Find the velocity first, the "slope".
m=(y2-y1)/(x2-x1)
m=(95.5-53)/(4-2)
m=21.25 ? Your equations are wrong, I'll double check with another set of points...
m=(350.5-308)/(16-14)
m=21.25 LOL, yep none of your lines has the correct slope.
Anyway...
y=21.25x+b, using any point we can solve for the y-intercept, or "b", (2, 53)
53=21.25(2)+b
53=42.5+b
10.5=b so the distance to the lighthouse as a function of time is:
y=21.25x+10.5
So essentially, you should fire your teacher or burn that book you are using :)
58^(1/2) is √58 if that is what you meant
The answer would be 8.7639
Remember that
If the given coordinates of the vertices and foci have the form (0,10) and (0,14)
then
the transverse axis is the y-axis
so
the equation is of the form
(y-k)^2/a^2-(x-h)^2/b^2=1
In this problem
center (h,k) is equal to (0,4)
(0,a-k)) is equal to (0,10)
a=10-4=6
(0,c-k) is equal to (0,14)
c=14-4=10
Find out the value of b
b^2=c^2-a^2
b^2=10^2-6^2
b^2=64
therefore
the equation is equal to
<h2>(y-4)^2/36-x^2/64=1</h2><h2>the answer is option A</h2>
Horizontal translation: 2 units right.
Vertical translation: 5 units up
Stretch/compression: stretched by a factor of 3.
Reflection: not reflected.
