Answer:
Step-by-step explanation:
![\displaystyle = \frac{x^2(y-2)}{3y} \\\\Put \ x = 3, \ y = -1\\\\= \frac{(3)^2(-1-2)}{3(-1)}\\\\= \frac{9(-3)}{-3} \\\\= 9 \\\\ \rule[225]{225}{2}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%3D%20%5Cfrac%7Bx%5E2%28y-2%29%7D%7B3y%7D%20%5C%5C%5C%5CPut%20%5C%20x%20%3D%203%2C%20%5C%20y%20%3D%20-1%5C%5C%5C%5C%3D%20%5Cfrac%7B%283%29%5E2%28-1-2%29%7D%7B3%28-1%29%7D%5C%5C%5C%5C%3D%20%5Cfrac%7B9%28-3%29%7D%7B-3%7D%20%5C%5C%5C%5C%3D%209%20%5C%5C%5C%5C%20%5Crule%5B225%5D%7B225%7D%7B2%7D)
Hope this helped!
<h3>~AH1807</h3><h3>Peace!</h3>
<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:
4x-11
Step-by-step explanation:
open parentheses: 4x-8-3
solve: 4x-11
Answer:
the last one
Step-by-step explanation:
