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I think so cuz I did this and got it right hopefully it is right good luck
You would distribute, 300 times 80 is 2400 and -3x times 1x is -3x so the equation would be y=2400-3x
Observe attached picture.
On picture we have:
A = height of flagpole = x ft
B = length of flagpole's shadow = 24 ft
C = height of sign = 6 ft
D = length of sign's shadow = 3 ft
When we draw a picture representing this problem we can also add another line marked in red. This way we can see that we have two right-angle triangles. We can see that both have same angle marked with α.
We can apply trigonometry rules to find height of flagpole.
From small triangle containing sign we can find tangens function:

Similarly we can do for large triangle containing flagpole:

We see that these two equations have same left sides. This means that their right sides must also be same:

We can solve for A:

Height of flagpole is 48 feet.
First, we should figure out the area of the entire face of the clock because we need that information to solve the problem. The formula for the area of a circle is A=pi*r^2. Since we know that r (radius) is equal to 11.25 feet, we can plug this in for r and solve for A: A=pi*11.25^2 which equals A=397.61 ft^2 rounded to the nearest hundredth.
Now, to find the area the hand sweeps over in 5 minutes, we should determine how much of the clock the hand sweeps over in 5 minutes. Think about it like this: since 5 minutes goes into 60 minutes 12 times (60/5=12), then 5 minutes is one twelfth of the clock's face. Therefore, we are going to divide the total area by 12 (397.61/12) to get 33.13 ft^2, so the answer is C.
I hope this helps.