Answer:
I think its A to the third plus 12 to the third power
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
Answer:
12
Step-by-step explanation:
(-2)(-3)(-3+5)
6(2)
12
Merry Christmas :)
Answer:
hi
Step-by-step explanation:
4- 3*1=3
6*1/100= 0.06
3+0.06= 3.06
5- 7*1= 7
3*1/10= 0.3
4*1/1,000= 0.004
7+0.3+0.004= 7.304
7- 3.3, 3.30
8- 9.3000, 9.30
9- 9.6, 9.600
10- 4.4000, 4.40
for #4-5 just put the number of 0s in the denominator of the fraction for the answer
I think the answer should be 0 but I’m not so sure!
