I’m assuming what you’re asking here is how to *factor* this expression. For that, let’s rearrange the expression into a more familiar form:
-c^2-4c+21
From here, we’ll factor out a -1 so that we have:
-(c^2+4c-21)
Let’s focus on the quadratic expression inside the parentheses. To find our factors (c + x)(c + y), we’ll need to find two terms x and y that multiply together to make -21 and add together to make 4. It turns out that the numbers -3 and 7 work out perfectly for that purpose (-3 x 7 = -21 and 7 + (-3) = 4), so substituting them in for x and y, we have:
(c + (-3))(c + 7)
(c - 3)(c + 7)
And adding back on the negative from a few steps earlier:
-(c - 3)(c + 7)
Answer:
The answer in the attached figure
Step-by-step explanation:
Let
x------> the abscissa
y-----> the ordinate
we know that
-----> given problem
The domain is ![[-1, 0, 1]](https://tex.z-dn.net/?f=%5B-1%2C%200%2C%201%5D)
so
For 
The point is 
For 
The point is 
For 
The point is 
The answer in the attached figure
Answer:
S(6, 10)
Step-by-step explanation:
1. The Midpoint (in this case, M) will always be halfway between both R and S (or other characters in some cases).
2. As M is only one X value away from R, it is only 1 X value from S as well, but in the other direction.
3. (7, 10) - (1, 0) = (6, 10)
The (7, 10) is the Midpoint Coordinates.
The (1, 0) is the distance from M to R.
The (6, 10) is the coordinates of S.
x=2. -5/6 have a great rest of your day
