Please show your work :)
1. What is the period of the function? y= 2 cos π x
2
1
4
4.John's clock is broken. The minute hand rotates around the clock correctly, but the hour hand is stuck in the three o'clock position. Suppose John first looks at the clock when the hands are
aligned and it shows 3:15. He looks at the clock again and sees that the hour and minute hands of the clock form the arms of an angle measuring -135 degrees. How many degrees could the minute
hand have rotated to reach its current position?
-495
32
-45
1. The period of the function y = 2cos(πx) is found by using the formula T = 2π/b, where b is the coefficient of x. In this case, b = π/2, so T = 2π/(π/2) = 4. Therefore, the period is 4.
2. To find the angle rotated by the minute hand, we first need to find the angle between the hour and minute hands at 3:15. At 3:15, the minute hand has rotated 15 minutes beyond the 12 o'clock position, which is equivalent to 90 degrees. The hour hand is still stuck at 3 o'clock, which means it is also pointing in the same direction as the 3 o'clock mark, or 90 degrees from the 12 o'clock position. The angle between the hour and minute hands is the difference between their positions, which is 0 degrees.
Next, we need to find the angle between the hour and minute hands when they form an angle measuring -135 degrees. Since the minute hand rotates 360 degrees in one hour, it rotates 30 degrees in 5 minutes (since 5 minutes is 1/12 of an hour). Therefore, the minute hand rotates 15 x 30 = 450 degrees in the time between 3:15 and the time when the hands form an angle of -135 degrees.
Let θ be the angle rotated by the minute hand from 3:15 to the current position. We can set up the equation:
θ - 450 - 135 = 360n
where n is an integer that represents the number of complete rotations made by the minute hand. Solving for θ, we get:
θ = 450 + 135 + 360n
θ = 585 + 360n
We want to find the angle rotated by the minute hand, so we only need to consider the fractional part of θ (i.e., the part that is less than 360 degrees). We can write:
θ = 585 + 360n + k
where k is the fractional part of θ. Since we know that the angle between the hour and minute hands is -135 degrees, we can set up another equation:
k - 135 = 360m
where m is an integer that represents the number of complete rotations made by the minute hand from the time when the hands form an angle of -135 degrees to the current position. Solving for k, we get:
k = 135 + 360m
Substituting this expression for k into the first equation, we get:
θ = 585 + 360n + 135 + 360m
θ = 720 + 360(n + m)
Since we only care about the fractional part of θ, we can write:
θ = 360p
where p is an integer that represents the number of complete rotations made by the minute hand from 3:15 to the current position. Solving for p, we get:
p = (θ/360) = [(585 + 360n + 135 + 360m)/360] = (5/8)n + (3/8)m + (13/8)
Since we want to find the angle rotated by the minute hand, we only need to consider the fractional part of p. We can write:
p = (5/8)n + (3/8)m + q
where q is the fractional part of p. We know that p is an integer, so q must be 0. Therefore, we get:
p = (5/8)n + (3/8)m
Since the minute hand rotates 30 degrees in 5 minutes, the total angle rotated by the minute hand from 3:15 to the current position is 30p. Substituting the expression for p, we get:
30p = (5/8)30n + (3/8)30m
30p = 75n + 45m
We want to find the value of 30p that is less than 360 degrees, so we can write:
30p = 360q + r
where q is an integer and r is the remainder when 30p is divided by 360. Substituting the expression for 30p, we get:
75n + 45m = 360q + r
Dividing both sides by 15, we get:
5n + 3m = 24q + r/15
Since r is less than 360, r/15 is less than 24. Therefore, the only integer values that 5n + 3m can take on are 0, 5, 10, 15, 20. We can check each of these cases to see if they give us a value of 30p that is less than 360:
- If 5n + 3m = 0, then 30p = 360q, which means the minute hand has not rotated at all. This is not possible, since we know that the minute hand has rotated from 3:15 to the current position.
- If 5n + 3m = 5, then 30p = 360q + 6, which means the minute hand has rotated 6 degrees. This is not the correct answer, since none of the choices given match this value.
- If 5n + 3m = 10, then 30p = 360q + 12, which means the minute hand has rotated 12 degrees. This is not the correct answer, since none of the choices given match this value.
- If 5n + 3m = 15, then 30p = 360q + 18, which means the minute hand has rotated 18 degrees. This is not the correct answer, since none of the choices given match this value.
- If 5n + 3m = 20, then 30p = 360q + 24, which means the minute hand has rotated 24 degrees. This is the correct answer, since -135 + 24 = -111 degrees, which matches one of the answer choices.
Therefore, the angle rotated by the minute hand is -24 degrees (or 336 degrees, since the minute hand rotates clockwise).
1. To find the period of the function y = 2cos(πx), we need to determine the length of one complete cycle of the cosine function. The general formula for the period of a cosine function is given by T = 2π/b, where b is the coefficient of x in the function.
In this case, the coefficient of x is π, so the period T = 2π/π = 2.
Therefore, the period of the function y = 2cos(πx) is 2.
4. Let's break down the problem to find the answer step-by-step.
If the hour hand is stuck at the three o'clock position, it means that it should be pointing directly at three on the clock face.
Since John first looks at the clock when the hands are aligned and it shows 3:15, it means that the minute hand is pointing at three while the hour hand is pointing at the 3 plus one-fourth of the way between three and four on the clock face.
To find the number of degrees the minute hand has rotated, we need to find the difference in position between the hour and minute hands in degrees.
Since each hour mark on the clock corresponds to 30 degrees and each minute mark corresponds to 6 degrees, the angle between the hour and minute hands is given by:
(1/4 of an hour) * (30 degrees/hour) - 15 minutes * (6 degrees/minute)
= (1/4) * 30 - 15 * 6
= 7.5 degrees - 90 degrees
= -82.5 degrees
Therefore, the angle formed between the hour and minute hands is -82.5 degrees.
Since John looks at the clock again and sees that the hour and minute hands form an angle measuring -135 degrees, we need to find out how many degrees the minute hand has rotated from its initial position.
To do this, we need to subtract the initial angle (-82.5 degrees) from the current angle (-135 degrees):
-135 degrees - (-82.5 degrees)
= -135 degrees + 82.5 degrees
= -52.5 degrees
Therefore, the minute hand could have rotated -52.5 degrees to reach its current position.