heureka

I think it must be sin4(x-3) insted of cos4(x-3) in the last step !

Yes!

Because: $$\int{\cos(4u)}\ du = \frac{1}{4}\sin(4u)$$

Two cars collide with each other. Before the collision, one car (m = 1300 kg) is going north at 30 m/s and the other car (m = 900 kg) is going south at 15 m/s. What is the momentum of the system made up of the two cars after the collision ?

$$\small{\text{ $ \boxed{v'_1 = v'_2 = \dfrac {(m_1\cdot v_1 - m_2\cdot v_2 )}{ (m_1+ m_2 ) } } \quad \begin{array}{rcl} m_1 &=& 1300\ \mathrm{kg} \\ v_1 &=& 30\ \frac{\mathrm{m} }{ \mathrm{s} } \end{array}\quad \begin{array}{rcl} m_2 &=& 900\ \mathrm{kg} \\ v_2 &=& 15\ \frac{\mathrm{m} }{\mathrm{s}} \end{array} $ }}\\\\\\ \small{\text{ $ v'_1 = v'_2 = \dfrac {(1300\cdot 30 - 900\cdot 15 )}{ (1300+ 900 ) } \ \dfrac{\mathrm{m} }{\mathrm{s}} $ }}\\\\ \small{\text{ $ v'_1 = v'_2 = \dfrac { 25500 }{ 2200 } \ \dfrac{ \mathrm{m}}{\mathrm{s}} = 11.5909090909\ \dfrac{ \mathrm{m} }{\mathrm{s} } $ }}\\\\ \small{\text{ $ m_1\cdot v'_1 + m_2\cdot v'_2 = 1300 * 11.591 + 900 * 11.591 = 25500\ \frac{\mathrm{kg}\cdot \mathrm{m}}{\mathrm{s}} $ }}$$

Find the radius of a sphere

$$\boxed{V=\dfrac{4}{3}\pi r^3 \qquad A=4\pi r^2}\\\\ r=\sqrt[3]{\dfrac{3V}{4\pi}} = \dfrac{1}{2}\sqrt{\dfrac{A}{\pi}}$$

Right triangle. Angle c= 90 degrees. Side A=5. Side B=15. Find angle b

$$\tan{(B)}}=\dfrac{15}{5}=3\\\\ B=\tan^{-1}{(3)}=71.5650511771\ensurement{^{\circ}}$$

$$\dfrac{BK}{KD} =\dfrac {1} { \dfrac{a}{b} }\cdot \left(1+\dfrac{d}{c} \right) \\\\\\ \dfrac{a}{b} = \dfrac{1}{2} \qquad \dfrac{d}{c} = \dfrac{2}{1} \\\\ \dfrac{BK}{KD} = 2\cdot \left( 1+2 \right) = 6 \\\\ \boxed{\dfrac{BK}{KD} = \dfrac{6}{1}}$$

In the figure:http://i57.tinypic.com/29ggy1u.jpg,D and E are the points lying on AC and AB respetively such that BE:AE =AD:DC=2:1.BD andCE intersects at K .

Find EK:KC ?

$$\\\small{\text{We set the angles:}}\\ \small{\text{angle ABD $= B$ }}\\ \small{\text{angle ADB $ = D $ }}\\ \small{\text{angle BKE = angle BKC $ = K $}}\\\\ \small{\text{We set the line segments:}}\\ \small{\text{line segment BE $= b$ }}\\ \small{\text{line segment AE $= a$ }}\\ \small{\text{line segment CD $= c$ }}\\ \small{\text{line segment DA $= d$ }}\\ \small{\text{line segment EK $= p$ }}\\ \small{\text{line segment KC $= q$ }}\\$$

I.

$$\dfrac{\sin{(B)}} {d} = \dfrac{\sin{(D)}} {a+b}\quad \Rightarrow \quad \dfrac{\sin{(B)}}{\sin{(D)}}= \dfrac{d}{a+b}$$

II.

$$\small{\text{$\sin{(180-D)} = \sin{D}$}}\\\\ \dfrac{\sin{(D)}} {q} = \dfrac{\sin{(K)}} {c}\quad \Rightarrow \quad q= c \cdot \dfrac{\sin{(D)}}{\sin{(K)}}$$

III.

$$\dfrac{\sin{(K)}} {b} = \dfrac{\sin{(B)}} {p}\quad \Rightarrow \quad p= b \cdot \dfrac{\sin{(B)}}{\sin{(K)}}$$

IV.

$$\dfrac{p}{q} = \dfrac{ b \cdot \dfrac {\sin{(B)}} {\sin{(K)}} } { c \cdot \dfrac{\sin{(D)}}{\sin{(K)}} } } =\dfrac {b} {c} \cdot \dfrac{\sin{(B)}}{\sin{(D)}} } =\dfrac {b} {c} \cdot \left(\dfrac{d}{a+b} \right) \\\\ = \dfrac {1} { \dfrac{c}{d} }\cdot \left(\dfrac {1}{ \dfrac{a+b}{b} } \right) = \dfrac {1} { \dfrac{c}{d} }\cdot \left(\dfrac {1}{ \dfrac{a}{b}+\dfrac{b}{b} } \right) = \dfrac {1} { \dfrac{c}{d} }\cdot \left(\dfrac {1}{ \dfrac{a}{b}+1 } \right)\\\\\\ \dfrac{c}{d} = \dfrac{a}{b} = \dfrac{1}{2} \\\\ \dfrac{p}{q} =2\cdot \left(\dfrac {1}{ \dfrac{1}{2}+1 } \right) = 2\cdot \left(\dfrac {1}{ \dfrac{3}{2} }\right) =2\cdot \dfrac{2}{3} \\\\ \boxed{\dfrac{p}{q} = \dfrac{4}{3}}$$

