Answer:
Step-by-step explanation:
<u>Given triangle LNK </u>
- m∠NLK = 72°
- m∠LNK = 58°
- m∠LKN = ?
<u>Interior angles sum to 180°</u>
- m∠NLK + m∠LNK + m∠LKN = 180°
- 72° + 58° + m∠LKN = 180°
- m∠LKN = 180° - 130°
- m∠LKN = 50°
The answer is none of the above
Recall the sum identity for cosine:
cos(a + b) = cos(a) cos(b) - sin(a) sin(b)
so that
cos(a + b) = 12/13 cos(a) - 8/17 sin(b)
Since both a and b terminate in the first quadrant, we know that both cos(a) and sin(b) are positive. Then using the Pythagorean identity,
cos²(a) + sin²(a) = 1 ⇒ cos(a) = √(1 - sin²(a)) = 15/17
cos²(b) + sin²(b) = 1 ⇒ sin(b) = √(1 - cos²(b)) = 5/13
Then
cos(a + b) = 12/13 • 15/17 - 8/17 • 5/13 = 140/221
Answer:
21 r16
Step-by-step explanation:
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).
_____
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.