heureka

If (x+1/x)^2=3, what is x^3+1/x^3

\(\begin{array}{|rcll|} \hline \left(x+\frac{1}{x}\right)^2 &=&3 \\ x+\frac{1}{x} &=& \sqrt{3} \\ \hline \end{array} \)

\(\small{ \begin{array}{|rclcll|} \hline (x+\frac{1}{x})^3 &=& \left(x+\frac{1}{x}\right)^2 \cdot \left(x+\frac{1}{x}\right) &=& 3\cdot \sqrt{3} \\ x^3+3x^2\cdot \frac{1}{x} +3x\cdot \frac{1}{x^2}+\frac{1}{x^3} &=& && 3\cdot \sqrt{3} \\ x^3+3x +3\cdot \frac{1}{x}+\frac{1}{x^3} &=& && 3\cdot \sqrt{3} \\ x^3+\frac{1}{x^3}+ 3x +3\cdot \frac{1}{x} &=& && 3\cdot \sqrt{3} \\ x^3+\frac{1}{x^3}+ 3\cdot(x + \frac{1}{x}) &=& && 3\cdot \sqrt{3} \quad & | \quad x + \frac{1}{x}=\sqrt{3} \\ x^3+\frac{1}{x^3}+ 3\cdot\sqrt{3} &=& && 3\cdot \sqrt{3} \\ x^3+\frac{1}{x^3} &=& && 3\cdot \sqrt{3}- 3\cdot\sqrt{3} \\ \mathbf{x^3+\frac{1}{x^3}} &\mathbf{=} & &&\mathbf{ 0 } \\ \hline \end{array} }\)

The general form of the equation of a circle is x2+y2−4x−8y−5=0.

What are the coordinates of the center of the circle?

A circle can be defined as the locus of all points that satisfy the equation

\((x-h)^2 + (y-k)^2 = r^2\)  ( Standard Form )

where r is the radius of the circle,
and h,k are the coordinates of its center.

The general Form is:

\(x^2+y^2 +ax+by+c = 0\)

Standard Form to general Form:

\(\begin{array}{|rcll|} \hline (x-h)^2 + (y-k)^2 &=& r^2 \\ x^2-2xh+h^2+y^2-2yk+k^2 &=& r^2 \\ x^2+y^2+x\cdot\underbrace{(-2h)}_{=a}+y\cdot\underbrace{(-2k)}_{=b}+\underbrace{h^2+k^2-r^2}_{=c} &=& 0 \\ \hline \end{array} \)

h,k and r ?

\(\begin{array}{|rcll|} \hline x^2+y^2+x\cdot\underbrace{(-2h)}_{=a}+y\cdot\underbrace{(-2k)}_{=b}+\underbrace{h^2+k^2-r^2}_{=c} &=& 0 \\\\ a &=&-2h\\ \color{red}h &\color{red}=& \color{red}-\frac{a}{2} \\\\ b &=&-2k\\ \color{red}k &\color{red}=& \color{red}-\frac{b}{2} \\\\ c &=&h^2+k^2-r^2\\ c &=&(-\frac{a}{2})^2+(-\frac{b}{2})^2-r^2\\ c &=& \frac{a^2+b^2}{4} -r^2\\ r^2 &=& \frac{a^2+b^2}{4} -c \\ \color{red}r &\color{red}=& \color{red} \sqrt{\frac{a^2+b^2}{4} -c} \\ \hline \end{array} \)

If we have a,b and c, we can calculate h,k and r:

\(\begin{array}{|lcll|} \hline \mathbf{x^2+y^2 +ax+by+c = 0} \\ h = -\dfrac{a}{2} \\ k = -\dfrac{b}{2} \\ r = \sqrt{\dfrac{a^2+b^2}{4} -c} \\ \hline \end{array} \)

\(a=-4\\ b=-8\\ c=-5\)

\(\begin{array}{|lcll|} \hline \mathbf{x^2+y^2 -4x-8y-5 = 0} \\\\ h = -\dfrac{-4}{2} \\ \mathbf{h = 2} \\\\ k = -\dfrac{-8}{2} \\ \mathbf{k = 4} \\\\ r = \sqrt{\dfrac{(-4)^2+(-8)^2}{4} -(-5)} \\ r = \sqrt{\dfrac{16+64}{4} +5} \\ r = \sqrt{20 +5} \\ r = \sqrt{25} \\ \mathbf{r = 5} \\ \hline \end{array}\)

The coordinates of the center of the circle is (2,4) and the radius is 5

consider the intersection of the function; x2+y-8=0 and x+y=2.

what is the greater of the x-coordinates of the points of intersection?

