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What is the surface area of a prism with a length of 15mm a height of 4 mm and a width of 3mm?
The surface area of a prism is twice the area of the end plus the area of all the sides. For a prism with a polygon cross-section the area of all the sides if given by the perimeter of the cross-section polygon multiplied by the length of the prism. Assuming you have a prism of cross-section that is a rectangle with sides 4 mm and 3 mm, and is 15 mm long (ie it is an oblong), its surface area is: → surface_area = 2 × (4 mm × 3 mm) + 2 × (4 mm + 3 mm) × 15 mm → surface area = 2 × 12 mm² + 14 mm × 15 mm = 234 mm²
Find the surface area of each solid round to the nearest tenth is necessary 15 in by6in?
find the surface area of each solid to the nearest tenth.(use n=3.14). a=10 mm b=14 mm c=14 mm
Find the surface area of a shot put with a diameter of 90 mm?
2700
What is the surface area of a 12mm cube?
6*12*12 = 864 square mm
What is the area of 8 mm 10 mm 10 mm?
If the shape is a triangle, then the area is approx 36.7 square mm.
What is the surface area of 5 mm 4 mm 2 mm rectangular prism?
76 square mm
What is the surface area of a rectangular prism that has side lengths of 3 mm and 5 mm and 4mm?
2x(3x5+3x4+4x5) = 94 sq mm
What is the surface area of a rectangular prism with a length of 3 mm a width of 5 mm and a height of 8 mm?
It is: 2(3*5)+2(5*8)+2(3*8) = 104 square mm
What is the surface area of a prism with a length of 15mm a height of 4 mm and a width of 3mm?
The surface area of a prism is twice the area of the end plus the area of all the sides. For a prism with a polygon cross-section the area of all the sides if given by the perimeter of the cross-section polygon multiplied by the length of the prism. Assuming you have a prism of cross-section that is a rectangle with sides 4 mm and 3 mm, and is 15 mm long (ie it is an oblong), its surface area is: → surface_area = 2 × (4 mm × 3 mm) + 2 × (4 mm + 3 mm) × 15 mm → surface area = 2 × 12 mm² + 14 mm × 15 mm = 234 mm²
What is the surface area of rectangular prism 2mm x6mmx2mm?
Surface area = 2*(2*6 + 6*2 + 2*2) = 2*(12 + 12 + 4) = 2*28 = 56 square mm.
What is the length of ta rectangular prism in mm?
9005493
What is the surface area of a cylinder if the diameter is 124 mm and the length is 170 mm?
A cylinder if the diameter is 124 mm and the length is 170 mm has a surface area of 90,377.34mm2
What is the area of a 1.89 L rectangular container?
There is no such thing as a rectangular container since a rectangle is a 2-dimensional concept whereas a container is 3-dimensional. It could be 12 cm x 12 cm x 13.1250 cm which has a surface area of 918 sq cm. or 1 cm x 1 cm x 1890 cm with a surface area of 7562 sq cm or 1 mm x 1 mm x 189 metres with a surface area of 75600.02 sq cm.
What is the surface area of a 100 mm dish?
The surface area of a 100 mm dish is approximately 314 square millimeters.
how to calculate surface area of ISMB 200?
To calculate the surface area of an ISMB 200, you need to know its dimensions. The ISMB 200 is an Indian Standard Medium Weight Beam, and its dimensions are: Height: 200 mm Width: 100 mm Flange thickness: 10.2 mm Web thickness: 5.7 mm Length: Typically supplied in standard lengths of 6, 9, 12, or 15 meters To calculate the surface area of an ISMB 200, you need to find the surface area of each of its parts separately and then add them up. The surface area of the flanges and the web can be calculated as follows: Surface area of each flange = Width × Flange thickness × 2 = 100 mm × 10.2 mm × 2 = 2040 mm² Surface area of the web = Height × Web thickness = 200 mm × 5.7 mm = 1140 mm² Total surface area of an ISMB 200 = Surface area of both flanges + Surface area of the web = (2040 mm² + 2040 mm²) + 1140 mm² = 5220 mm² Therefore, the surface area of an ISMB 200 is 5220 square millimeters.
L6.9 mm w8.2 mm h14mm by using Surface Area?
Total surface area = 2*(LW + WH + HL) = 538.76 sq mm.
What is the surface area of a jewel case that measures 140 mm 120 mm and 9 mm?
Assuming that the jewel case is in the shape of a cuboid, the total surface area is 38280 sq mm.
