Two pieces of space junk are following the same orbit path, 1 is 1000 miles behind the other. the front piece is traveling 17000 miles per hour and the other is traveling 17500 per hour. how many miles will the front piece have to travel before the second piece catch up

34000 miles

Step-by-step explanation:

Given that,

front piece of space junk is traveling at 17000 miles/hr

space junk which is behind is traveling at 17500 miles/hr

difference between the two space junk = 1000miles

suppose distance which front space junk have traveled when it meets = x

so distance which behind space junk have travelled when it meets = x + 1000

Formula which we will use

(Because at the point that the two pieces of space junk meet, both

will have traveled for the same length of time)

x / (17000) = (x + 1000/ (17500))

x / (17000) -  (x + 1000 / (17500) = 0

x / (17000) - x/17500 - 2/35=0

x / (17000) - x/17500 = 2/35

Taking LCM

x(17500) - x(17000) = 297500 000 * 2 /35

500x = 297500 000*2/35

x = 297500 000 *2/ 35(500)

x = 34000 miles

Thus, 34,000 miles is the distance the front piece of space junk travels before the second piece catches up to it

the number of possible outcomes that do not show a 1 on the first spinner and show the number 4 on the second spinner is 2.

spinner 1 spinner 2

1                     1

1                     2

1                     3

1                     4

2                   1

2                   2

2                   3

2                   4   = not 1, and 4

3                   1

3                   2

3                   3

3                   4 = not 1, and 4

only 2 outcomes have passed the given condition.

step-by-step explanation: