First, let's calculate the horizontal and vertical components of the wind speed (W) and the airplane speed (A), knowing that south is a bearing of 270° and northeast is a bearing of 45°:


Now, let's add the components of the same direction:

To find the resultant bearing (theta), we can use the formula below:

The angle -86° is equivalent to -86 + 360 = 274°.
Therefore the correct option is b.
<span>Solve for "x":
a/x = x/4a
---
x^2 = 4a^2
x = 2a</span>
The work has one step: divide the expression by 3.
.. -4 > x
Because the symbol is "greater than", the circle at x=-4 is an open circle.
I think x equals 54
Hope this helps
Simply simplify:
7a³(7a²)-4a^6 ⇒ 49a^6 - 4a^6 ⇒ 45a^6
Hope that helps!