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What is the value of st÷(6r) if r=5, s=32, and t=45
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First of all, when I do all the math on this, I get the coordinates for the max point to be (1/3, 14/27). But anyway, we need to find the derivative to see where those values fall in a table of intervals where the function is increasing or decreasing. The first derivative of the function is 
. Set the derivative equal to 0 and factor to find the critical numbers. 
, so x = -3 and x = 1/3. We set up a table of intervals using those critical numbers, test a value within each interval, and the resulting sign, positive or negative, tells us where the function is increasing or decreasing. From there we will look at our points to determine which fall into the "decreasing" category. Our intervals will be -∞<x<-3, -3<x<1/3, 1/3<x<∞. In the first interval test -4. f'(-4)=-13; therefore, the function is decreasing on this interval. In the second interval test 0. f'(0)=3; therefore, the function is increasing on this interval. In the third interval test 1. f'(1)=-8; therefore, the function is decreasing on this interval. In order to determine where our points in question fall, look to the x value. The ones that fall into the "decreasing" category are (2, -18), (1, -2), and (-4, -12). The point (-3, -18) is already a min value.
The ground, the building and the line of sight between the ground camera and the top of the building form right triangle with hypotenuse equal to 45 feet and angle adjacent to this hypotenuse and ground - 50°. The building is an opposite leg to the given angle, then you should write sine function:
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The answer to the nearest tenth of a foot is 34.5 feet