Let's evaluate ~
The given expression is :
plug the value of x as 3
Therefore, the required value is 20
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:
x=3, z=-1, y=6
Step-by-step explanation:
you need to make the coefficients of each variable equal to each other then subtract 2 equations.
this one: multiply -4 in 1st
4x+4y+4z=32
-4x+4y+5z=7
_____________
8x+0-z=25
we have 2 equations with 2 variables now
2x+2z=4
8x-z=25
Given:
To find:
The quadrant in which 
lie.
Solution:
Quadrant concept:
In Quadrant I, all trigonometric ratios are positive.
In Quadrant II, only 
and 
are positive.
In Quadrant III, only 
and 
are positive.
In Quadrant IV, only 
and 
are positive.
We have,
Here, is negative and is also negative. It is possible, if 
lies in the Quadrant IV.
Therefore, the correct option is D.
