Answer:C
Therefore: In Quadrant I, cos(θ) > 0, sin(θ) > 0 and tan(θ) > 0 (All positive). For an angle in the second quadrant the point P has negative x
Step-by-step explanation:
Answer:
Yes, no verticle line passes through two graphed points
Step-by-step explanation:
A function is when 1 input only has 1 output. None of the inputs have different outputs. So, this is a function.
Well us as kids are constantly on the computer and phones so google classroom is a great way to keep us connected with our teachers because we can communicate with them and they with us by posting assignments. Therefore we are less likely to forget what they told us in the classroom because it is just one click away.
Answer:
the answer is C..............look at the picture i sent
A graphing calculator shows four (4) zeros of
f(x) = sin(2x -9°) -cos(x +30°)
in the range 0° ≤ x ≤ 360°
Solutions are
x ∈ {23°, 129°, 143°, 263°).
_____
You can make use of a couple of trig identities to rearrange this equation.
cos(x) = sin(x+90°)
sin(a) -sin(b) = 2*cos((a+b)/2)*sin((a-b)/2)
So
sin(2x -9°) -sin(x +120°) = 2*cos((3x +111°)/2)*sin((x -129°)/2) = 0
The cosine factor will be zero for
(3x +111°)/2 = n*180° +90°
3x -69° = n*360°
x = 23° +n*120° . . . . . . for any integer n
The sine factor will be zero for
(x -129°)/2 = n*180°
x = 129° +n*360° . . . . . for any integer n
Combined, these solutions give the ones listed above in the range 0..360°.