Quadratic f(x) = (x -h)² +k has vertex (h, k) and axis of symmetry x=h. When k is negative, the number of real solutions is 2, because both branches of the function cross the x-axis.
In your equation, h = -3 and k = -8.
Axis of symmetry: x = -3
Vertex: (-3, -8)
Number of real solutions: 2
Answer:
4
Step-by-step explanation:
Given
← evaluate the denominator
= 
← perform the division
= 4
Answer:
n < - 3 or n > - 2
Step-by-step explanation:
Inequalities of the type | x | > a , have solutions of the form
x < - a or x > a
Then
2n + 5 < - 1 or 2n + 5 > 1
Solve both inequalities
2n + 5 < - 1 ( subtract 5 from both sides )
2n < - 6 ( divide both sides by 2 )
n < - 3
OR
2n + 5 > 1 ( subtract 5 from both sides )
2n > - 4 ( divide both sides by 2 )
n > - 2
Solution is n < - 3 or n > - 2
Answer:
1.5
Step-by-step explanation:
Answer:
What are the options?
Step-by-step explanation:
