<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:
Step-by-step explanation:
a^2 + b^2 = c^2
where
a and b are the legs
and
c is the hypotenuse
.
the problem gives the "legs" because the LONGEST leg is always the hypotenuse.
a^2 + b^2 = c^2
1071^2 + 1840^2 = c^2
1147041 + 3385600 = c^2
4532641 = c^2
13/8 or 1 5/8
3/4 becomes 7/8 when add it is 13/8
204 /6 = 34
29 +31 +33 + 35 +37 +39 = 207
4th number = 35
Around this time many people find home prices going up dramatically. Let's start with the X factor. If we know the starting is $120,000, the we know the initial velocity. To find the final but the very simple equation one may use is; P=(120000)^x\0.5
Cheers
