Answer:
3 favorable outcomes
There are three paths that end up with spinning the same color twice
Answer:
$y=\frac{-2}{9}x+\frac{61}$
Step-by-step explanation:
For two lines to be parallel to each other, it means they have to have the same slope. So, the question is really asking for the line through $(8, 5)$ that has the same slope as the line shown.
We can start by finding the slope of the line shown. Recall that given two points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$, the slope of the line between them is $\frac{y_{2}-y_{1}}{x_{2}-x{1}}$, or more commonly remembered as rise over run. Plugging in the points $(-3, 1)$ and $(6, -1)$ into the equation for the slope, we see that the slope of the line shown is $\frac{-1-1}{6-(-3)}=\frac{-2}{9}$.
We recall that the point-slope form of a line is such that given a point $(x_{1}, y_{1})$ and the slope $m$, the equation for a line with that slope through the given point is $y-y_{1}=m(x-x_{1})$(Note: divide by $x-x_{1}$. What does that remind you of?). Plugging in our point and slope, we have that the equation for the line we want is $y-5=\frac{-2}{9}(x-8)$. Expanding the right-hand side gives $y-5=\frac{-2}{9}x+\frac{16}{9}$.
Adding $5$ to both sides gives the desired $\boxed{y=\frac{-2}{9}x+\frac{61}}$
Answer: The solution is the set of all real numbers (there are infinitely many solutions).
The reason why this is the case is because |x| is never negative. The smallest it can get is 0, which is larger than -3. That applies to |x-2| as well. So |x-2| is ALWAYS larger than -3 no matter what you pick for x. The smallest |x-2| can get is 0 and that happens when x = 2. Otherwise, the result is some positive value which is larger than -3.
So that's why |x-2| > -3 has infinitely many solutions. We can replace x with any real number we want, and the inequality would be true.
Substitute for the values of x and then, solve the equation so using one another.
The additional zeros are the complex conjugates of the two complex zeros. That gives the required 6. I worked out the polynomial in case you want it