Answer:
11/13
Step-by-step explanation:
198/234
(198/18)/(234/18)
11/13
Answer:
is there an image or something idk the vid
Quadratic equation: ax² + bx + c =0
x' = [-b+√(b²-4ac)]/2a and x" = [-b-√(b²-4ac)]/2a
6 = x² – 10x ; x² - 10x -6 =0
(a=1, b= - 10 and c = - 6
x' = [10+√(10²+4(1)(-6)]/2(1) and x" = [10-√(10²+4(1)(-6)]/2(1)
x' =5+√31 and x' = 5-√31
Answer:
Option (a) is correct.
The system of equation becomes

Step-by-step explanation:
Given : Equation 
We have to construct a system of equations that can be used to find the roots of the equation 
Consider the given equation 
To construct a system of equation put both sides of the given equation equal to a same variable.
Let the variable be "y", Then the equation 
becomes,
Thus, The system of equation becomes

Option (a) is correct.