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:
The answer to your question is: ∠K = 138°
Step-by-step explanation:
m∠N = 42°
m∠K = ?
The sum of all the internal angles in a quadrangle equals 360°
then
∠N + ∠L + ∠M + ∠K = 360°
∠N = ∠ L and ∠M = ∠K
So, 2 ∠N + 2 ∠K = 360
Substitution
2 (42) + 2 ∠K = 360
84 + 2 ∠K = 360
2 ∠K = 360 - 84
2 ∠K = 276
∠K = 276 / 2
∠K = 138°
Given that ∠B ≅ ∠C.
to prove that the sides AB = AC
This can be done by the method of contradiction.
If possible let AB =AC
Then either AB>AC or AB<AC
Case i: If AB>AC, then by triangle axiom, Angle C > angle B.
But since angle C = angle B, we get AB cannot be greater than AC
Case ii: If AB<AC, then by triangle axiom, Angle C < angle B.
But since angle C = angle B, we get AB cannot be less than AC
Conclusion:
Since AB cannot be greater than AC nor less than AC, we have only one possibility. that is AB =AC
Hence if angle B = angle C it follows that
AB = AC, and AB ≅ AC.
Y = -2/3x + 7.....the slope here is -2/3. A perpendicular line will have a negative reciprocal slope. All tht means is take the original slope, flip it, and change the sign. So we take -2/3....flip it making it -3/2.....change the sign making it 3/2. So ur perpendicular line will have a slope of 3/2.
y = mx + b
slope(m) = 3/2
(-5,6)....x = -5 and y = 6
now sub and find b, the y int
6 = 3/2(-5) + b
6 = -15/2 + b
6 + 15/2 = b
12/2 + 15/2 = b
27/2 = b
so ur perpendicular equation is : y = 3/2x + 27/2 <==
Percent markup=amountmarkedup/original times 100
amountmarkedup=25-13.5.5=11.5
original=13.5
percent markup=11.5/13.5 times 100=0.85 times 100=85% markup