Answer:
Answer: 84 meters squared. (i.e. 84m^2).
Step-by-step explanation:
Area of the path = total area of path and plot - plot area only
First, let’s find the area of the plot:
20m x 20m = 400m^2
The path is 1m wide and runs all the way around, so, it describes an ‘outer square’ of 22m x 22m (2 x 1m larger because there are 1m of the path on each side).
The area of this ‘outer square’ is equivalent to the ‘total area of path and plot’
22m x 22m = 484m^2
So:
Area of the path = total area of path and plot - plot area only
= 484 - 400 = 84m^2
The answer is C) 160.
We know this because if mA = 50, we know that mC must also be 50. This is due to the fact that AB = BC. This leaves us with mB as 80 since the angles of a triangle always have to equal 180.
Now knowing this, it is easy to find the arc lengths in degrees. When you have a transcribed triangle, all we are going to do here is double the angle of the triangle to get the arc measure.
mB = 80
80*2 = 160
The answer to the question is 2
Answer:
1,665.25 km
Step-by-step explanation:
The domain of this equation is the lowest x value to the highest x value. For example: lowest is -4 and highest is 3. You would write it as [-4,3]
For this equation, the domain is (-infinity,infinity) because there is no lowest or highest x value.