Answer:
option C
Step-by-step explanation:
option C is the correct answer of this question.
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107.75mThis is a trigonometric ratio problem. the angle of depression (10) is opposite the height (19m) and the distance is adjacent to the distance (x). opposite ad adjacent sides to the angle is a tangent problem.

plus in values, we know.

switch tan(10) and x using the products property.

plug in to calulator to get answer
x=107.75
X = height of pole (in meters)
With respect to the 50 degree angle, the side x is the opposite leg. It is the leg furthest from the reference angle. The hypotenuse is 5 meters.
The trig function sine ties together the opposite and hypotenuse
sin(angle) = opposite/hypotenuse
sin(50) = x/5
5*sin(50) = x .... multiply both sides by 5
x = 5*sin(50)
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Since x = 5*sin(50) isn't listed as an answer choice, let's try using cosine. We can't use it right away because we don't know the adjacent side. What we can do is change the reference angle. The missing angle of the triangle is 90-50 = 40 degrees. Let's make the 40 degree angle the reference angle
So x is now the adjacent side with respect to the 40 degree reference angle. The hypotenuse is always the longest side. The hypotenuse stays at 5.
cos(angle) = adjacent/hypotenuse
cos(40) = x/5
5*cos(40) = x
x = 5*cos(40)
This expression is listed. The answer is choice B
The equation for a parabola with directrix y = –p and focus (0, p)
The distance between vertex and directrix is P
Also the distance between vertex and focus is also P
Focus (0,p) so focus lies on y axis
Directrix line is paralled to focus point
So Directex crosses the x axis .
Hence line of symmetry lies is the y-axis.
The graph is attached below.
9514 1404 393
Answer:
Step-by-step explanation:
The extrema will be at the ends of the interval or at a critical point within the interval.
The derivative of the function is ...
f'(x) = 3x² -4x -4 = (x -2)(3x +2)
It is zero at x=-2/3 and at x=2. Only the latter critical point is in the interval. Since the leading coefficient of this cubic is positive, the right-most critical point is a local minimum. The coordinates of interest in this interval are ...
f(0) = 2
f(2) = ((2 -2)(2) -4)(2) +2 = -8 +2 = -6
f(3) = ((3 -2)(3) -4)(3) +2 = -3 +2 = -1
The absolute maximum on the interval is f(0) = 2.
The absolute minimum on the interval is f(2) = -6.