The numbers in this problem are ordered pairs, which are points on a graph. These are (10, 20), (-10, 20), (-10, -10), and (10, -10). To find the area and perimeter of this shape, you must first find the distance between each point.
Distance between (10, 20) and (-10, 20): Since the y-value remains the same here, we just have to find the difference in x-values. This means 10 - (-10) A negative being subtracted is the same as a positive being added. That means 10 - (-10) is the same as 10 + 10. 10 + 10 = 20, so the distance between (10, 20) and (-10, 20) is 20 units.
Distance between (-10, 20) and (-10, -10): The x-values are the same here so just find the difference between the y-values. 20 - (-10) = 20 + 10 = 30 The distance between the (-10, 20) and (-10, -10) is 30 units.
Distance between (-10, -10) and (10, -10): The y-values are the same so just find the difference between the x-values. 10 - (-10) = 10 + 10 = 20 The distance between (-10, -10) and (10, -10) is 20 units.
Distance between (10, -10) and (10, 20): The x-values are the same so find the difference between the y-values. 20 - (-10) = 20 + 10 = 30 The distance between (10, -10) and (10, 20) is 30 units.
So now we know the side lengths of the room are 20 units, 30 units, 20 units, and 30 units.
To find the perimeter, add all the side lengths together. 20 + 30 + 20 + 30 = 100 The perimeter of the room is 100 units.
To find the area, multiply the length by the width. The length is 20 units and the width is 30 units. 20 • 30 = 600 The area of the room is 600 units.
First subtract 1 1/3 from 3 1/3 to find what is left over from her first batch of browns, 2. Then subtract 1 1/3 from 2 and get 2/3. 2/3 is 1/2 of 1 1/3, so she can make 1 1/2 more batches of brownies.
If two red marbles are removed, 1 red is returned. The number of reds is reduced by 1, and the number of blues is unchanged.
If two blue marbles are removed, 1 red is returned. The number of reds is increased by 1, and the number of blues is decreased by 2.
If one of each is removed, one blue is returned. The number of reds is reduced by 1, and the number of blues is unchanged.
So, at each step, the number of blue marbles is unchanged or reduced by 2. That is, it only changes by an even number. The number of blues is initially odd, so can never reach zero.