1. -6 ≤ x < -1 . . . . conjunction
2. x ≤ 6 . . or . . 10 ≤ x . . . . disjunction
3. 7 ≤ x < 12 . . . . conjunction
4. x < -9 . . or . . -3 ≤ x . . . . disjunction
5. 2 ≤ x ≤ 5 . . . . conjunction
6. x ≤ 54 . . or . . 66 ≤ x . . . . disjunction
7. 39 < x ≤ 43 . . . . conjunction
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Your problem statement provided no letters.
Answer:
1 hour
Step-by-step explanation:
I find these easiest to work by considering the initial difference in distance and the speed at with that gap is closing.
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The gap is 15 miles, the distance the first ship is from harbor when the second ship starts.
The rate of closure is the difference in the speeds of the two ships:
60 mph -45 mph = 15 mph
Then the closure time is ...
time = distance/speed
time = (15 mi)/(15 mi/h) = 1 h
It will take the second ship 1 hour to catch up to the first ship.
Answer:
15/22
Step-by-step explanation:
Small is the blank which is th answer
Sorry that is wrong, you distribute the 2 to the h and the 8 so it would be 2h-16-h. that would be h-16 on the left side,