I’m assuming ‘dives further’ means to go directly down
the angle of elevation of the ship from the submarine is equal to the angle of depression of the submarine from the ship, if we assume the sea level is perpendicular to ‘directly down’.
let both of these angles to be = $ when the submarine is at A and ¥ when the submarine is at B (excuse the lack of easily accessible variables as keys)
then this become a simple trig problem:
A)
Let O be the position of of the ship, and C be the original position of the submarine.
therefore, not considering direction
|OC| = 1.78km = 1780m
|CA| = 45m
these are the adjacent and opposite sides of a right angled triangle.
But tan($) = opp/adj = |CA|/|OC| = 45/1780
therefore $ = arctan(45/1780) which is roughly 1.45 degrees,
B)
similarly, noting that |CB| = |CA| + |AB| = 45 + 62 = 107m
tan(¥) = 107/1780
¥ = arctan(107/1780) which is roughly 3.44 degrees
There are 2 solutions. the highest degree of a polynomial is 2
I believe that the answeres are either (The functions have the same shape. The Y intercept of y= |x| is 0, and the y-intercept of the second function is -9) or ( the two functions are the same) my best guess is (The functions have the same shape. The Y intercept of y= |x| is 0, and the y-intercept of the second function is -9)
<h2>33 km</h2>
3.3cm = 0.000033km (divide by 100,000.
So, 0.000033km/yr.
0.000033 × 1,000,000
=33km
