Answer:
a) 297°
b) 4.52 minutes
Step-by-step explanation:
a) Consider the attached figure. The boat's actual path will be the sum of its heading vector BA and that of the current, vector AC. The angle of BA north of west has a sine equal to 5/11. That is, the heading direction measured clockwise from north is ...
270° + arcsin(5/11) = 297°
__
b) The "speed made good" is the boat's speed multiplied by the cosine of the angle between the boat's heading and the boat's actual path. That same value can be computed as the remaining leg of the right triangle with hypotenuse 11 and leg 5.
boat speed = √(11² -5²) = √96 ≈ 9.7980 . . . . miles per hour
Then the travel time will be ...
time = distance/speed
(3900 ft)×(1 mi)/(5280 ft)×(60 min)/(1 h)/(9.7980 mi/h) ≈ 4.523 min
A(x) = 17 - 8x. Hope this helps!
Answer:
C (Y = -5/8x - 10)
Step-by-step explanation:
first find the slope (y₂ - y₁) ÷ (x₂ - x₁)
(5 - (-20)) ÷ (-24 - 16)
-5/8 = slope (m)
now that we know the slope is -5/8 we can rule out A and B
now to find the y intercept just substitute a point in for x and y
5 = (-5/8)(-24) + b
multiply
5 = 15 + b
subtract 15 from both sides tp isolate b
-10 = b
I think it is B sorry if I’m wrong
U(x) = f(x).(gx)
v(x) = f(x) / g(x)
Use chain rule to find u(x) and v(x).
u '(x) = f '(x) g(x) + f(x) g'(x)
v ' (x) = [f '(x) g(x) - f(x) g(x)] / [g(x)]^2
The functions given are piecewise.
You need to use the pieces that include the point x = 1.
You can calculate f '(x) and g '(x) at x =1, as the slopes of the lines that define each function.
And the slopes can be calculated graphycally as run / rise of each graph, around the given point.
f '(x) = slope of f (x); at x = 1, f '(1) = run / rise = 1/1 = 1
g '(x) = slope of g(x); at x = 1, g '(1) = run / rise = 1.5/ 1 = 1.5
You also need f (1) = 1 and g(1) = 2
Then:
u '(1) = f '(1) g(1) + f(1) g'(1) = 1*2 + 1*1.5 = 2 + 1.5 = 3.5
v ' (x) = [f '(1) g(1) - f(1) g(1)] / [g(1)]^2 = [1*2 - 1*1.5] / (2)^2 = [2-1.5]/4 =
= 0.5/4 = 0.125
Answers:
u '(1) = 3.5
v '(1) = 0.125
