The sum of the percent frequencies for all classes will always equal:
1 answer:
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The number of tiles needed to cover the surface area of the box excluding the bottom is: 1,046 tiles.
<h3>What is the Surface Area of a Box?</h3>The surface area of a box is the area surrounding all its faces. A box has 6 rectangular faces. Therefore, the total surface area of the box equals the sum of all 6 rectangular faces.
<h3>What is the Surface Area of a Box?</h3>SA = 2(lw + lh + hw), where:
- l = length
- w = width
- h = height of the box
The image attached below shows the box Dmitri wants to cover. Since the bottom of the box would be excluded, therefore:
The surface area to be covered = surface area of the box - area of the bottom rectangular face
The surface area to be covered = 2(lw + lh + hw) - (l)(w)
l = 26
w = 15
h = 8
Substitute
The surface area to be covered = 2(l×w + lh + hw) - (l)(w) = 2·(15·26+8·26+8·15) - (26)(15) =
The surface area to be covered = 1436 - 390 = 1,046 cm
Area of one tile = 1 cm square
Number of tiles needed = 1,046/1
Number of tiles needed = 1,046 tiles.
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Let
x---------> the length of the rectangle
y--------> the width of the rectangle
we know that
A=84 ft²
[area of rectangle]=x*y-----> 84=x*y-----> equation 1
x=y-5------> equation 2
substitute 2 in 1
84=[y-5]*y-----> 84=y²-5y---------> y²-5y-84=0
using a graph tool-----> to resolve the second order equation
see the attached figure
the solution is
y=12
x=y-5-----> x=12-5-----> x=7
the answer is
the length of the rectangle is 7 ft
the width of the rectangle is 12 ft
x---------> the length of the rectangle
y--------> the width of the rectangle
we know that
A=84 ft²
[area of rectangle]=x*y-----> 84=x*y-----> equation 1
x=y-5------> equation 2
substitute 2 in 1
84=[y-5]*y-----> 84=y²-5y---------> y²-5y-84=0
using a graph tool-----> to resolve the second order equation
see the attached figure
the solution is
y=12
x=y-5-----> x=12-5-----> x=7
the answer is
the length of the rectangle is 7 ft
the width of the rectangle is 12 ft