Using the greatest common factor, it is found that the greatest dimensions each tile can have is of 3 feet.
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- The widths of the walls are of <u>27 feet, 18 feet and 30 feet.</u>
- <u>The tiles must fit the width of each wall</u>, thus, the greatest dimension they can have is the greatest common factor of 27, 18 and 30.
To find their greatest common factor, these numbers must be factored into prime factors simultaneously, that is, only being divided by numbers of which all three are divisible, thus:
27 - 18 - 30|3
9 - 6 - 10
No numbers by which all of 9, 6 and 10 are divisible, thus, gcf(27,18,30) = 3 and the greatest dimensions each tile can have is of 3 feet.
A similar problem is given at brainly.com/question/6032811
Answer: V = (12in - 2*x)*(8 in - 2*x)*x
Step-by-step explanation:
So we have a rectangular cardboard sheet, and we cut four squares of side length x in each corner so we can make a box.
Remember that for a box of length L, width W and height H, the volume is:
V = L*W*H
In this case, the length initially is 12 inches, but we remove (from each end) x inches of the length, then the length of the box will be:
L = 12 in - 2*x
For the width we have a similar case:
W = 8in - 2*x
And te height of the box will be equal to x, then:
H = x
This means that the volume is:
V = (12in - 2*x)*(8 in - 2*x)*x
Here we can see the connection between the cutout and the volume of the box
Answer:
My sister and her sister and my mom are so adorable I love her and
Step-by-step explanation: was the name wrong I was talking to him about it haha is that a little
Kidsss was a great night out and this is what I
