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.
Steve set the equation equal to 50 which is the entire length of JL. instead he should've set the equation equal to 25, since it's half of 50 and would represent the midpoint. he should've said 2x + 5 = 25
Answer:
976 cm^2.
Step-by-step explanation:
The surface area = 2 * area of the triangular side + 2 * area of the rectangular sides + area of the rectangular base.
= 2 + 1/2 * 24 * 14 + 2 * 25 * 10 + 14*10
= 336 + 500 + 140
= 976 cm^2.
Answer:
767.76 mL
Step-by-step explanation:
625.05 + `142.71 = 767.76
