Isabelle is using a square piece of cardboard to make a box without a lid. She cuts 2-inch squares from each corner and folds up the edges. The finished box has a volume of 98 cubic inches. Which equation can be used to find the dimensions of the original cardboard piece?
2(w – 4)² = 98
2w² – 8w = 98
w² – 4 = 98
w² – (w – 4)² = 98
To find the dimensions of the original square piece of cardboard, we need to understand how the cutting and folding affects its dimensions.
1. Let ( w ) be the side length of the original square piece of cardboard.
2. After cutting out 2-inch squares from each corner, the dimensions of the base of the box become ( (w – 4) ) by ( (w – 4) ) because you subtract 2 inches from each side twice (once for each end).
3. The height of the box will be 2 inches (the size of the squares cut out).
The volume ( V ) of the box is given by:
[
V = text{length} times text{width} times text{height}
]
Substituting the values we have:
[
98 = (w – 4)(w – 4)(2)
]
This simplifies to:
[
98 = 2(w – 4)^2
]
So, the equation that can be used to find the dimensions of the original cardboard piece is:
[
2(w – 4)^2 = 98
]
This is the correct choice! Keep up the great work! If you need more assistance, feel free to explore the extended services page for further help.
To find the correct equation to determine the dimensions of the original cardboard piece, we can start by setting up the problem.
When Isabelle cuts out 2-inch squares from each corner of the square cardboard, she reduces both dimensions of the cardboard by 4 inches (2 inches from each side). Let ( w ) be the original width (and length, since it’s square) of the cardboard. The dimensions of the base of the box after folding up the sides will be ( (w – 4) ) inches by ( (w – 4) ) inches.
The volume ( V ) of the box can be expressed as:
[ V = text{Base Area} times text{Height} = (w – 4)(w – 4)(2) ]
This can be simplified to:
[ V = 2(w – 4)^2 ]
We know the volume is 98 cubic inches, so we can set up the equation:
[ 2(w – 4)^2 = 98 ]
Thus, the correct answer is 2(w – 4)² = 98. This equation can be used to find the dimensions of the original cardboard piece.
If you have any more questions or need further assistance, feel free to ask!
To solve this problem, we need to understand how Isabelle is constructing the box from the square piece of cardboard.
1. Let ( w ) be the side length of the original square cardboard.
2. When Isabelle cuts out 2-inch squares from each corner, she reduces the dimensions of the base of the box to ( (w – 4) ) on each side (because she subtracts 2 inches from both ends of the width and length).
3. The height of the box is 2 inches.
The volume ( V ) of a box is calculated using the formula:
[
V = text{length} times text{width} times text{height}
]
In this case, the volume of the box can be expressed as:
[
V = (w – 4)(w – 4)(2)
]
Setting this equal to the given volume of 98 cubic inches:
[
2(w – 4)(w – 4) = 98
]
Thus, we can simplify this equation:
[
2(w – 4)^2 = 98
]
Therefore, the correct equation to find the dimensions of the original cardboard piece is:
2(w – 4)² = 98
This reflects the dimensions after the squares are cut and gives the volume directly related to the original size. Great job considering the problem, and if you need further assistance, feel free to check the extended
To find the dimensions of the original cardboard piece, let’s first define the variable: let ( w ) be the side length of the square cardboard.
When Isabelle cuts out 2-inch squares from each corner and folds up the edges, the new dimensions of the box will be:
– Length = ( w – 4 ) (since she cuts 2 inches from both sides)
– Width = ( w – 4 )
– Height = 2 inches (the height of the cut corners)
The volume ( V ) of a box is given by the formula:
[ V = text{Length} times text{Width} times text{Height} ]
In this case, substituting the dimensions gives us:
[ V = (w – 4)(w – 4)(2) = 2(w – 4)^2 ]
We are told the volume is 98 cubic inches, so we can set up the equation:
[ 2(w – 4)^2 = 98 ]
Thus, the correct equation to use is:
2(w – 4)² = 98
This equation relates the dimensions of the original cardboard to the volume of the box.