Rom

Look at a regular polygon of n sides.

The area is given by

$$A_n=\dfrac s 2 a n$$

s is the length of a side

a is the length of the apothem

n is the number of sides

I leave it to you to work out that for a given radius r

$$\dfrac s 2 = r \sin\left(\dfrac \pi n\right)$$

$$a=r \cos\left(\dfrac \pi n\right)$$

so

$$A_n=r^2 n\sin\left(\dfrac \pi n\right)\cos\left(\dfrac \pi n\right)$$

You can show that

$$\displaystyle{\lim_{n\to\infty}}n \sin\left(\frac \pi n\right)\cos\left(\frac \pi n\right)=\pi$$

so

$$\displaystyle{\lim_{n\to \infty}}A_n=\pi r^2$$

This limit corresponds to a regular polygon of increasingly many but smaller sides which taken to infinity represents a circle and thus you get

$$A_{circle}=\pi r^2$$

this question has been answered

$$P V = n R T$$

$$V1 = \dfrac {n R T}{P}$$

$$V2 = \dfrac {n R (2 T)}{(2P)}=\dfrac{n R T}{P}=V1$$

$$V2=V1$$

$${\mathtt{36}}{\mathtt{\,\times\,}}{\mathtt{75}} = {\mathtt{2\,700}}$$

$$\lambda_1=1+2\imath$$

$$\begin{bmatrix}2-\lambda_1 &-5 \\ 1 &-\lambda_1 \end{bmatrix} \Rightarrow \begin{bmatrix}1-2\imath &-5 \\ 1 &-1-2\imath\end{bmatrix}\Rightarrow$$

$$\begin{bmatrix}1 &-1-2\imath\\ 1-2\imath &-5 \end{bmatrix}\Rightarrow
\begin{bmatrix}1 &-1-2\imath \\0 &0\end{bmatrix}$$

$$x-(1+2\imath)y=0$$

$$v_1=\begin{bmatrix}1 \\1+2\imath\end{bmatrix}$$

$$\lambda_2=1-2\imath$$

$$\begin{bmatrix}2-\lambda_2 &-5 \\ 1 &-\lambda_2 \end{bmatrix} \Rightarrow \begin{bmatrix}1+2\imath &-5 \\ 1 &-1+2\imath\end{bmatrix}\Rightarrow$$

$$\begin{bmatrix}1 &-1+2\imath\\ 1+2\imath &-5 \end{bmatrix}\Rightarrow
\begin{bmatrix}1 &-1+2\imath \\0 &0\end{bmatrix}$$

$$x-(1-2\imath)y=0$$

$$v_2=\begin{bmatrix}1 \\1-2\imath\end{bmatrix}$$

the tangent function

The square root of -1220 isn't defined on the real numbers.

On the complex numbers it is given as

$$\sqrt{-1220}=\imath \sqrt{1220}=2\imath \sqrt{305}$$

where

 $$\imath = \sqrt{-1}$$

The easiest way to answer this is to look at the energy at the top and bottom and note that since there is no friction this value must be the same.  Energy is conserved.

E = PE + KE ; potential energy + kinetic energy

PE = m g h  ; mass x gravity acceleration (9.8m/s^2) x height

KE = 1/2 m v^2 ; v is velocity

Let the bottom of the slide be at height 0, and the top at height 32.3m.

At the top PE = 32.3 mg, at the bottom PE = 0.

At the top the kinetic energy is 0, at the bottom it is 1/2 m v^2

so 32.3 mg = 1/2 m v^2

32.3g = 1/2 v^2

633.08 = v^2

v=25.16 m/s

Wild Roger Rollercoaster (mass of 75 kg) is travelling on a new ride called the Riddler's Revenge. On a 5.1 m hill he is moving along at 32.6 m/s, then slows down to 24.6 m/s on the second hill. How high is the second hill?

The way to solve this is by noting that energy on this ideal roller coaster is conserved.

The total energy of the car is the sum of its potential and kinetic energies.

The potential energy at a point at heigh h is given by  $$m g h$$

m - mass of the car

g - the gravitational accleration $$9.8 m/s^2$$ 

h - the height

The kinetic energy at a given speed is given by $$\frac 1 2 m v^2$$

m - the mass

v - the speed

At the first point the total energy is 

$$(75)(9.8)(5.1)+\frac 1 2 (75)(32.6)^2=43602 Joules$$

At the second point the car has this same total energy but less of it is kinetic and more of it is potential.

$$43602=(75)(9.8)h_2+\frac 1 2 (75)(24.6)^2\\
43602-22693.5=735 h_2\\
h_2=28.45m$$