Which of the following are exterior angles , check all that apply?
1 answer:
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From the problem, the vertex = (0, 0) and the focus = (0, 3)
From the attached graphic, the equation can be expressed as:
(x -h)^2 = 4p (y -k)
where (h, k) are the (x, y) values of the vertex (0, 0)
The "p" value is the difference between the "y" value of the focus and the "y" value of the vertex.
p = 3 -0
p = 3
So, we form the equation
(x -0)^2 = 4 * 3 (y -0)
x^2 = 12y
To put this in proper quadratic equation form, we divide both sides by 12
y = x^2 / 12
Source:
http://www.1728.org/quadr4.htm
From the attached graphic, the equation can be expressed as:
(x -h)^2 = 4p (y -k)
where (h, k) are the (x, y) values of the vertex (0, 0)
The "p" value is the difference between the "y" value of the focus and the "y" value of the vertex.
p = 3 -0
p = 3
So, we form the equation
(x -0)^2 = 4 * 3 (y -0)
x^2 = 12y
To put this in proper quadratic equation form, we divide both sides by 12
y = x^2 / 12
Source:
http://www.1728.org/quadr4.htm
Answer:
Step-by-step explanation:
The given system is:
Since I prefer to use smaller numbers I'm going to divide both sides of the first equation by 3 and both sides of the equation equation by 6.
This gives me the system:
We could solve the first equation for and replace the second
with that.
Let's do that.
Subtract on both sides:
So we are replacing the second in the second equation with
which gives us:
Now recall the first equation we arranged so that was the subject. I'm referring to
.
We can now find given that
using the equation
.
Let's do that.
with
:
So the solution is (8,-1).
We can check this point by plugging it into both equations.
If both equations render true for that point, then we have verify the solution.
Let's try it.
with
:
is a true equation so the "solution" looks promising still.
with
:
is also true so the solution has been verified since both equations render true for that point.