9) an aircraft carrier left hawaii and traveled
toward dry dock at an average speed of 15
mph. a submarine left two hours later and
traveled in the opposite direction with an
average speed of 15 mph. find the
number of hours the submarine needs to
travel before the vessels are 300 mi. apart.

The Time taken by submarines to travel before the vessel is 11 hours

Step-by-step explanation:

Given as :

The average speed of aircraft carrier = 15 mph

The average speed of submarine = 15 mph

The distance between them = 300 mile

The time taken by aircraft = T hour

The time taken by submarine = T + 2  hour

Now, Speed =

So, Distance =  Speed × Time

Now,

300 miles = 15 × T + 15 × ( T + 2)

Or, 300 = 15 T + 15 T + 30

or, 300 - 30 = 30 T

Or, 270 = 30 T

∴ T =

I,e T = 9 hours

So, T + 2 = 9 + 2 = 11 hours

Hence The Time taken by submarines to travel before the vessel is 11 hours Answer

the equation cos(35) =a/25 can be used to find the length of bc what is the length of bc? round to the nearest tenth

step-by-step explanation: