<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:
x = 9cm
Step-by-step explanation:
This is a right triangle so we can use the Pythagorean theorem
a^2 + b^2 = c^2 where a and b are the legs and c is the hypotenuse
6^2 + x^2 = ( sqrt(117) ^2)
36 + x^2 = 117
Subtract 36 from each side
x^2 = 117 -36
x^2 = 81
Take the square root of each side
sqrt(x^2) = sqrt(81)
x = 9
Answer: ( 5, -8)
Step-by-step explanation:
Answer:
c(2) = -10
Step-by-step explanation:
The first equation says that the first term of the sequence is -20.
The second equation is saying that to find any term of the sequence, add 10 to the previous term.
c(2) = c(2-1) + 10
c(2) = c(1) + 10
c(2) = -20 + 10 = -10
