What is the surface area of a rectangular prism whose dimensions are 334 dm×412 dm×5 dm in square decimeters?(1 point)

Question

Responses

11614 dm2
116 Start Fraction 1 over 4 End Fraction dm squared

8438 dm2
84 Start Fraction 3 over 8 End Fraction dm squared

5818 8 dm2
58 Start Fraction 1 over 8 End Fraction dm squared

16 dm

Answer 1

11614 dm2

Answer 2

Apply the technique of using nets to find the surface area of a turtle’s tank, which measures 33.93 inches long by 15.81 inches wide by 17.69 inches tall. What is the surface area of the tank if the tank does not include a top cover? Round the answer to the nearest hundredth.(1 point)
Responses

9,489.51 in2
9,489.51 in squared

1,416.33 in2
1,416.33 in squared

2,832.67 in2
2,832.67 in squared

2,296.23 in2

Answer 3

9,489.51 in2 (rounding off to the nearest hundredth)

Answer 4

A shipping box has a length of 12.5 inches, a height of 9.5 inches, and a width of 3.0 inches. What is the surface area of the box? Hint: Draw a net.(1 point)
Responses

369.5 square inches
369.5 square inches

341 square inches
341 square inches

184.75 square inches
184.75 square inches

511 square inches

Answer 5

To find the surface area of the box, we calculate the area of each side separately and then add them together.

Given dimensions:
Length = 12.5 inches
Height = 9.5 inches
Width = 3.0 inches

1. Front and back:
- Area of front = Length * Height = 12.5 * 9.5
- Area of back = Length * Height = 12.5 * 9.5
Total for front and back = 2 * (12.5 * 9.5)

2. Top and bottom:
- Area of top = Length * Width = 12.5 * 3.0
- Area of bottom = Length * Width = 12.5 * 3.0
Total for top and bottom = 2 * (12.5 * 3.0)

3. Sides:
- Area of side1 = Height * Width = 9.5 * 3.0
- Area of side2 = Height * Width = 9.5 * 3.0
Total for sides = 2 * (9.5 * 3.0)

Now, add the areas calculated above to get the total surface area.

Surface area of the box = Total for front and back + Total for top and bottom + Total for sides
Surface area = 2 * (12.5 * 9.5) + 2 * (12.5 * 3.0) + 2 * (9.5 * 3.0)
Surface area = 237.5 + 75 + 57
Surface area = 369.5 square inches

Therefore, the surface area of the box is 369.5 square inches.

Answer 6

Jeni is painting a flattened shoebox for a school project. The shoebox measures 72.5 cm x 47.5 cm x 104 cm. What is one area that would be identified on the box’s net?(1 point)
Responses

358,150 cm2
358,150 cm squared

31,847.5 cm2
31,847.5 cm squared

7,540 cm2
7,540 cm squared

224 cm2

Answer 7

To find the area that would be identified on the shoebox's net, you can find the area of each side of the shoebox first, and then identify one of those areas. For a shoebox, there are six sides: top, bottom, front, back, left, and right.

Given dimensions:
Length = 72.5 cm
Width = 47.5 cm
Height = 104 cm

1. Top and bottom:
- Area of the top and bottom = Length * Width = 72.5 cm * 47.5 cm

2. Front and back:
- Area of the front and back = Length * Height = 72.5 cm * 104 cm

3. Left and right:
- Area of the left and right side = Width * Height = 47.5 cm * 104 cm

Now, calculate the areas based on the given dimensions:
- Top and bottom area = 72.5 cm * 47.5 cm = 3443.75 cm²
- Front and back area = 72.5 cm * 104 cm = 7540 cm²
- Left and right area = 47.5 cm * 104 cm = 4940 cm²

One area that would be identified on the shoebox's net is 7540 cm².

Therefore, the correct response is 7,540 cm2 (7,540 cm squared).