Y = |x² - 3x + 1|
y = x - 1
|x² - 3x + 1| = x - 1
|x² - 3x + 1| = ±1(x - 1)
|x² - 3x + 1| = 1(x - 1) or |x² - 3x + 1| = -1(x - 1)
|x² - 3x + 1| = 1(x) - 1(1) or |x² - 3x + 1| = -1(x) + 1(1)
|x² - 3x + 1| = x - 1 or |x² - 3x + 1| = -x + 1
x² - 3x + 1 = x - 1 or x² - 3x + 1 = -x + 1
- x - x + x + x
x² - 4x + 1 = -1 or x² - 2x + 1 = 1
+ 1 + 1 - 1 - 1
x² - 4x + 1 = 0 or x² - 2x + 0 = 0
x = -(-4) ± √((-4)² - 4(1)(1)) or x = -(-2) ± √((-2)² - 4(1)(0))
2(1) 2(1)
x = 4 ± √(16 - 4) or x = 2 ± √(4 - 0)
2 2
x = 4 ± √(12) or x = 2 ± √(4)
2 2
x = 4 ± 2√(3) or x = 2 ± 2
2 2
x = 2 ± √(3) or x = 1 ± 1
x = 2 + √(3) or x = 2 - √(3) or x = 1 + 1 or x = 1 - 1
x = 2 or x = 0
y = x - 1 or y = x - 1 or y = x - 1 or y = x - 1
y = (2 + √(3)) - 1 or y = (2 - √(3)) - 1 or y = 2 - 1 or y = 0 - 1
y = 2 - 1 + √(3) or y = 2 - 1 - √(3) or y = 1 or y = -1
y = 1 + √(3) or y = 1 - √(3) (x, y) = (2, 1) or (x, y) = (0, -1)
(x, y) = (2 ± √(3), 1 ± √(3))
The solution (0, -1) can be made by one function (y = x - 1) while the solution (2 ± √(3), 1 ± √(3)) can be made by another function (y = |x² - 3x + 1|). So the solution (2, 1) can be made by both functions, making the two solutions equal.
Answer:
I need how much is it per one mile <em>or</em> how much does is cost every 5 mins at least
The approximate middle value
Answer: 0.4
Step-by-step explanation:
i think
Answer:
x<5
Open circle at 5 going to the left
x >1
Open circle at 1 going to the right
Step-by-step explanation:
7x -19 < 16
Add 19 to each side
7x -19 +19< 16+19
7x< 35
Divide by 7
7x/7 < 35/7
x<5
Open circle at 5 going to the left
9+3x>12
Subtract 9 from each side
9-9 +3x >12-9
3x >3
3x/3 >3/3
x >1
Open circle at 1 going to the right
