The first step is to convert the depreciation rate into a decimal. \(\displaystyle 13\%=0.13\). Now lets recall the exponential decay model. \(\displaystyle y=C(1-r)^t\), where \(\displaystyle C\) is the starting amount of money, \(\displaystyle r\) is the annual rate of decay, and \(\displaystyle t\) is time (in years). After substituting, we get

\(\displaystyle y=20000(1-0.13)^8\)

\(\displaystyle y=20000(0.87)^8\)

\(\displaystyle y=20000\cdot 0.3282117\)

\(\displaystyle y= 6564.23\)

To convert degrees to radians, we need to remember the following formula.

\(\displaystyle \mbox{degrees}\cdot \frac{\pi}{180}\).

Now lets substitute for degrees.

\(\displaystyle 320\cdot \frac{\pi}{180}=\frac{16\pi}{9}\)

Since these are complementary angles, we can set up the following equation.

\(\displaystyle x^2+4+10x-4=180\)

\(\displaystyle x^2+10x=180\)

\(\displaystyle x^2+10x-180=0\)

Now we will use the quadratic formula to solve for \(\displaystyle x\).

\(\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\)

\(\displaystyle x=\frac{-10\pm \sqrt{10^2-4(1)(-180)}}{2(1)}\)

\(\displaystyle x=\frac{-10\pm \sqrt{100+720}}{2}\)

\(\displaystyle x=\frac{-10\pm \sqrt{820}}{2}\)

\(\displaystyle x=\frac{-10\pm \sqrt{4\cdot 205}}{2}\)

\(\displaystyle x=\frac{-10\pm 2\sqrt{ 205}}{2}\)

\(\displaystyle x=-5\pm \sqrt{ 205}\)

Note, however, that the measure of an angle cannot be negative, so \(\displaystyle x=-5- \sqrt{ 205}\) is not a viable answer. The correct answer, then, is \(\displaystyle x=-5+ \sqrt{ 205}\)

For finding the equation of a line, we will be using point-slope form, which is

\(\displaystyle y-y_0=m(x-x_0)\), where \(\displaystyle m\) is the slope, and \(\displaystyle (x_0, y_0)\) is a point. 

\(\displaystyle m=\frac{y_2-y_1}{x_2-x_1}=\frac{4-3}{-10-0}=\frac{1}{-10}=-\frac{1}{10}\)

We will pick the point \(\displaystyle (0,3)\)

\(\displaystyle y-3=-\frac{1}{10}(x-0)\)

\(\displaystyle y=-\frac{1}{10}x+3\)

If we picked the point \(\displaystyle (-10,4)\)

\(\displaystyle y-4=-\frac{1}{10}(x+10)\)

\(\displaystyle y=-\frac{1}{10}x-1+4\)

\(\displaystyle y=-\frac{1}{10}x+3\)

We get the same result

Since we want a line that is parallel, we will have the same slope as the line \(\displaystyle (-3)\). We can use point slope form to create an equation.

\(\displaystyle y-y_0=m(x-x_0)\), where \(\displaystyle m\) is the slope and \(\displaystyle (x_0, y_0)\) is a point.

\(\displaystyle y-5=-3(x-(-2))\)

\(\displaystyle y-5=-3(x+2)\)

\(\displaystyle y-5=-3x-6\)

\(\displaystyle y=-3x-1\)

We define the variables as \(\displaystyle w = \mbox{width}\) and \(\displaystyle l = \mbox{length} = w + 2\).

We substitute these values into the equation for the area of a rectangle and get \(\displaystyle A_{\mbox{rectangle} }= wl = w(w + 2) = 80\). 

\(\displaystyle w^2 +2w-80=0\)

\(\displaystyle (w - 8)(w + 10) = 0\)

\(\displaystyle w = 8\) or \(\displaystyle w = -10\)

Lengths cannot be negative, so the only correct answer is \(\displaystyle w = 8\). If \(\displaystyle w = 8\), then \(\displaystyle l = 10\). 

Therefore, \(\displaystyle \mbox{perimeter }= 2w + 2l = 16 + 20 = 36\:\mbox{m}\).

Subtract \(\displaystyle 9x\) and \(\displaystyle 72\) from the both sides to get \(\displaystyle -6x = 48\). 

Divide both sides by \(\displaystyle -6\), to get

 \(\displaystyle x=\frac{48}{-6}=-8\)

In order to figure out what the equation of the image is, we need to find the vertex. From the graph we can determine that the vertex is at \(\displaystyle (1,2)\). We can use vertex form to solve for the equation of this graph.

Recall vertex form,

\(\displaystyle y=a(x-h)^2+k\), where \(\displaystyle h\) is the \(\displaystyle x\) coordinate of the vertex, and \(\displaystyle k\) is the \(\displaystyle y\) coordinate of the vertex.

Plugging in our values, we get

\(\displaystyle y=a(x-1)^2+2\)

To solve for \(\displaystyle a\), we need to pick a point on the graph and plug it into our equation.

I will pick \(\displaystyle (-1,10)\).

\(\displaystyle 10=a(-1-1)^2+2\)

\(\displaystyle 10=a(-2)^2+2\)

\(\displaystyle 10=4a+2\)

\(\displaystyle 8=4a\)

\(\displaystyle a=\frac{8}{4}=2\)

Now our equation is

\(\displaystyle y=2(x-1)^2+2\)

Let's expand this,

\(\displaystyle y=2(x^2-2x+1)+2\)

\(\displaystyle y=2x^2-4x+2+2\)

\(\displaystyle y=2x^2-4x+4\)

All we need to do is remember the quotient rule for exponents.

\(\displaystyle \frac{a^n}{a^m}=a^{n-m}\)

We apply this to each term and we get the following.

\(\displaystyle \frac{10a^2b^{-3}c^3}{5bc^2}=2a^2b^{-3-1}c^{3-2}=2a^2b^{-4}c^1=\frac{2a^2c}{b^4}\)

The first step is to cancel out the denominator by multiplying both sides by 7:

\(\displaystyle 7\cdot \frac{x+3}{7}=3\cdot7\)

\(\displaystyle x+3=21\)

Subtract 3 from both sides to get \(\displaystyle x\) by itself:

\(\displaystyle x+3-3=21-3\)

\(\displaystyle x=18\)