\(\begin{array}{|lrcll|} \hline (1) & x^2+y-8 &=& 0 \\ & y &=& 8-x^2 \\\\ (2) & x+y&=& 2 \quad & | \quad y = 8-x^2 \\ & x+8-x^2 &=& 2 \\ & x^2 - x - 6 &=& 0 \\ & x &=& \frac{1\pm \sqrt{1-4\cdot(-6)} }{2} \\ & x &=& \frac{1\pm \sqrt{1+24} }{2} \\ & x &=& \frac{1\pm 5 }{2} \\ \\ & x_1 &=& \frac{1+5 }{2} \\ & x_1 &=& \frac62 \\ &\mathbf{ x_1 } & \mathbf{=} & \mathbf{3} \\\\ & x_2 &=& \frac{1-5 }{2} \\ & x_2 &=& -\frac{4}{2}\\ &\mathbf{ x_2 } & \mathbf{=} & \mathbf{-2} \\ \hline \end{array}\)

The greater of the x-coordinates of the points of intersection is 3

4x^2-y^2+2y-1=0 ?? is two straight line at Cartesian level

\(\begin{array}{|rcll|} \hline 4x^2-y^2+2y-1 &=& 0 \\ 4x^2-(y^2-2y+1) &=& 0 \quad & | \quad y^2-2y+1 = (y-1)^2 \\ 4x^2-(y-1)^2 &=& 0 \\ (y-1)^2 &=& 4x^2 \quad & | \quad \text{square root both sides} \\ y-1 &=& \pm 2x \\ y &=& \pm 2x + 1 \\\\ \text{Line 1 }: y &=& 2x + 1 \\ \text{Line 2 }: y &=& -2x + 1 \\ \hline \end{array}\)

Given : sin\alpha+sin\beta=1 , cos\alpha+cos\beta=1

Find the value of sin\alpha-cos\beta

Given:

\(sin\alpha+sin\beta=1,\ cos\alpha+cos\beta=1\)

Find the value of

\(sin\alpha-cos\beta\)

\(\begin{array}{|lrcll|} \hline (1) & \sin(\alpha) + \sin(\beta) &=& 1 \\ (2) & \cos(\alpha) + \cos(\beta) &=& 1 \\ \hline (1)-(2): & \sin(\alpha) + \sin(\beta) - \Big(\cos(\alpha) + \cos(\beta) \Big) &=& 0 \\ & \sin(\alpha) + \sin(\beta) -\cos(\alpha) - \cos(\beta) &=& 0 \\ & \underbrace{\sin(\alpha) -\cos(\alpha)}_{=\sqrt{2}\sin(\alpha-45^{\circ})} &=& \underbrace{\cos(\beta) - \sin(\beta)}_{=-\sqrt{2}\sin(\beta-45^{\circ})} \\ & \sqrt{2}\sin(\alpha-45^{\circ}) &=& -\sqrt{2}\sin(\beta-45^{\circ}) \\ & \sin(\alpha-45^{\circ}) &=& - \sin(\beta-45^{\circ}) \\ & \sin(\alpha-45^{\circ}) &=& \sin\Big(-(\beta-45^{\circ})\Big) \\ & \sin(\alpha-45^{\circ}) &=& \sin(45^{\circ}-\beta ) \\ & \alpha-45^{\circ} &=& 45^{\circ}-\beta \\ & \alpha &=& 90^{\circ}-\beta \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline && \sin(\alpha) - \cos(\beta) \\ & =& \sin(90^{\circ}-\beta) - \cos(\beta) \\ & =& \cos(\beta) - \cos(\beta) \\ & =& 0 \\ \hline \end{array}\)

Proof:

\(\begin{array}{|rcll|} \hline \sin(\alpha) + \sin(\beta) &=& 1 \quad | \quad \alpha = 90^{\circ}-\beta \\ \sin(90^{\circ}-\beta) + \sin(\beta) &=& 1 \\ \cos(\beta) + \sin(\beta) &=& 1 \quad | \quad \cos(\beta) + \sin(\beta)=\sqrt{2}\sin(\beta+45^{\circ}) \\ \sqrt{2}\sin(\beta+45^{\circ}) &=& 1 \\ \sin(\beta+45^{\circ}) &=& \frac{1}{\sqrt{2}} \\ \beta+45^{\circ} &=& \arcsin( \frac{1}{\sqrt{2}} ) + n\cdot 360^{\circ} \qquad n \in \mathbb{Z} \\ \beta+45^{\circ} &=& 45^{\circ} + n\cdot 360^{\circ} \qquad n \in \mathbb{Z} \\ \beta &=& 90^{\circ} + n\cdot 360^{\circ} \qquad n \in \mathbb{Z} \\\\ \alpha &=& 90^{\circ}-\beta \\ \alpha &=& 90^{\circ}-(90^{\circ} + n\cdot 360^{\circ}) \qquad n \in \mathbb{Z} \\ \alpha &=& 0^{\circ} + n\cdot 360^{\circ} \qquad n \in \mathbb{Z} \\ \hline \end{array}\)

