Answer:
Real values of x where x < -1
Step-by-step explanation:
Above the x-axis, the function is positive.
The function is decreasing when the gradient is negative.
The function has a positive
coefficient, therefore the vertex is a local minimum;
This means the gradients are negative before the vertex and positive after it;
To meet the conditions therefore, the function must be before the vertex and above the x-axis;
This will be anywhere before the x-intercept at x = -1;
Hence it is when x < -1.
Answer: I'm not sure if I'm right but i think it is A.) or maybe C.) I really don't know it could be any So please don't get mad at me if you get it wrong because of me
Step-by-step explanation:
Answer:
x1 = t, x2 = -t and x3 = 0
Step-by-step explanation:
Given the system of equation
x1 + x2 + x3 = 0 .... 1
x1 + x2 + 9x3 = 0 .... 2
Subtract both equation
x3 - 9x3 = 0
-8x3 = 0
x3 = 0
Substitute x3 = 0 into equation 1
x1 + x2 + 0 = 0
x1+x2 = 0
x1 = -x2
Let t = x1
t = -x2
x2 = -t
Hence x1 = t, x2 = -t and x3 = 0
Answer:
x
Step-by-step explanation:
