A) c=18
0.5c^2 +13=94
subtract 13 from both sides
0.5c^2 = 81
Square root of both sides to get c alone
0.5c=9
double c to make it just c
c=18
B) x=8
(x-2)^2+1=37
Subtract one from each side
(x-2)^2=36
square root each side
x-2=6
add the 2 over
x=8
<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>
(0,-3) (1, -1) (2, 1) ....
No.
The distance is |-5 -3| = 8 units. (Y-coordinates are the same, so the distance is measured entirely in the x-direction. Distance is non-negative.)
Step-by-step explanation:
