Answer:
![d = \sqrt{(x2 - x1) + (y2 - y1)} \\ d = \sqrt{ (- 3 + 2) + (1 + - 4)} \\ d = \sqrt{ - 1 + - 3 } \\ d = \sqrt{ - 4 } \\ d = - 2](https://tex.z-dn.net/?f=d%20%3D%20%20%5Csqrt%7B%28x2%20-%20x1%29%20%2B%20%28y2%20-%20y1%29%7D%20%20%5C%5C%20d%20%3D%20%20%5Csqrt%7B%20%28-%203%20%2B%202%29%20%2B%20%281%20%2B%20%20-%204%29%7D%20%20%5C%5C%20d%20%3D%20%20%5Csqrt%7B%20-%201%20%2B%20%20-%203%20%7D%20%20%5C%5C%20d%20%3D%20%20%5Csqrt%7B%20-%204%20%7D%20%20%5C%5C%20d%20%3D%20%20-%202)
Step-by-step explanation:
<h2><em><u>hope </u></em><em><u>it </u></em><em><u>helps</u></em><em><u> you</u></em><em><u><</u></em><em><u>3</u></em></h2>
Answer:
![y=(x-3)^2-6](https://tex.z-dn.net/?f=y%3D%28x-3%29%5E2-6)
Step-by-step explanation:
![y=x^2-6x+3](https://tex.z-dn.net/?f=y%3Dx%5E2-6x%2B3)
This is written in the standard form of a quadratic function:
![y=ax^2+bx+c](https://tex.z-dn.net/?f=y%3Dax%5E2%2Bbx%2Bc)
where:
- ax² → quadratic term
- bx → linear term
- c → constant
You need to convert this to vertex form:
![y=a(x-h)^2+k](https://tex.z-dn.net/?f=y%3Da%28x-h%29%5E2%2Bk)
where:
To find the vertex form, you need to find the vertex. For this, use the equation for axis of symmetry, since this line passes through the vertex:
![x=-\frac{b}{2a}](https://tex.z-dn.net/?f=x%3D-%5Cfrac%7Bb%7D%7B2a%7D)
Using your original equation, identify the a, b, and c terms:
![a=1\\\\b=-6\\\\c=3](https://tex.z-dn.net/?f=a%3D1%5C%5C%5C%5Cb%3D-6%5C%5C%5C%5Cc%3D3)
Insert the known values into the equation:
![x=-\frac{(-6)}{2(1)}](https://tex.z-dn.net/?f=x%3D-%5Cfrac%7B%28-6%29%7D%7B2%281%29%7D)
Simplify. Two negatives make a positive:
![x=\frac{6}{2} =3](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B6%7D%7B2%7D%20%3D3)
X is equal to 3 (3,y). Insert the value of x into the standard form equation and solve for y:
![y=3^2-6(3)+3](https://tex.z-dn.net/?f=y%3D3%5E2-6%283%29%2B3)
Simplify using PEMDAS:
![y=9-18+3\\\\y=-9+3\\\\y=-6](https://tex.z-dn.net/?f=y%3D9-18%2B3%5C%5C%5C%5Cy%3D-9%2B3%5C%5C%5C%5Cy%3D-6)
The value of y is -6 (3,-6). Insert these values into the vertex form:
![(3_{h},-6_{k})\\\\y=a(x-3)^2+(-6)](https://tex.z-dn.net/?f=%283_%7Bh%7D%2C-6_%7Bk%7D%29%5C%5C%5C%5Cy%3Da%28x-3%29%5E2%2B%28-6%29)
Insert the value of a and simplify:
![y=(x-3)^2-6](https://tex.z-dn.net/?f=y%3D%28x-3%29%5E2-6)
:Done
Answer:
(C)Angle MRN and Angle NRO
Step-by-step explanation:
A diagram of the given problem is drawn and attached below.
From the diagram:
Option A: Angle PRL and Angle LRM = Angle PRM which is not 180 degrees.
Option B: Angle ORP and Angle MRN are vertically opposite angles and are therefore not linear pairs.
Option C: Angle MRN and Angle NRO are on the line ORM and therefore forms a linear pair.
<u>Therefore, the correct option is C.</u>
It's the factor theorem.
P(-4)
Equal to 0