This is it, i control it!!!

Solution:

If you don't see it clearly,

By using the formula,|60H-11M|/2=the angle between the hands of the clock,𝑤ℎ𝑒𝑟𝑒 𝐻=𝐻𝑜𝑢𝑟𝑠 𝑎𝑛𝑑 𝑀=𝑀𝑖𝑛𝑢𝑡𝑒s.

In this problem,H=6 so,

|(60*6)-11M|/2=0 ,zero is the angle between the hands of the clock because of,if they overlap,then,they will not move, resulting to the zero angle.

|360-11M|=0

360=11M→M=360/11≈32.72 minutes

Let X be the number of minutes. The hour hand travels $\frac{x}{2}$ minutes, but since it is located after 6 o'clock but before 7 o'clock, we have to add 6*30 = 180 degrees. The minute hand travels 6x degrees.

Setting $\frac{x}{2} + 180 = 6x$ gives $x=\frac{360}{11}$ which is about $\boxed{32.72}$ minutes.

Consider the rate at which the two hands sweep around the clock:

1. The hour hand completes 360 degrees every 12 hours, or 12 60 60 seconds. So, it advances at a rate of 360/(12 60 60) degrees per second, which is 1/120 degrees per second.

2. The minute hand completes 360 degrees every hour, or 60 60 seconds. So it advances at a rate of 360/(60 60) degrees per second, which is 1/10 degree per second.

Now consider the relative rates of the two hands:

If the time is 8 o'clock, the minute hand is separated from the hour hand, and as it advances towards the hour hand, the hour hand moves away from it. The problem then becomes equivalent to one in which the hour hand remains stationary, and the minute hand advances upon the hour hand at a reduced rate which is the relative rate of the two hands. Because of the two rates at which the hands are moving as well as their movement being in the same direction (both clockwise!), the equivalent problem is one in which the minute hand advances on the hour hand at the slower rate of (1/10 - 1/120) degrees per second, or 11/120 degrees per second.

We then need to know how long in this equivalent problem it would take for the minute hand to travel 180 degrees, since that is the separation between the two hands if they start at 6 o'clock.

The required time is 180/(11/120) seconds or 180*120/11 seconds. This is (1963 + 7/11) seconds.

1963 seconds is 32 minutes and 43 seconds. So, the total time it takes for the minute hand to reach the stationary hour hand is 32 minutes and (43 + 7/11) seconds.

Since the time at which the hands started moving is 6 o'clock, this means that the two hands will meet at 32 minutes and (43 + 7/11) seconds past 6. This is also the time at which they will meet in the original question, since the two problems are equivalent.

So, to answer the question, the hands of the clock will be together 32 minutes and (43 + 7/11) seconds after 6 o'clock, or at the time of 32 minutes and (43 + 7/11) seconds past 6.

the hour hand moves at a speed of 5 ''minutes" per hour while the minute hand moves at a speed of 60 "minutes" per hour

let the time taken for both hands to come together be x hours.

at 6 o' clock, the hour hand is at 30 "minutes" while the minute hand is at 0"minutes"

so, 30 + 5x=60x

that gives us x=3/55 hours that is approximately 32.72 minutes