Boat A leaves a dock headed due east at 2:00PM traveling at a speed of 9 mi/hr. At the same time, Boat B leaves the same dock tr
aveling due south at a speed of 15 mi/hr. Find an equation that represents the distance d in miles between the boats and any time t in hours.
1 answer:
Answer: 
Step-by-step explanation:
Given
Speed of boat A is 
Speed of boat B is 
Both are moving perpendicular to each other
Distance traveled by Boat A 
Distance traveled by Boat B 
Distance between them is given by Pythagoras theorem

Distance between them is 
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Answer:
167
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Answer:
It would be 5.
Step-by-step explanation:
Hope this help!
3x-10+x+8=90
4×-2=90
4x=92
X=23
One angle is 31. 23+8
The second one is 59. 3 (23)-10
Answer:
D
Step-by-step explanation:
One number will be positive, and one will be negative