Dimension of one of the floors of one room that David wants to install tiles is 18feet long by 12 feet wide
Then
Area of the above room = 18 * 12 square feet
= 216 square feet
Dimension of the floor of the other room that David wants to install tiles is 24 feet long and 16 feet wide
Then
Area of the other room = 24 * 16 square feet
= 384 square feet
Then
The total square feet of the
rooms that David wants to install tiles = 216 + 384
= 600 square feet
Cost of the tile that covers 1 square feet = $5
Cost of the 4 tiles that cover 4 square feet = $17
Then
Area that can be covered with 4 square feet of tiles = 600/4 square feet
= 150 square feet
Minimum cost of covering
the two rooms that David wants to install tiles = 150 * 17 dollars
= 2550 dollars
So the minimum cost of installing the tiles on the two floors of David's two rooms is $2550. I hope the procedure is simple enough for you to understand.
The answer to the question is D
-4 because it is closer to the 0
The correct answer for this problem is 6,500 in one month
Answer:
the radius
Step-by-step explanation:
the correct answer is the radius
