<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>
6,051m = 6,000m + 51m
1km=1,000 metres
--------------------------------
Therefore:
6,051 metres = 6,000/1,000km + 51/1,000km
=6,051/1,000km
=6.051km
Answer:
The equation you are given is a quadratic. The standard form of a quadratic is y = a(x-h)2 + k where (h,k) is the vertex of the graph, which is a parabola. Vertically moving the graph 4 units upward means that you are moving k +4 units.
y = a(x-h)2 + k standard form
y = 5x2 - 4 original equation
y = 5(x-0)2 - 4 re-written in standard form
h = 0 k = -4
Four (4) units up is k + 4--->-4 + 4 = 0.
Therefore, f(x) = 5x2 + 0--->f(x) = 5x2.
Step-by-step explanation:
hope this helps
plz mark brainliest
C - 10 = g
L - 5 = C
g = 5
then altogether he finds
40 bugs
Answer:
7.8
Step-by-step explanation:
