Solving this problem needs the distance formula of point to a line.
The formula is:
distance = | a x + b y + c | / √ (a^2 + b^2)
So we are given the equation: y = 2 x + 4
Rewriting this would be: y – 2 x – 4 = 0 -> a = -2, b = 1, c = -4
We are also given the points:
(-4, 11) = (x, y)
Plugging it in the distance formula at points (x, y):
distance = | -2 * -4 + 1 * 11 + -4 | / √ [(- 2)^2 + (1)^2]
= 15 / √ (5)
= 6.7
So the tree is approximately 6.7 feet away from the zip line.
Answer: 21
Step-by-step explanation:
Answer:
D.
Step-by-step explanation:
If she connects it to any other point the angle would be more then 30 degrees.
<em>Greetings from Brasil</em>
From radiciation properties:
![\large{A^{\frac{P}{Q}}=\sqrt[Q]{A^P}}](https://tex.z-dn.net/?f=%5Clarge%7BA%5E%7B%5Cfrac%7BP%7D%7BQ%7D%7D%3D%5Csqrt%5BQ%5D%7BA%5EP%7D%7D)
bringing to our problem
![\large{6^{\frac{1}{3}}=\sqrt[3]{6^1}}](https://tex.z-dn.net/?f=%5Clarge%7B6%5E%7B%5Cfrac%7B1%7D%7B3%7D%7D%3D%5Csqrt%5B3%5D%7B6%5E1%7D%7D)
<h2>∛6</h2>