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This can be re-written as (y - 10)² = x + 6,  so y = 10 ± √(x + 6)

This is a parabola.

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First find the price before sales tax:

P = $20000/1.072

The difference between this and the MSRP must equal nine times the cost of the remote keyless entry (4-way power passenger seat + keyless entry), so:

9*K = P - 17980  where K is cost of keyless entry.

9*K = 20000/1.072 - 17980

$${\mathtt{K}} = {\frac{\left({\frac{{\mathtt{20\,000}}}{{\mathtt{1.072}}}}{\mathtt{\,-\,}}{\mathtt{17\,980}}\right)}{{\mathtt{9}}}} \Rightarrow {\mathtt{K}} = {\mathtt{75.190\: \!713\: \!101\: \!160\: \!862\: \!4}}$$

or K ≈ $75

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If we exclude complex numbers, then the domain of f(x) is from 0 to infinity, because we can't take the square root of a negative number.

For the same reason the domain of g(x) is from negative infinity to 1.

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How to change negative exponents

$$\boxed{\
\left( \dfrac{a}{b} \right)^{-n}
=\left( \dfrac{b}{a} \right)^{n}
\ }$$

Example:

$$a=2 \quad b=1\quad n=3\\\\
\left( \dfrac{2}{1} \right)^{-3}
=2^{-3}
=\left( \dfrac{1}{2} \right)^{3}
=\dfrac{1}{2^3}$$

How do you turn 8 into six through multiplying

$$\textcolor[rgb]{1,0,0}{\not8}\cdot \dfrac {\textcolor[rgb]{0,0,1}{6}}{\textcolor[rgb]{1,0,0}{\not8}}=\textcolor[rgb]{0,0,1}{6}\\$$

5x^2=100

$$\begin{array}{rcl}
5\cdot x^2 &=& 100 \qquad| \quad :5 \\
x^2 &=& 20 \\
x^2 &=& 20 \qquad | \quad \sqrt\\
x &=& \sqrt {20} \\
x &=& \sqrt {4\cdot 5} \\
x &=& \sqrt {4} \cdot \sqrt {5} \qquad | \qaud \sqrt{4} = 2\\
x &=& 2 \cdot \sqrt {5} \\
\end{array}$$

cosA = 3/5

A = 53.130102354156 degrees

This is really more of a division problem

First we should set up an algebraic equation. We need to figure out what number we can multiply 8 by to get 6.

So, $${\mathtt{8}}{\mathtt{\,\times\,}}{\mathtt{x}} = {\mathtt{6}}$$ where six is the number we need to find

Next divide both sides of the equation by 8 so that x is left alone on one side of the equation

$${\frac{\left({\mathtt{8}}{\mathtt{\,\times\,}}{\mathtt{x}}\right)}{{\mathtt{8}}}} = {\frac{\left({\mathtt{6}}\right)}{{\mathtt{8}}}}$$   --->   $${\mathtt{x}} = {\frac{{\mathtt{6}}}{{\mathtt{8}}}}$$

simplify your fraction 

$${\frac{{\mathtt{6}}}{{\mathtt{8}}}} = {\frac{{\mathtt{3}}}{{\mathtt{4}}}}$$

So $${\mathtt{x}} = {\frac{{\mathtt{3}}}{{\mathtt{4}}}}$$

So to turn 8 into 6 through multiplying, you will multiply it by 3/4

Multiply Each side of the equation by 3

$${\mathtt{3}}{\mathtt{\,\times\,}}\left({\mathtt{4}}{w}\right) = {\mathtt{3}}{\mathtt{\,\times\,}}\left({\frac{{\mathtt{2}}}{{\mathtt{3}}}}\right)$$    --->   $${\mathtt{12}}{w} = {\mathtt{2}}$$

Divide both sides by 12 to solve for w

$${\frac{\left({\mathtt{12}}{w}\right)}{{\mathtt{12}}}} = {\frac{\left({\mathtt{2}}\right)}{{\mathtt{12}}}}$$   --->   $${\mathtt{w}} = {\frac{{\mathtt{2}}}{{\mathtt{12}}}}$$

Simplify

$${\frac{{\mathtt{2}}}{{\mathtt{12}}}} = {\frac{{\mathtt{1}}}{{\mathtt{6}}}}$$

Your final anwer is

$${\mathtt{w}} = {\frac{{\mathtt{1}}}{{\mathtt{6}}}}$$

There are lot of videos on this type of thing.

I have never bothered with synthetic division because you cannot use it in as many circumstances as you can with normal algebraic long division.

easy the answer is obviously 2349. DUH!!!!!!!!!!!!!!!!!!!!!!!!

NOTE that your earlier hyperbola had a minus sign between the squared terms

and

The ellipse as a positive sign.   If the numbers on  the botton were the same then it would be a circle. 

A circle is a special case of an elipse because it has a double focii instead of 2 separate focii. 

Effectively a cirlce has one centre and other ellipses have 2 centres.

if it is x^2 it will be a up or down parabola.

if it is y^2 it will be a left or right parabola

if the coefficient of the squared term is positive then it will open in the positive direction

if the coefficient of the squared term is negative then it will open in the negative direction 

Oh and Welcome to the forum Tyler :)

$$\\4x^2 + 7y^2 + 32x - 56y + 148 = 0 \\\\
4x^2+ 32x + 7y^2 - 56y =-148 \\\\
4(x^2+8x)+7(y^2-8y)=-148\\\\
4(x^2+8x+16)+7(y^2-8y+16)=-148+4*16+7*16\\\\
4(x+4)^2+7(y-4)^2=28\\\\
\frac{4(x+4)^2}{28}+\frac{7(y-4)^2}{28}=1\\\\
\frac{(x+4)^2}{7}+\frac{(y-4)^2}{4}=1\\\\
\frac{(x+4)^2}{(\sqrt7)^2}+\frac{(y-4)^2}{2^2}=1\\\\$$

This is an ellipse centre (-4,+4)

The ends of the major axis are   $$(-4-\sqrt7,4) and (-4+\sqrt7,4)$$

Then ends of the minor axis are  (-4,4-2) (-4,4+2)  that is    $$(-4,2) and (-4,6)$$

references

You can just use the divide option the calculator will put the calcul into a fraction

360 degrees

We are not here to do all your homework for you.

And this looks like it is a test!

just divide 9 by 6 and move the decimal 2 to the right.

150%

Need help on all of them pls 

How can i help you my lord?