The problem says that <span>Brandon sights a helicopter above a building that is 200 feet away at an angle of elevation of 30 degrees. So, you can calculate the height asked, by following this procedure:
</span>
Tan(α)=Opposite leg/Adjacent leg
α=30°
Opposite leg=x
Adjacent leg=200 feet
When you substitute these values into the formula above (Tan(α)=Opposite leg/Adjacent leg), you have:
Tan(α)=Opposite leg/Adjacent leg
Tan(30°)=x/200
You must clear "x":
x=200xTan(30°)
Therefore, the value of "x" is:
x=115 feet
<span>
How high above the ground the is the helicopter?
The answer is: 115 feet</span>
Actually, distributive property isn't needed to combine this. You can simply subtract 7 from 13 which will give you 6. The solution is 6c.
The nth taylor polynomial for the given function is
P₄(x) = ln5 + 1/5 (x-5) - 1/25*2! (x-5)² + 2/125*3! (x-5)³ - 6/625*4! (x - 5)⁴
Given:
f(x) = ln(x)
n = 4
c = 3
nth Taylor polynomial for the function, centered at c
The Taylor series for f(x) = ln x centered at 5 is:

Since, c = 5 so,

Now
f(5) = ln 5
f'(x) = 1/x ⇒ f'(5) = 1/5
f''(x) = -1/x² ⇒ f''(5) = -1/5² = -1/25
f'''(x) = 2/x³ ⇒ f'''(5) = 2/5³ = 2/125
f''''(x) = -6/x⁴ ⇒ f (5) = -6/5⁴ = -6/625
So Taylor polynomial for n = 4 is:
P₄(x) = ln5 + 1/5 (x-5) - 1/25*2! (x-5)² + 2/125*3! (x-5)³ - 6/625*4! (x - 5)⁴
Hence,
The nth taylor polynomial for the given function is
P₄(x) = ln5 + 1/5 (x-5) - 1/25*2! (x-5)² + 2/125*3! (x-5)³ - 6/625*4! (x - 5)⁴
Find out more information about nth taylor polynomial here
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we need the equation or graph to figure that out
If the triangle has two congruent sides.