EK:KC = 4 : 3

What is the wavelength in meters of ultraviolet light with ν = 2.50 x 1015 s-1 ?

$$\\\small{\text{Light moves with a speed $ c \,=\, 299792458\ \frac{\text{m}}{\text{s}} $ }}\\ \small{\text{We denote wavelength by $\lambda = \mathrm{wavelength}$}}\\ \small{\text{We denote frequency by $\nu=\mathrm{frequency} $}}\\ \small{\text{Frequency is measured in Hertz $ = Hz = s^{-1}.$}}\\ \\ \small{\text{ $ \boxed{ \lambda =\dfrac{c}{\nu} }\qquad \nu = 2.50 \cdot 10^{15}\ s^{-1} $ }}\\\\ \small{\text{ $ \lambda = \dfrac{2.99792458\cdot 10^8\ \dfrac{m}{s} } {2.50 \cdot 10^{15}\ s^{-1} } $ }}\\\\ \small{\text{ $ = 1.199169832 \cdot 10^{-7}\ m $ }}\\ \small{\text{ $ = 119.9169832 \cdot 10^{-9}\ m $ }}\\ \small{\text{ $ = 119.9\ nm $ }}$$

A ball rolls on the top of a stairway with a horizontal velocity of 2.0 m/s.

The steps are each 20.0 cm high and 20.0 cm wide.

Which step will the ball hit first ?

$$\\\small{\text{horizontal move: $\boxed{ x=v_0\cdot t} \; \Rightarrow \; t =\dfrac{x}{v_0} \qquad v_0=2\ \frac{m}{s}$ }}\\\\ \small{\text{vertical move: $\boxed{ y=\frac{g}{2}\cdot t^2 } \quad t =\dfrac{x}{v_0} \; \Rightarrow \; y(x)=\dfrac{g}{2}\cdot \dfrac{x^2}{v_0^2} \qquad g=10\ \frac{m}{s^2}$ }}\\\\ \small{\text{$x=y(x) ?:\qquad x=\dfrac{g}{2}\cdot \dfrac{x^2}{v_0^2} \; \Rightarrow \; x=\dfrac{2\cdot v_0^2 }{g} $ }}\\ \small{\text{ $ x=\dfrac{2\cdot (2\cdot \frac{m}{s})^2 }{10\cdot \frac{m}{s^2}} \qquad x=\frac{8}{10}\cdot m = 0.8\ m \qquad 0.8\ m = 4 * 0.2\ m $ }}$$

The ball will hit first step 4.

Use the definition of the derivative to find the derivative $$\small{\text{$f(x)=x^2-2$}}$$

$$\\\lim \limits_{h \to 0} \left[ \frac{f\left( x+h \right) - f\left( x \right)}{h} \right] \small{\text{$\quad f(x+h) = (x+h)^2-2 \qquad f(x) = x^2-2 $ }}\\\\ \lim \limits_{h \to 0} \left[ \frac{\left( (x+h)^2-2 \right) - \left( x^2-2 \right)}{h} \right] = \lim \limits_{h \to 0} \left[ \frac{ x^2+2*x*h+h^2-2 - x^2+2}{h} \right] = \lim \limits_{h \to 0} \left[ \frac{ 2*x*h+h^2}{h} \right] \\\\ = \lim \limits_{h \to 0} \left[ \frac{ 2*x*h}{h}+\frac{ h^2}{h} \right] = \lim \limits_{h \to 0} \left[ 2*x+ h \right] = 2x$$

I is a parabola with (1,-18) as its vertex and opening upwards touches (4,0) which is one of the roots , a straight line y=-x+11 pass through the parabola at P and Q , find the coordinates of P and Q.

$$\boxed{Parabola: y=a(x-x_v)^2+y_v }\qquad x_v=1 \quad y_v=-18\\ a? \quad x = 4 \quad y = 0 \\ 0=a(4-1)^2-18\\\\ a=\dfrac{18}{3^2}=\frac{18}{9}=2\\\\ \boxed{Parabola: y=2(x-1)^2-18 \quad line: y=-x+11}$$

The cut:

$$y=-x+11\\ y+x=11 \\ x = 11-y \\ x-1 = 10-y$$

We set x-1 = 10 - y in Parabola:

$$y=2(x-1)^2-18 \\ y=2(10-y)^2-18 \\ y=2(100-20y+y^2)-18\\ y=200-40y+2y^2-18\\ \boxed{2y^2-41y+182=0}\\ y_{1,2}=\dfrac{41 \pm \sqrt{41^2-4*2*182} }{2*2}=\dfrac{41 \pm \sqrt{225} }{4}=\dfrac{41 \pm 15 }{4}\\\\ y_1=\frac{56}{4}=14 \qquad y_2 = \frac{26}{4}=6.5\\ x_1 = 11-y_1 =11-14=-3 \qquad x_2=11-y_2=11-6.5=4.5$$

Point P = (-3, 14) and Point Q = (4.5, 6.5)