- Periodic
- Circular
- Sinusoidal
- Systemic
- Trigonometric

The solution is #4 because all of the others are descriptions of Trigonometric Functions

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# Which One Does Not Belong – Trigonometric

# Which One Does Not Belong – Inequalities

# Which One Does Not Belong – Misc

# Which One Does Not Belong – Misc

# Which One Does Not Belong – Quadratic and Inequality

# Which One Does Not Belong – Quadratics

# Which One Does Not Belong – Exponential

# Which One Does Not Belong – Geometry

# Which One Does Not Belong – Geometry

# Which One Does Not Belong – Misc

- Periodic
- Circular
- Sinusoidal
- Systemic
- Trigonometric

The solution is #4 because all of the others are descriptions of Trigonometric Functions

- 5x – 4 > 11
- 7 – 12x > 19
- -x + 10 > 20
- 10 + 2x < x
- 12 – 2x < -5x

The solution is #1 because all of the others simplify to be x <

- 2x
^{2}(x – 1) – x - 4x – 12(2x
^{2}+ 2) – 2x^{2} - 2x – 5x(5x
^{4}– 4) - 5x
^{4}(x – 3) + 2x - 2x – x(5x
^{4 }-1)

The solution is #2 because all of the other are odd degree polynomials

- f(g(x))
- f ° g (x)
- g(f(x))
- g(x) ° f(x)
- g(x)f(x)

The solution is #5 because all of the others are forms of function composition

- y > 2x
^{2}– 3x - y < -2x
^{2}– 3x - 2x
^{2}+ y < 5x - 3x + y > x
^{2} - y > -x
^{2}+ 5x

The solution is #5 because all of the other inequalities have a feasible region inside the parabola

- (x + 3)(x – 1)
- (x + 8)(x – 6)
- (x – 3)(x + 1)
- (x – 5)(x + 7)
- (x – 7)(x + 9)

The answer is #3 because all of the others have a middle term of two when multiplied

- (2)(2)(2)
- 2
^{3} - 2(2)
^{2} - 4 + (2)
^{2} - 4
^{2}

The solution is #5 because all of the others are equal to 8

- Corresponding Angles
- Alternate Interior Angles
- Same Side Interior Angles
- Alternate Exterior Angles
- Vertical Angles

The solution is #3 because all of the other angle pairs are congruent

- SSS
- SAS
- AAA
- ASA
- HL

The solution is #3 because all of the others are theorems to prove triangle congruence

- (x
^{2}+ 2x + 1) ÷ (x + 2) - (-2x
^{2}+ 3x – 1) ÷ (x + 2) - (2x
^{2}+ 3x + 1) ÷ (x + 1) - (x
^{2}+ 2x – 1) ÷ (x + 1) - (-2x
^{2}+ 3x + 1) ÷ (x + 1)

The solution is #3 because all of the others have remainders/are not factors