Answer: y=1/2x+5/2
Step-by-step explanation:
y-3=1/2 (x-1)
y-3(-3)=1/2x-1/2(+3)
y=1/2x-1/2+6/2
y=1/2x+5/2
hope this helped!!
We have that
using a graph tool
see the attached figure
The horizontal asymptote of this function is at <span>y=3</span><span>.
So,
the range of this function is from </span><span>
(−∞,3<span>
)</span></span>
Answer:
<h3>
</h3>
Step-by-step explanation:
Apply cross product property
⇒
Multiply the numbers
⇒
Hope I helped!
Best regards!!
<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>
