Answer:
3rd Option is correct.
Step-by-step explanation:
Given Equation:
x² - 16x + 12 = 0
First We need to find solution of the given equation.
x² - 16x + 12 = 0
here, a = 1 , b = -16 & c = 12






Now,
Option 1).
( x - 8 )² = 144
x - 8 = ±√144
x - 8 = ±12
x = 8 + 12 = 20 and x = 8 - 12 = -4
Thus, This is not correct Option.
Option 2).
( x - 4 )² = 4
x - 4 = ±√4
x - 4 = ±2
x = 4 + 2 = 6 and x = 4 - 2 = 2
Thus, This is not correct Option.
Option 3).
( x - 8 )² = 52
x - 8 = ±√52
x - 8 = ±2√13
x = 8 + 2√13 and x = 8 - 2√13
Thus, This is correct Option.
Option 4).
( x - 4 )² = 16
x - 4 = ±√116
x - 4 = ±4
x = 4 + 4 = 8 and x = 4 - 4 = 0
Thus, This is not correct Option.
Therefore, 3rd Option is correct.
Answer:
o.2222
Step-by-step explanation:
just divide one by five
hope that helps
Step-by-step explanation:
First you find out the area of the square.
Then you find the area of the circle.
After this you add the Two area together.
Answer:
y=-2x-4
Step-by-step explanation:
Given that,
Slope, m = -2
It passes through the point (1,-6).
We need to find the equation of line that passes throgh the given point. It can be calculated as:

Put y₁ = -6, x₁ = 1 and m = -2

Hence, the equation of line is y=-2x-4.