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.
<h2>
Answer:</h2>
<u>The expression is</u><u> 5(40+80)</u>
<h2>
Step-by-step explanation:</h2>
The sum of 40 and 80 can be written as
(40+80)
and when we take the 5 times of the sum
we can write as
5(40+80)
33.3% is the answer, I think
20/60 ---> 1/3, covert this to a percentage and it equals 33.3%
base = 3*3 = 9 sq. in.
side = 1/2 *3*6 = 9 sq. in.
4 sides x 9 = 36 + 9 (base) = 45 in^2
Answer is A
Answer:
uhh
Step-by-step explanation:
