Answer:
C. 178−−−√ m
Step-by-step explanation:
Given the following :
v = final velocity (in m/s)
u = initial velocity (in m/s)
a = acceleration (in m/s²)
s = distance (in meters).
Find v when u is 8 m/s, a is 3 m/s², and s is 19 meters
Using the 3rd equation of motion :
v^2 = u^2 + 2as
v^2 = 8^2 + 2(3)(19)
v^2 = 64 + 114
v^2 = 178
Take the square root of both sides :
√v^2 = √178
v = √178
I assume there are some plus signs that aren't rendering for some reason, so that the plane should be

.
You're minimizing

subject to the constraint

. Note that

and

attain their extrema at the same values of

, so we'll be working with the squared distance to avoid working out some slightly more complicated partial derivatives later.
The Lagrangian is

Take your partial derivatives and set them equal to 0:

Adding the first three equations together yields

and plugging this into the first three equations, you find a critical point at

.
The squared distance is then

, which means the shortest distance must be

.
Answer:
d
Step-by-step explanation:
The answe of this question is d
Simplified is 325 and not simplified is 325.561
Answer:
the screen is really blurry sorry bout that
Step-by-step explanation: