Option A. All the real values of x where x < -1
Procedure
Solve the inequality:
(x -3)(x+1)>0
That happens in two cases.
1) When both factors >0
x-3>0 and x+1>0
x>3 and x >-1
The intersection is x >3
2) When both factors <0
x-3<0 and x+1<0
x<3 and x<-1
the intersection is x<-1.
We have obtained that the function is positive for the intervals x < -1 and x > 3. But in one of those intervals the function is decresing and in the other is increasing.
You can recognize that the function given is a parabola and, because the coefficient of the quadratic term is positive, the parabola opens upward. Then the function is decreasing in the first interval and increasing in the second interval.
Step-by-step explanation:
Subtracting equation (2) from equation (1)
Substituting d = - 3 in equation (1), we find:
Hence, first term is zero.
Answer:
Range: (-∞,∞)
Domain: (-∞,∞)
Step-by-step explanation:
Remember that if there are arrows that the function continues.
Answer:
yes jess does have enough for her sandbox
Step-by-step explanation:
I did this test to and jess does indeed have enough for the sandbox
X - y = 5,
x = 5 + y
substitute (x = 5 + y) into the first equation,
5 + y + 2y = -1
3y = -6
y = -2,
substitute y = -2 into (x - y = 5),
x + 2 = 5
x = 3
Hence, x = 3, y = -2
