Analog Clock

The hour hand moves  360 degrees/(12 hours x 60 min)  degrees each minute= 0.5 degrees/min

    9 hours 49 min = 589 minutes     or 294.5 degrees

The minute hand moves  360degrees/60 min = 6 degrees / min

    49 minutes x 6 degrees/min = 294 degrees

294.5 - 294 = .5 degrees between the hands at 9 49

The hands will be atop each other at what time?

270 + .5 x  = 6x      x = 270/5.5 = 49.0909090909090909        

                                                                    or 9:49 .090909*60 = 9:49:05.45454

At 9 o'clock......hands are separated by 1/4 * 360  =  90°

And, at this time.....let the minute hand be at 0°

And let the hour hand be at 270°

In 49 minutes.....the hour hand moves 360/12 * 49/60   = 24.5°

So.....the hour hand is at   [270 + 24.5]   =  294.5°

And the minute hand has moved  [ 0 +  49/ 60 * 360 ]  = 294°

So....they are separated by 0.5°  =  (1/2) of a degree

To find the exact time when they are together....

The hour hand moves at  30/60  =  .5 degrees per miniute = (1/2)° per minute

The minute hand moves at (360/60) = 6° per minute

So....we need to solve this

294.5 + (1/2)M  =  294 +(6)M       where M is the number of minutes sfter 9:49

Subtract 294 and (1/2)M from both sides

.5  =  5.5M      divide both sides by 5.5

.5 / 5.5  = M   =  5/55 =   1/11

So....they will be together 1/11 of a minute ater 9:49 ≈ 5.45 seconds after 9:49

To answer the first problem, we can use the clock formula, which is \(|30H-5.5M|\). Therefore, plugging the values in, we yield with \(|30(9)-269.5|=270-269.5=\boxed{.5}\)