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.
Answer:
1 5/12
Step-by-step explanation:
12 1/3 - 10 11/12
We need to get a common denominator of 12
12 1/3 * 4/4 - 10 11/12
12 4/12 - 10 11/12
We need to borrow from the 12 (the whole number) because the 2nd fraction is bigger than the first
12 becomes 11 and the 1 becomes 12/12
11+ (12/12 + 4/12) - 10 11/12
11 16/12 - 10 11/12
Subtract the whole numbers
11-10 =1
Subtract the fractions
16/12 - 11 /12 = 5 /12
We are left with 1 5/12
Answer:
-1
Step-by-step explanation:
Hope this helps!
Answer:
the answer for this question is 18
