An observer (O) is located 500 feet from a school (S). The observer notices a bird (B) flying at a 39° angle of elevation from his line of sight. How high is the bird flying over

the school? You must show all work and calculations to receive full credit. (10 points)

404.9 ft

Step-by-step explanation:

The information given is illustrated in the diagram attached below. We have a right triangle.

We have the following:

Angle of elevation = 39°

Adjacent side = 500 ft

Opposite side, which is how high the bird is flying over the school = h

Apply the following trigonometric ratios:

Plug in the values

Multiply both sides by 500

404.9 = h (nearest tenth)

✅The bird is flying at about 404.9 ft high over the school.

x =-.20752

step-by-step explanation:

4^(x+2)=12

4^(x+2)=12

take the log base 4 on each side

log 4 (4^(x+2) = log 4(12)

we know log b(a^y) = y log b (a)

(x+2) log 4(4) = log 4 (12)

x+2 = log 4 (12)

we know   logb c = loga c/loga b

we want to convert to base 10   so a = 10

by default, we do not write the 10   c = 4   and b = 10

log4 10 = log 12/log 4

log4 10 = log 12/ log 4

x+2 = log 12/ log 4

subtract 2 from each side

x +2 -2 = log 12/ log 4 -2

x = 1.79248125 -2

x =-.20752