Answer:
(A)
Step-by-step explanation:
The standard form of the equation of a circle is given as:

Simplifying the above given equation, we get




which is the required general form of the equation.
Hence, option A is correct.
2x - 4y + 1z = 11 ⇒ 2x - 4y + 1z = 11
1x + 2y + 3z = 9 ⇒ <u>1x + 2y + 3z = 9</u>
3x + 5z = 12 1x - 2y - 2z = 2
1x - 2y - 2z = 2
2x - 4y + 1z = 11 <u>-2x + 2y - 2z = -3</u>
1x + 2y + 3z = 9 ⇒ 1x + 2y + 3z = 9 x - 4z = 1
3x + 5z = 12 ⇒ <u>3x + 5z = 12</u> x - 4z + 4z = 1 + 4z
-2x + 2y - 2z = -3 x = 1 + 4z
<u /> 1 + 4z - 2y - 2z = 2<u />
1 - 2y + 4z - 2z = 2
1 - 2y + 2z = 2
<u>- 1 - 1</u>
-2y + 2z = 1
-2y + 2z - 2z = 1 - 2z
<u>-2y</u> = <u>1 - 2z</u>
-2 -2
y = -0.5 + z
x + 2(-0.5 + z) - 2z = 2
x - 1 + z - 2z + 2 = 2
x - 1 + z = 2
<u> + 1 + 1</u>
x + z = 3
x - x + z = 3 - x
z = 3 - x
<u />
A perpendicular bisector cuts a line segment directly in half. a property of perpendicular bisectors is that any point on the line of the bisector will be equidistant (the same distance from) both end points.
Ex, If point X was 5 ft away from one endpoint, it would be 5 feet away from the other as well.
Answer:
(b) 5k+1
Step-by-step explanation:
https://www.symbolab.com/solver/step-by-step/simplify%20%5Cfrac%7B30k%5E%7B2%7D%2B6k%7D%7B5k%2Bk%7D