1. 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?
-45
-325
-405
-495
2. 5. If sin0 =√2/2 which could not be the value of 0.
225
45
135
405
3. For which value of 0 is cot 0 equal 0
pi
2pi
pi/
0
3 answers
pi/2
Can you please answer Question 1- 3
1. 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?
-45
-325
-405
-495
2. 5. If sin0 =√2/2 which could not be the value of 0.
225
45
135
405
3. For which value of 0 is cot 0 equal 0
pi
2pi
pi/
0
1. 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?
-45
-325
-405
-495
2. 5. If sin0 =√2/2 which could not be the value of 0.
225
45
135
405
3. For which value of 0 is cot 0 equal 0
pi
2pi
pi/
0
1. The answer is -45 degrees.
To solve this problem, we need to use the fact that the minute hand rotates 360 degrees in 60 minutes, or 6 degrees per minute. Since John first looked at the clock when the hands were aligned at 3:15, we know that the minute hand had rotated 15 minutes * 6 degrees/minute = 90 degrees from the 12 o'clock position.
Now, let x be the number of degrees the minute hand has rotated from the 12 o'clock position to reach its current position. Then, the hour hand has rotated (3 + x/360) * 30 degrees clockwise from the 12 o'clock position (since it is stuck at the 3 o'clock position).
At the second observation, the hour and minute hands form an angle of -135 degrees, or 225 degrees counterclockwise from the 3 o'clock position. Therefore, we have:
(3 + x/360) * 30 - x = 225
90 - x/12 - x = 225
-x/12 = 135
x = -45
So the minute hand has rotated -45 degrees from the 12 o'clock position to reach its current position.
2. The answer is 405.
We know that sin 45 degrees = √2/2, so one possible value of 0 is 45 degrees. However, sin is a periodic function with period 360 degrees, so we can add or subtract any multiple of 360 from 45 degrees to get other values of 0 that satisfy sin 0 = √2/2.
45 + 360k and 315 + 360k (where k is an integer) are all possible values of 0 that satisfy sin 0 = √2/2.
However, 405 is not a possible value of 0 that satisfies this equation.
3. The answer is pi/2.
Recall that cot 0 = cos 0 / sin 0. Therefore, cot 0 = 0 when cos 0 = 0 (since sin 0 cannot be 0).
Cosine is equal to 0 at all odd multiples of pi/2. Therefore, the possible values of 0 that satisfy cot 0 = 0 are:
pi/2, 3pi/2, 5pi/2, ...
The only value of 0 in the interval [0, 2pi) is pi/2.
To solve this problem, we need to use the fact that the minute hand rotates 360 degrees in 60 minutes, or 6 degrees per minute. Since John first looked at the clock when the hands were aligned at 3:15, we know that the minute hand had rotated 15 minutes * 6 degrees/minute = 90 degrees from the 12 o'clock position.
Now, let x be the number of degrees the minute hand has rotated from the 12 o'clock position to reach its current position. Then, the hour hand has rotated (3 + x/360) * 30 degrees clockwise from the 12 o'clock position (since it is stuck at the 3 o'clock position).
At the second observation, the hour and minute hands form an angle of -135 degrees, or 225 degrees counterclockwise from the 3 o'clock position. Therefore, we have:
(3 + x/360) * 30 - x = 225
90 - x/12 - x = 225
-x/12 = 135
x = -45
So the minute hand has rotated -45 degrees from the 12 o'clock position to reach its current position.
2. The answer is 405.
We know that sin 45 degrees = √2/2, so one possible value of 0 is 45 degrees. However, sin is a periodic function with period 360 degrees, so we can add or subtract any multiple of 360 from 45 degrees to get other values of 0 that satisfy sin 0 = √2/2.
45 + 360k and 315 + 360k (where k is an integer) are all possible values of 0 that satisfy sin 0 = √2/2.
However, 405 is not a possible value of 0 that satisfies this equation.
3. The answer is pi/2.
Recall that cot 0 = cos 0 / sin 0. Therefore, cot 0 = 0 when cos 0 = 0 (since sin 0 cannot be 0).
Cosine is equal to 0 at all odd multiples of pi/2. Therefore, the possible values of 0 that satisfy cot 0 = 0 are:
pi/2, 3pi/2, 5pi/2, ...
The only value of 0 in the interval [0, 2pi) is pi/2.