Answer:
12r/t
Step-by-step explanation:
Answer:
B
Step-by-step explanation:
Using the determinant to determine the type of zeros
Given
f(x) = ax² + bx + c ( a ≠ 0 ) ← in standard form, then the discriminant is
Δ = b² - 4ac
• If b² - 4ac > 0 then 2 real and distinct zeros
• If b² - 4ac = 0 then 2 real and equal zeros
• If b² - 4ac < 0 then 2 complex zeros
Given
f(x) = (x - 1)² + 1 ← expand factor and simplify
= x² - 2x + 1 + 1
= x² - 2x + 2 ← in standard form
with a = 1, b = - 2, c = 2, then
b² - 4ac = (- 2)² - (4 × 1 × 2) = 4 - 8 = - 4
Since b² - 4ac < 0 then the zeros are complex
Thus P(x) has no real zeros
Answer:
C
Step-by-step explanation:
-1.5x=-2.55
1.5x=2.55
x=2.55/1.5
x=1.7
1b,
2b,
3a, which grade is this?
Answer:

Step-by-step explanation:
The form of equation of line given in the problem is the point-slope form of a line. That is given by:

We need
and m (denoted by boxes)
is the y coordinate of the first set of points.
The first coordinate pair is (9,7), so
would be 7

Now, the slope (m).
It has formula

So, x_1 = 9
y_1 = 7
x_2 = 4
y_2 = -8
Substituting, we get the slope to be:

Hence, the equation of the line in point-slope is:
