Answer:
fglbef bgkfv dfg hti gri efgkrtgm ruigermfui5ygmruf thiptofktuigkrlv
Step-by-step explanation:
We have that
<span>A' (2,1)
C(-2,2)-------> </span>Using the transformation-----> C' (-2+2,2+1)----> C' (0,3)
with
A' (2,1) and C' (0,3)
find the distance <span>C'A'
d=</span>√[(y2-y1)²+(x2-x1)²]----> d=√[(3-1)²+(0-2)²]----> d=√8----> 2√2 units
the answer is
the distance C'A' is 2√2 units
Answer: 1) (-∞, -6) U (-3, ∞)
2) (-∞, -4) U (2, ∞)
3) (-∞, -3) U (8, ∞)
<u>Step-by-step explanation:</u>
Find the zeros. Since the a-value is positive, the curve will be positive to the left of the leftmost zero and to the right of the rightmost zero. + - +
←---|----|--→
1) y = x² + 9x + 18
y = (x + 3)(x + 6)
0 = (x + 3)(x + 6)
0 = x + 3 0 = x + 6 + -- +
x = -3 x = -6 ←------|-----------|--------→
-6 -3
Positive Interval: (-∞, -6) U (-3, ∞)
2) y = x² + 2x - 8
y = (x + 4)(x - 2)
0 = (x + 4)(x - 2)
0 = x + 4 0 = x - 2 + -- +
x = -4 x = 2 ←------|-----------|--------→
-4 2
Positive Interval: (-∞, -4) U (2, ∞)
3) y = x² - 5x - 24
y = (x + 3)(x - 8)
0 = (x + 3)(x - 8)
0 = x + 3 0 = x - 8 + -- +
x = -3 x = 8 ←------|-----------|--------→
-3 8
Positive Interval: (-∞, -3) U (8, ∞)
F (x) = 2 × 2 (- 2)
f(x) = 2 × - 4
f(x) = - 8
