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<span>N(t) = 16t ; Distance north of spot at time t for the liner.
W(t) = 14(t-1); Distance west of spot at time t for the tanker.
d(t) = sqrt(N(t)^2 + W(t)^2) ; Distance between both ships at time t.
Let's create a function to express the distance north of the spot that the luxury liner is at time t. We will use the value t as representing "the number of hours since 2 p.m." Since the liner was there at exactly 2 p.m. and is traveling 16 kph, the function is
N(t) = 16t
Now let's create the same function for how far west the tanker is from the spot. Since the tanker was there at 3 p.m. (t = 1 by the definition above), the function is slightly more complicated, and is
W(t) = 14(t-1)
The distance between the 2 ships is easy. Just use the pythagorean theorem. So
d(t) = sqrt(N(t)^2 + W(t)^2)
If you want the function for d() to be expanded, just substitute the other functions, so
d(t) = sqrt((16t)^2 + (14(t-1))^2)
d(t) = sqrt(256t^2 + (14t-14)^2)
d(t) = sqrt(256t^2 + (196t^2 - 392t + 196) )
d(t) = sqrt(452t^2 - 392t + 196)</span>
Answer:
Solving the given formula for v2 gives us:
Step-by-step explanation:
Solving an equation for a particular variable means that the variable has to be isolated on one side of the equation.
Given equation is:
Multiplying both sides by t2-t1
Adding v1 to both sides of the equation
Hence,
Solving the given formula for v2 gives us:
For this you use A=Pe^(rt)
A=furutre amount
P=present amount
e=natural base
r=rate in decimal
t=times in years
A=?
P=5000000
r=0.04
t=30
A=5,000,000e^(0.04*30)
A=5,000,000e^(1.2)
A=16600584.613683
Answer:8498
Step-by-step explanation:
(7*1000)+(7*200)+(7*10)+(7*4)
7000+1400+70+28
8498