A trapezoid has an area of 341 square miles.

One base is 8 miles long.

The height measures 22 miles.

What is the length of the other base ?

\(\begin{array}{|rcll|} \hline A &=& \left(\frac{x+8}{2} \right) \cdot 22 \\ 341 &=& \left(\frac{x+8}{2} \right) \cdot 22 \\ 341 &=& (x+8) \cdot 11 \quad &| \quad : 11 \\ 31 &=& x+8 \quad &| \quad - 8 \\ 31-8 &=& x \\ 23 &=& x \\ \hline \end{array} \)

The other base is 23 miles long.

Find the remainder when 100^100 is divided by 7

\(\begin{array}{|rcll|} \hline && 100^{100} \pmod{7} \quad & |\quad \boxed{ 100 \equiv 2 \pmod {7} } \\ &\equiv & 2^{100} \pmod{7} \quad & |\quad \text{because } gcd(2,7) = 1\ : \\ && & |\quad 2^{\phi(7)} \equiv 1 \pmod {7} \quad \phi(7) = 6 \quad \phi() \text{ is the Euler's totient function}\\ && & |\quad \boxed{2^{6} \equiv 1 \pmod {7}} \\ &\equiv &2^{100} \pmod{7} \quad & | \quad 100 = 6\cdot 16 +4 \\ &\equiv &2^{6\cdot 16 +4} \pmod{7} \\ &\equiv & (2^{6} )^{16}\cdot 2^4 \pmod{7} \\ &\equiv & 1^{16}\cdot 2^4 \pmod{7} \\ &\equiv & 2^4 \pmod{7} \\ &\equiv & 16 \pmod{7} \\ &\equiv & 2 \pmod{7} \\ \hline \end{array}\)

John is eating jellybeans.In his bag of jellybeans he has 2 red, 2 green and 1 black jellybean left.

Find the probability that the next 2 jellybeans he eats are both red?

Method 1:

probability  2 red \( = \frac{2}{5} \cdot \frac{1}{4} = \frac{2}{20} = \frac{1}{10} = 0.1\quad (10 \% )\)

Method 2:

probability  2 red

\(\begin{array}{|rcll|} \hline &= & \frac{\binom{2}{2}_{red}\binom{2}{0}_{green}\binom{1}{0}_{black}}{\binom52} \\ &=& \frac{1\cdot 1 \cdot 1 } { \frac52\cdot \frac41} \\ &=& \frac{2}{20} \\ &=& \frac{1}{10} \\ &=& 0.1\quad (10 \% ) \\ \hline \end{array}\)

x + y − z = −5

x + 2y + 3z = −12

2x − y − 13z = 13

\(\text{Let Matrix } A = \begin{pmatrix} 1 & 1 & -1 \\ 1 & 2 & 3 \\ 2 & -1 & -13 \end{pmatrix}\)

\(\text{Let } \vec{r} = \begin{pmatrix} -5 \\ -12 \\ 13 \end{pmatrix}\)

\(\text{Let } \vec{x} = \begin{pmatrix} x \\ y \\ z \end{pmatrix}\)

\(\text{Let Matrix } A^{-1} = \begin{pmatrix} -23 & 14 & 5 \\ 19 & -11 & -4 \\ -5 & 3 & 1 \end{pmatrix}\)

Then

\(\begin{array}{|rcll|} \hline A\vec{x} &=& \vec{r} \\ \vec{x} &=& A^{-1}\vec{r} \\ \vec{x} &=& \begin{pmatrix} -23 & 14 & 5 \\ 19 & -11 & -4 \\ -5 & 3 & 1 \end{pmatrix}\cdot \begin{pmatrix} -5 \\ -12 \\ 13 \end{pmatrix} \\ \vec{x} &=& \begin{pmatrix} 12 \\ -15 \\ 2 \end{pmatrix} \\ \hline \end{array} \)

x =  12

y = -15

z =  2