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# How many cubes each of length 2p meters can be packed into the rectangular container?

A rectangular container has length 15p meters, width 12p meters, and height 6p meters.

## Research, Knowledge and Information :

### a sealed rectangular container has a length 15p meters, width ...

a sealed rectangular container has a length ... cubes each of lenght 2p meters can be packed ... we want to know how many cubes fit into the rectangular ...

### A container has l= 15p meters, w= 12p meters, and h= 6p ...

A container has l= 15p meters, ... each of length 2p meters, can be packed into the container? ... How many cubes, each of length 2p meters, can be packed into the ...

### Small cubes with edge lengths of 1/4 inch will be packed into ...

How many small cubes can be packed in the rectangular ... How many small cubes can be packed in the ... The cube is then cut into cubes each with edge of length 1 ...

### Volume with Fractional Edge Lengths and Unit Cubes

Volume with Fractional Edge Lengths and Unit Cubes. ... How many small toys can be packed into the larger ... The height of the box is 3.1 meters, and the length is 2 ...

### How many .5 cm x.5 cm x .5 cm cubes are needed to completely ...

The side length of each cube is ... Small cubes with edge lengths of 1/4 inch will be packed into the right rectangular ... How many small cubes can be packed in ...

### How many 3-cm cubes can be packed in a rectangular box ...

240 cubes. Each edge of ... How many 3-cm cubes can be packed in a rectangular box ... 2 inches by 1 inch that can be packed into a cube shaped box ...

### How Many Cubes? - inside mathematics

A solid figure which can be packed without gaps or overlaps using n unit ... each cube only one ... How Many Cubes? How Many Cubes? Mean: 4.28, S.D.: 2.52. 0 500 1000 ...

### Cuboids, Rectangular Prisms and Cubes - Maths Resources

Cuboids, Rectangular Prisms and Cubes. ... and each face is a square. ... Area = 2 × Width × Length + 2 × Length × Height + 2 × Width × Height .

## Suggested Questions And Answer :

### How many cubes each of length 2p meters can be packed into the rectangular container?

??????? yu meen a BOX ??????? 15*12*6=1080 kubes but start point =2p, so divide bi 2^3=8 1080/8=135 but 1 side=15, not a multipel av 2, so kan oenlee yuze 14*12*6...1008 1008/8=126

### b.Choose an appropriate length, width, and height for your package so that it can fit the product that you’re shipping. Using these dimensions, what is the ratio of surface area to volume?

SA/Vol=2(LW+LH+HW)/LWH=2(1/H + 1/W + 1/L). If all three dimensions are the same (cube) then the above expression comes to 6/X where X is the side length. This is the same as 6X^2/X^3, which is the area of 6 square faces divided by the volume.  To answer the question we need to know the dimensions of what you're packing. The object doesn't have to be a regular shape, because all you need is to work out the the widest, longest and deepest (tallest) parts of the object. If you were doing this practically you could take two books (for example) and measure how far apart they would need to be in parallel to accommodate the length, width and height separately. Remember a cube gives the maximum amount of volume, so if you were packing many small objects into one box, they would probably fit best into a cube. So there's no answer to your question until you have something to pack!

