac{4}{0.2\times \dfrac{4}{6\times \dfrac{3}{3\times \dfrac{3.06}{\dfrac{6}{\dfrac{3}{955\times 6}}}}}} &=& 112352.\overline{9411764705882352}\ (\text{period} 16) \\ \hline \end{array}\)

972 modulo 16

\(\begin{array}{|rcll|} \hline && 972 \text{ modulo } 16 \\ &\equiv & 972 - 60\cdot 16 \text{ modulo } 16 \\ &\equiv & 972 - 960 \text{ modulo } 16 \\ &\equiv & 12 \text{ modulo } 16 \\ \hline \end{array}\)

integral( (x+4) / (x^2-5x+6) ) dx

\(\begin{array}{rcl} && \int { \frac{x+4} {x^2-5x+6} \ dx} \quad & | \quad x^2-5x+6 = (x-3)(x-2) \\ &=& \int { \frac{x+4} {(x-3)(x-2)} \ dx} \\ \end{array} \)

Partial fraction decomposition:

\(\begin{array}{|lrcll|} \hline & \frac{x+4} {(x-3)(x-2)} &=& \frac{A}{x-3} + \frac{B}{x-2} \\ & x+4 &=& A\cdot (x-2) + B\cdot (x-3) \\ x = 2: & 2+4 &=& A\cdot (2-2) + B\cdot (2-3) \\ & 6 &=& 0 + B\cdot (-1) \\ & 6 &=& B\cdot (-1) \\ & \mathbf{B} &\mathbf{=}& \mathbf{-6} \\ \\ x = 3: & 3+4 &=& A\cdot (3-2) + B\cdot (3-3) \\ & 7 &=& A\cdot (1) + B\cdot (0) \\ & 7 &=& A\cdot (1) \\ & \mathbf{A} &\mathbf{=}& \mathbf{7} \\\\ & \mathbf{\frac{x+4} {(x-3)(x-2)}} &\mathbf{=}& \mathbf{\frac{7}{x-3} - \frac{6}{x-2}} \\ \hline \end{array} \)

\(\begin{array}{rcl} && \mathbf{ \int { \frac{x+4} {x^2-5x+6} \ dx} } \\ &=& \int { \frac{x+4} {(x-3)(x-2)} \ dx} \\ &=& \int { \Big(\frac{7}{x-3} - \frac{6}{x-2} \Big) \ dx} \\ &=& 7\cdot \int { \frac{1}{x-3} \ dx} -6\int { \frac{1}{x-2} \ dx} \\ &\mathbf{=}& \mathbf{7\cdot \ln(|x-3|) -6\cdot \ln(|x-2|) +c }\\ \end{array} \)