<u>Answer:</u>
<u>Answer:a. Since 20° is in the first quadrant, the reference angle is 20° .</u>
<u>Answer:a. Since 20° is in the first quadrant, the reference angle is 20° .b. Reference Angle: the acute angle between the terminal arm/terminal side and the x-axis. The reference angle is always positive. In other words, the reference angle is an angle being sandwiched by the terminal side and the x-axis. It must be less than 90 degree, and always positive.</u>
<u>Answer:a. Since 20° is in the first quadrant, the reference angle is 20° .b. Reference Angle: the acute angle between the terminal arm/terminal side and the x-axis. The reference angle is always positive. In other words, the reference angle is an angle being sandwiched by the terminal side and the x-axis. It must be less than 90 degree, and always positive.c. The rays corresponding to supplementary angles intersect the unit circles in points having the same y-coordinate, so the two angles have the same sine (and opposite cosines).</u>
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Answer:
Step-by-step explanation:
I see you're in college math, so we'll solve this with calculus, since it's the easiest way anyway.
The position equation is
That equation will give us the height of the rock at ANY TIME during its travels. I could find the height at 2 seconds by plugging in a 2 for t; I could find the height at 12 seconds by plugging in a 12 for t, etc.
The first derivative of position is velocity:
v(t) = -3.72t + 15 and you stated that the rock will be at its max height when the velocity is 0, so we plug in a 0 for v(t):
0 = -3.72t + 15 and solve for t:\
-15 = -3.72t so
t = 4.03 seconds. This is how long it takes to get to its max height. Knowing that, we can plug 4.03 seconds into the position equation to find the height at 4.03 seconds:
s(4.03) = -1.86(4.03)² + 15(4.03) so
s(4.03) = 30.2 meters.
Calculus is amazing. Much easier than most methods to solve problems like this.
Answer:
14 is what percent of 92? = 15.22.
Step-by-step explanation:
A, perhaps,plz check the definition of functions
