Suppose an airplane is flying 180 mph with a heading 25° east of south (295° measured counterclockwise from due east) when it encounters a 35 mph wind blowing steadily in a direction 15° north of east (15° measured counterclockwise from due east). Use →v to represent the speed vector of the airplane and →w to represent the speed vector of the wind. Start by drawing a head-to-tail configuration for →v and →w and the resultant vector →v+→w.

Write both →v and →w in component form. (Your answers should be in the form "< #, # >".)

→v=

→w=

Write the resultant vector →v+→w in component form.
→v+→w=

Now, write the resultant vector →v+→w in polar form. (You do not need to enter a degree symbol in your answer.)
→v+→w=

Your answer to part (c) tells us that the plane is actually moving at _ mph in the direction_  ° counterclockwise of due east.
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765238725478629038640962136509762139056074

step-by-step explanation:

64087235609746097215074621039754021037

x coordinate is none

step-by-step explanation:

step-by-step explanation:

part a. the differences are decreasing, so we know the sequence is not arithmetic. the ratios are constant (=3/4), so we know the sequence is geometric.

part b. from part a, we know the common ratio is 3/4, so each term is 3/4 of the one before. the initial term is the number of gallons at h=1.

g[h] = (3/4)·g[h-1]

g[1] = 384 . . . . initial value required for a recursive definition

part c. the general formula for a geometric sequence is

a[n] = a[1]·r^(n-1)

we have a[1] = 384, r = 3/4, and the term number given by h. the formula that follows is

g[h] = 384·(3/4)^(h-1)

part d. filling in the value and doing the arithmetic, we have

g[12] = 384·(3/4)^11 ≈ 384·0.042235136 ≈ 16.2183 . . gallons

part e. each gallon is 4 quarts, so

16.2183 gal × 4 qt/gal = 64.8732 qt