heureka

Solve for the indicated variable

1. x4-4x2+2=0 ; 0

$\small{\text{$ \begin{array}{rcl} x^4-4x^2 +2 &=& 0 \qquad | \qquad x^2 = z \qquad x = \pm\sqrt{z}\\ z^2 -4z+2 &=& 0 \\ z_{1,2} &=& \dfrac{4\pm\sqrt{16-4\cdot 2} }{2} \\ z_{1,2} &=& \dfrac{4\pm\sqrt{8} }{2} \qquad | \qquad \sqrt{8} = \sqrt{4\cdot 2} = 2\sqrt{2}\\ z_{1,2} &=& \dfrac{4\pm2\sqrt{2} }{2} \\ z_{1,2} &=& 2 \pm\sqrt{2} \\ z_1 &=& 2 +\sqrt{2} \\ z_2&=& 2 -\sqrt{2} \\\\ x_{1,2} &=& \pm\sqrt{z_1} \\ x_{1,2} &=& \pm\sqrt{2 +\sqrt{2}}\\ \mathbf{ x_1 }&\mathbf{=}& \mathbf{\sqrt{2 +\sqrt{2}}}\\ \mathbf{x_2 }&\mathbf{=}& \mathbf{-\sqrt{2 +\sqrt{2}}}\\\\ x_{3,4} &=& \pm\sqrt{z_2} \\ x_{3,4} &=& \pm\sqrt{2 -\sqrt{2}}\\ \mathbf{x_3} &\mathbf{=}& \mathbf{\sqrt{2 -\sqrt{2}}}\\ \mathbf{x_4} &\mathbf{=}& \mathbf{-\sqrt{2 -\sqrt{2}}}\\\\ \end{array} $}}$

Factor completely

1. 5cos2x-5sin2x+cosx+sinx

$\small{\text{$ \begin{array}{rcl} & & 5\cos^2{(x)}-5\sin^2{(x)}+\cos{(x)}+\sin{(x)} \\ &=& 5 [ \cos^2{(x)}-\sin^2{(x)} ] + [ \cos{(x)}+\sin{(x)} ] \\ &=& 5 [ \cos{(x)}-\sin{(x)} ] [ \cos{(x)}+\sin{(x)} ]+ [ \cos{(x)}+\sin{(x)} ] \\ \mathbf{ 5\cos^2{(x)}-5\sin^2{(x)}+\cos{(x)}+\sin{(x)} }& \mathbf{=} & \mathbf{ [ \cos{(x)}+\sin{(x)} ] [ 5 ( \cos{(x)}-\sin{(x)} ) +1 ] }\\ \end{array} $}}$

Simplify each expression

1. (x1/2-x1/3) / x1/6

$\small{ \begin{array}{rcl} \dfrac{ x^\frac12-x^\frac13 } { x^\frac16 } \\ &=& \dfrac{ x^\frac12 } { x^\frac16 } - \dfrac{ x^\frac13 } { x^\frac16 }\\ &=& x^\frac12 x^{-\frac16 } - x^\frac13 x^{-\frac16} \\ &=& x^\frac36 x^{-\frac16 } - x^\frac13 x^{-\frac16} \\ &=& x^\frac26 - x^\frac13 x^{-\frac16} \\ &=& x^\frac13 - x^\frac13 x^{-\frac16} \\ \dfrac{ x^\frac12-x^\frac13 } { x^\frac16 } &=& x^\frac13 ( 1 - x^{-\frac16} )\\ \end{array} }$

2. (1-(sinx+cosx)2) / 2sinx

$\small{\text{$ \begin{array}{rcl} \dfrac{ 1-[\sin{(x)}+\cos{(x)}]^2 } { 2\sin{(x)} } \\ &=& \dfrac{ 1-[\sin^2{(x)}+2\cdot \sin{(x)} \cos{(x)} +\cos^2{(x)}] } { 2\sin{(x)} } \qquad | \qquad \sin^2{(x)}+ \cos^2{(x)} = 1\\ &=& \dfrac{ 1-[1+2\cdot \sin{(x)} \cos{(x)} ] } { 2\sin{(x)} } \\ &=& \dfrac{ 1-1-2\cdot \sin{(x)} \cos{(x)} } { 2\sin{(x)} } \\ &=& \dfrac{ -2\cdot \sin{(x)} \cos{(x)} } { 2\sin{(x)} } \\ \mathbf{\dfrac{ 1-[\sin{(x)}+\cos{(x)}]^2 } { 2\sin{(x)} } } & \mathbf{=}& \mathbf{-\cos{(x)} } \\ \end{array} $}}$

