f(x) > 0 - I and II quadrant
f(x) < 0 - III and IV quadrant
Look at the picture.
A. F(x) < 0 on the interval x < 0. TRUE
B. F (x) > 0 on the interval x <0. FALSE
C. F (x) < 0 on the interval 0 < x < 1. TRUE
D. F (x) > 0 on the interval 0 < x < 1. FALSE
E. F (x) < 0 on the interval 1 < x < 3. FALSE
F. F (x) > 0 on the interval 1 < x < 3. TRUE
G. F (x) < 0 on the interval x > 3. TRUE
H. F (x) > 0 on the interval x > 3. FALSE
Hey, using y2-y1/x2-x1 you know that slope has to be -1/4
Answer:
See below.
Step-by-step explanation:
The given equation is in vertex form. We can convert it to standard form as follows:
y = a(x - h)^2 + k
y = a(x^2 - 2hx + h^2) + k
y = ax^2 - 2ahx + (ah^2 + k) (answer).
Answer:
First Quartile Q1 = 9.9525
Step-by-step explanation:
For a standard normal distribution,
First quartile Q1 = μ - 0.675 σ
From the question mean μ = 10.56
Standard deviation σ = 0.9
Plugging these values into the first quartile equation, we have;
Q1 = 10.56 -0.675(0.9)
Q1 = 10.56 - 0.6075
Q1 = 9.9525
It is already simplified to the furthest extent