Answer: 84.1m
Step-by-step explanation:
Let the height of the flagpole be represented by x.
First and foremost, we have to know that the cosine of 18° = 0.9511.
The equation to get the height will then be:
0.9511 = 80/x
Cross multiply
(0.9511 × x) = 80
0.9511x = 80
Divide both side by 0.9511
0.9511x/0.9511 = 80/0.9511
x = 84.1
The height of the flagpole is 84.1m
Answer:
(2x+5)(x−2)
Step-by-step explanation:
Factor 2x2+1x−10
2x2+x−10
=(2x+5)(x−2)
Step-by-step explanation:
$13.50-$1.50=12
-$1.50
12÷.60=
20 miles to airport
In rotation, the angle through which the plane figure is rotated is called the point of a figure through a specified angle and direction about a fixed point. Identify the corresponding vertices of the rotation.
Answer: The second derivative would be
![f''(x)=5e^{-x}-50e^{-5x}](https://tex.z-dn.net/?f=f%27%27%28x%29%3D5e%5E%7B-x%7D-50e%5E%7B-5x%7D)
Step-by-step explanation:
Since we have given that
![f(x)=5e^{-x}-2e^{-5x}](https://tex.z-dn.net/?f=f%28x%29%3D5e%5E%7B-x%7D-2e%5E%7B-5x%7D)
We will find the first derivative w.r.t. 'x'.
So, it becomes,
![f'(x)=-5e^{-x}+10e^{-5x}](https://tex.z-dn.net/?f=f%27%28x%29%3D-5e%5E%7B-x%7D%2B10e%5E%7B-5x%7D)
Then, we will find the second derivative w.r.t 'x'.![f''(x)=5e^{-x}+10\times -5e^{-5x}\\\\f''(x)=5e^{-x}-50e^{-5x}](https://tex.z-dn.net/?f=f%27%27%28x%29%3D5e%5E%7B-x%7D%2B10%5Ctimes%20-5e%5E%7B-5x%7D%5C%5C%5C%5Cf%27%27%28x%29%3D5e%5E%7B-x%7D-50e%5E%7B-5x%7D)
Hence, the second derivative would be
![f''(x)=5e^{-x}-50e^{-5x}](https://tex.z-dn.net/?f=f%27%27%28x%29%3D5e%5E%7B-x%7D-50e%5E%7B-5x%7D)