Factor as indicated

1.  $(2x+1)^{\frac32}x^{\frac12}+(2x+1)^{\frac52}x^{-\frac{1}{2} } = (2x+1)^{\frac32}x^{-\frac{1}{2}}(........)$

$\small{ \begin{array}{rcl} (2x+1)^{\frac32}x^{\frac12}+(2x+1)^{\frac52}x^{-\frac{1}{2} } \\ &=& (2x+1)^{\frac32}x^{\frac{2-1}{2} } + (2x+1)^{\frac{2+3}{2}}x^{-\frac{1}{2} }\\ &=& (2x+1)^{\frac32}x^{\frac22}x^{-\frac{1}{2} } + (2x+1)^{\frac22} (2x+1)^{\frac{3}{2}}x^{-\frac{1}{2} }\\ &=& (2x+1)^{\frac32}x\cdot x^{-\frac{1}{2} } + (2x+1)\cdot (2x+1)^{\frac{3}{2}}x^{-\frac{1}{2} }\\ &=& (2x+1)^{\frac32} x^{-\frac{1}{2} } \left[ x + (2x+1) \right] \\ \mathbf{ (2x+1)^{\frac32}x^{\frac12}+(2x+1)^{\frac52}x^{-\frac{1}{2} } } &\mathbf{=} & \mathbf{(2x+1)^{\frac32} x^{-\frac{1}{2} } \left( 3x+1\right) }\\ \end{array} }$

The position of a squirrel running in a park is given by r⃗   = [(0.280m/s)t+(0.0360m/s2)t2]i+ [ (0.0190m/s3)t3 ] j

a) At 5.01 s , how far is the squirrel from its initial position?

b) At 5.01 s , what is the magnitude of the squirrel's velocity?

c) At 5.01 s , what is the direction (in degrees counterclockwise from +x-axis) of the squirrel's velocity?

Express your answer to three significant figures and include the appropriate units.

$\begin{array}{rcl} \vec{r} &=& \left[ 0.280\ \frac{m}{s}\cdot t+0.0360\ \frac{m}{s^2}\cdot t^2 \right] \vec{i}+ \left[ 0.0190\ \frac{m}{s^3}\cdot t^3 \right] \vec{j}\\ \\ \vec{r} &=& \dbinom{x(t)}{y(t)}= \begin{pmatrix} 0.280\ \frac{m}{s}\cdot t+0.0360\ \frac{m}{s^2}\cdot t^2 \\ 0.0190\ \frac{m}{s^3}\cdot t^3 \end{pmatrix}\\\\\\ \text{a) } t = 5.01~s\\ x(5.01~s)&=&0.280\ \frac{m}{s}\cdot (5.01~s)+0.0360\ \frac{m}{s^2}\cdot (5.01~s)^2 = 2.30640360000~m\\ y(5.01~s)&=&0.0190\ \frac{m}{s^3}\cdot (5.01~s)^3= 2.38927851900~m\\ r(5.01~s)&=&\sqrt{2.30640360000^2~m^2+2.38927851900^2~m^2} = 3.32086576173~m \end{array}$

The squirrel is 3.32 m from its initial position(0,0)

$\begin{array}{rcl} \vec{v} &=& \dbinom{v_x}{v_y} = \begin{pmatrix} \frac{d\ x(t)}{dt} \\ \frac{d\ y(t)}{dt} \end{pmatrix} = \begin{pmatrix} 0.280\ \frac{m}{s}+2\cdot 0.0360\ \frac{m}{s^2}\cdot t \\3\cdot 0.0190\ \frac{m}{s^3}\cdot t^2 \end{pmatrix}\\\\ \text{b) } t = 5.01~s\\ v_x(5.01~s)&=&0.280\ \frac{m}{s}+2\cdot 0.0360\ \frac{m}{s^2}\cdot (5.01~s) = 0.64072~\frac{m}{s}\\ v_y(5.01~s)&=&3\cdot 0.0190\ \frac{m}{s^3}\cdot (5.01~s)^2= 1.4307057~\frac{m}{s}\\ v(5.01~s)&=&\sqrt{0.64072^2~(\frac{m}{s})^2+1.4307057^2~({\frac{m}{s}})^2} = 1.56762269645~\frac{m}{s}\\ \end{array}$

The magnitude of the squirrel's velocity is 1.57 m/s

$\begin{array}{rcl} \text{c) direction } &=& \arctan{ \left( \frac{v_y}{v_x} \right) } \\ &=& \arctan{ \left( \frac{1.4307057~\frac{m}{s}}{0.64072~\frac{m}{s}} \right) } \\ &=& \arctan{ \left(2.23296556998 \right) } \\ &=& 65.8754973481^{\circ} \end{array}$

The direction (in degrees counterclockwise from +x-axis) of the squirrel's velocity is 65.9 degrees

sin(x)+tan(x)=sin(x)(......)

