Please help me with this math problem.
1 answer:
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The given quadratic describes a parabola that opens upward. Its one absolute extreme is a minimum that is found at x = -3/2. The value of the function there is
(-3/2 +3)(-3/2) -1 = -13/4
The one relative extreme is a minimum at (-1.5, -3.25).
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For the parabola described by ax² +bx +c, the vertex (extreme) is found where
x = -b/(2a)
Here, that is x=-3/(2·1) = -3/2.
(-3/2 +3)(-3/2) -1 = -13/4
The one relative extreme is a minimum at (-1.5, -3.25).
_____
For the parabola described by ax² +bx +c, the vertex (extreme) is found where
x = -b/(2a)
Here, that is x=-3/(2·1) = -3/2.
Answer: see proof below
<u>Step-by-step explanation:</u>
Given: cos 330 =
Use the Double-Angle Identity: cos 2A = 2 cos² A - 1
Proof LHS → RHS:
LHS cos 165
Double-Angle: cos (2 · 165) = 2 cos² 165 - 1
⇒ cos 330 = 2 cos² 165 - 1
⇒ 2 cos² 165 = cos 330 + 1
Given:
Divide by 2:
Square root:
Scratchwork:
Since cos 165 is in the 2nd Quadrant, the sign is NEGATIVE
LHS = RHS
