Alan

Distance A has travelled north at time t is  y = 30t + (1/2)*30 miles or y = 30t + 15 miles

Distance B has travelled east at time t is x = 35t

Use Pythagoras to get distance between them:  

\(D=\sqrt{x^2+y^2}=\sqrt{(35t)^2+(30t+15)^2}=\sqrt{2125t^2+900t+225} \text{ miles}\)

When t=3,  \(D=\sqrt{2125*9+900*3+225}=148.49 \text{ miles}\)

Oops! This is the distance between them 3 hours after B leaves port.  See heureka's reply for the distance 3 hours after A leaves port.

4((5(x-3)+2)-3(2(x+5)-9)) → 4((5x-15+2)-3(2x+10-9)) → 4(5x - 13 -3(2x+1)) → 4(5x - 13 - 6x - 3) → 4(-x - 16) → -4x - 64

When a is negative.

They are constants so they are doing neither!

f(3) is just -2*e^3 = -2*2.71828^3 = -40.171

You just replace x in the function by 3, you don't multiply the result by 3.

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As many times as you like!!  

If you mean how many times can you subtract until you reach zero, then it's seven times.

Look at this as \(\frac{8}{49}\times \frac{7}{9} \rightarrow \frac{8}{7\times9}\rightarrow \frac{8}{63}\)

Note that in the first step I've used the fact that 49 = 7*7 and cancelled one of the 7s with the 7 in the numerator.

It is just 0.42        

Think of 42 as 42.0 Dividing by 100 is like moving the decimal point two digits to the left.

Just type in the numerator followed by / followed by the denominator.  e.g.   2/3 

I hope this agrees with the text-book answer!