A submarine is moving parallel to the surface of the ocean at a depth of 626 m. It begins a constant ascent so that it will reach the surface after travelling a distance of 4420 m.
a) What angle of ascent, to the nearest tenth of a degree, did the submarine make? (3
marks)
b) How far did the submarine travel horizontally, to the nearest metre, during its ascent to
the surface? (3 marks)

a) the angle of ascent is 8.2°

b) the horizontal distance traveled is 4375 m

Step-by-step explanation:

depth of ocean = 626 m

distance traveled in the ascent = 4420 m

This is an angle of elevation problem with

opposite side to the angle = 626 m

hypotenuse side = 4420 m

a) angle of ascent ∅ is gotten from

sin ∅ = opp/hyp = 626/4420

sin ∅ = 0.142

∅ = 0.142

∅ = 8.2°  this is the angle of ascent of the submarine.

b) The horizontal distance traveled will be gotten from Pythagoras theorem

= +

The horizontal distance traveled will be the adjacent side of the right angle triangle formed by these distances

= +

adj =

adj = 4375 m  this is the horizontal distance traveled.

step-by-step explanation:

a.   since the profit made in 2012 is equal to the profit made in 2014, the profit is           symmetric about the line y = 120,000.

step-by-step explanation:

given that megan owns a clothing boutique.

the profits of 2010, 11, are recorded and shown as follows:

years since     2010 0     1     2     3       4       5

profits made in 1000's   39       84     111     120   111     84

we find that profit gradually increases form 2010 to 2013 upto 120,   then starts decreasing.

the profits for 2012 and 2014 are the same while that of 2011 and 2015 are also the same.

this shows that there is symmetry about the line profit = 120000

hence answer is:

a.   since the profit made in 2012 is equal to the profit made in 2014, the profit is           symmetric about the line y = 120,000.

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