$\sin{(x)} +\tan{(x)} \\ =\sin{(x)} + \frac{ \sin{(x)} } { \cos{(x)}} \\ =\sin{(x)}\cdot \left (1 + \frac{ 1 } { \cos{(x)}} \right) \\$

32C5=32!/(32-5)!5!

$32C5\\ =\binom{32}{5}\\ =\frac{32}{5}\cdot \frac{31}{4}\cdot \frac{30}{3}\cdot \frac{29}{2}\cdot \frac{28}{1}\\ =\frac{24165120}{120}\\ \mathbf{32C5=201376}$

30!*2/20

$30! \cdot \frac{2}{20} = 30! \cdot \frac{1}{10} = 30! \cdot \frac{1}{2\cdot 5}$

Factorisation of 30!

$\small{ \text{Exponent prime Number 2 } = [\frac{30}{2}]+[\frac{30}{2^2}]+[\frac{30}{2^3}]+[\frac{30}{2^4}]=15+7+3+1=26\\ \text{Exponent prime Number 3}= [\frac{30}{3}]+[\frac{30}{3^2}]+[\frac{30}{3^3}]=10+3+1=14\\ \text{Exponent prime Number 5}= [\frac{30}{5}]+[\frac{30}{5^2}]=6+1=7\\ \text{Exponent prime Number 7}= [\frac{30}{7}]=4\\ \text{Exponent prime Number 11}= [\frac{30}{11}]=2\\ \text{Exponent prime Number 13}= [\frac{30}{13}]=2\\ \text{Exponent prime Number 17}= [\frac{30}{17}]=1\\ \text{Exponent prime Number 19}= [\frac{30}{19}]=1\\ \text{Exponent prime Number 23}= [\frac{30}{23}]=1\\ \text{Exponent prime Number 29}= [\frac{30}{29}]=1\\ }\\\\ \mathbf{30! = 2^{26}\cdot 3^{14}\cdot 5^7\cdot 7^4 \cdot 11^2\cdot 13^2 \cdot 17\cdot 19\cdot 23 \cdot 29}$

$\small{ \begin{array}{rcl} 30! \cdot \frac{2}{20} \\ &=& 30! \cdot \frac{1}{2\cdot 5} \\ &=& \frac{2^{26}\cdot 3^{14}\cdot 5^7\cdot 7^4 \cdot 11^2\cdot 13^2 \cdot 17\cdot 19\cdot 23 \cdot 29 }{2\cdot 5 }\\ 30! \cdot \frac{2}{20}&=& 2^{25}\cdot 3^{14}\cdot 5^6\cdot 7^4 \cdot 11^2\cdot 13^2 \cdot 17\cdot 19\cdot 23 \cdot 29 \\ 30! \cdot \frac{2}{20}&=&26525285981219105863630848000000 \end{array} }$

Ann is X years old. David is 3 years younger than ann. Ken is twice as old as Ann. The total of their ages is 25. How old is ann?

$\text{Ann } = x \\ \text{David } = x - 3\\ \text{Ken } = 2 \cdot x \\ \text{Ann + David + Ken } = 25\\ \begin{array}{rcl} x + (x-3) + 2 \cdot x &=& 25\\ 4x - 3 &=& 25\\ 4x &=& 28 \\ \mathbf{x} & \mathbf{=} & \mathbf{7} \end{array}\\ \text{Ann is 7 years old } \\ \text{David is 4 years old }\\ \text{Ken is 14 years old}$

prove that

A) sin60°cos30°-sin30°cos60°=sin30°

$\sin{(60^{\circ})}\cdot \cos{(30^{\circ})} - \cos{(60^{\circ})}\cdot \sin{(30^{\circ})} = \sin{( 60^{\circ} - 30^{\circ} )} = \sin{( 30 ^{\circ} )}$