### how many 1/2" candies will fit in a 41/2" x 4" bowl?

Assuming that 4" is the height of the bowl, we can conclude that it is ellipsoidal, like a rugby football, with the shorter axis (x axis) being 4.5" in diameter and its longer axis (y axis) needing to be determined from the given figures. Two domes are cut off each end of the ellipsoid leaving a 3.5" diameter circular top and a 3" diameter circular base. The general equation of the ellipse that will be used to model the bowl is x^2/2.25^2+y^2/a^2 (2.25" being the "radius" of the x axis, and a the radius of the y axis). The ellipse has its centre at the x-y origin. To find a we work out in terms of a where the base of the top dome has a width of 3.5" and where the base of the bottom dome has a width of 3". Putting x=1.75 (half of 3.5) in the equation of the ellipse, we get y=sqrt(a^2(1-1.75^2/2.25^2))=4asqrt(2)/9. Similarly for the base dome: y=asqrt(1-1.5^2/2.25^2)=-asqrt(5)/3, negative because it's below the centre or origin. The difference between these two y values is the height of the bowl, 4", so 4asqrt(2)/9+asqrt(5)/3=4, and a=36/(4sqrt(2)+3sqrt(5))=2.9114 approx.  The volume of the bowl is found by considering thin discs of radius x and thickness dy so that we can use the integral of (pi)x^2dy (the volume of a disc) between the y limits imposed by the heights of the two domes. x^2=5.0625(1-y^2/a^2). Because a is a complicated expression, we'll just use the symbol for it. We can also write 5.0625 as 81/16. The integrand becomes (81(pi)/16)(1-y^2/a^2)dy, with limits y=-asqrt(5)/3 and 4asqrt(2)/9. The integrand is simply the sum of the volumes of the discs between the bases of the two domes. Integration gives us (81(pi)/16)(y-y^3/(3a^2)) between the limits, which I calculated to be 53.39 cu in approx. How many candies fit into this volume? We know the length of the candies is 1/2", but we don't know any other dimensions, so we don't know the volume of each candy. Also, the alignment of candies will improve the number of candies fitting into the bowl, but if this cannot be arranged, we have to assume they will be randomly orientated. We could make the assumption that they are cone-shaped with height 0.5" and base diameter 0.25" (radius 0.125" or 1/8") giving them a volume of (1/3)(pi)0.125^2*0.5=0.0082 cu in each approximately. So divide this into 53.39 and we get about 6,526, with very little space between the candies. A more realistic figure can be obtained by finding out how many candies fill a cubic inch when tightly packed. Two will fit lengthwise into an inch. Think of the candies as cuboids 1/4" square and 1/2" long (volume=1/4*1/4*1/2=1/32 cu in). That gives you 32 in a cubic inch, over 1,700 in a bowl. The cuboid is a sort of container that will hold just one candy and allow it some lengthwise freedom of movement. At the other extreme think of the candies as 1/2" cubes (volume=1/8 cu in) then only 8 will fit into a cubic inch and only about 427 will fit into the bowl. But the candies have greater freedom of movement in a cubic container. Let's say you can get 100 into one cubic inch packed tightly, then you would get about 5,340 in the bowl. Get a small box and pack in as many candies as you can. Measure the volume of the box (length*width*height). If N is the number of candies, then each has an average effective volume of (box volume)/N. Use this number to divide into the volume of the bowl.

### a plastic container company design a container with a lid to hold one piece of pie

Let the radius of the circular pie be r and the thickness t. Also let the number of slices be N. We can then find a formula for the dimensions of the container. The length of the isosceles triangle forming the base and lid of the container is the height of the triangle, which is equal to r. The angle of the isosceles triangle is 360/N. To find the length of the isosceles triangle's base, drop a perpendicular from the apex of the triangle on to its base. The perpendicular will bisect the apex angle, so we have two back to back right-angled triangles with one angle equal to 360/2N=180/N. The half-length of the base of the triangle=rtan(180/N). So the base of the triangle is 2rtan(180/N). We then need to decide how many slices we are going to put into the container. Let S be the number of slices. So the height of the container is St at least to accommodate the slices. Now we can calculate the area of the prism. It has 3 rectangular sides and 2 triangles for top and bottom. The long side of the container has length L given by r/L=cos(180/N) so L=r/cos(180/N).  Rectangular areas: StL=rSt/cos(180/N) (two of these), base * thickness=2rtan(180/N)*St=2rSttan(180/N). Triangular base and lid: r^2tan(180/N) (two of these). The total area is the sum of these 5 shapes: 2r(St(sec(180/N)+tan(180/N))+2rtan(180/N)). If the container is to hold the entire pie then S=N. Now put in some figures: r=23cm, t=0.8cm, N=S=6. 180/N=30. tan30=0.5774, cos30=0.8660. St=4.8cm. More to follow...

### if i have a cube with a volume of 216, what is the side length?

V = a^3 216 = a^3 a = 3thrt(216) a = 6 216 = 6^3 216 - 6*6*6 216 = 36*6 216 = 216 a rectangular prizm  8*4*9 = 216                                   8*2*18 = 216

There are 180=6*6*5 cubes in all. To find out how many cubes will have no wax at all, reduce the sides of the large cuboid (rectangular prism) by two cubes for each face. This will leave a core of cubes 4*4*3=48. These cubes will not be covered in wax. So the remaining 180-48=132 cubes will have at least one face covered by wax. (It may help to think of a cube side 3. This contains 27 unit cubes, but only the central cube is unexposed and (3-2)^3=1. The dimensions of the faces are each reduced by 2.)

### If the volume of a cube is 216 cubic meters, what is the length (in meters) of each side?

The side length is the cube root of 216=6m, because 6*6*6=216.

### If the volume of a cube is 216 cubic meters, what is the length (in meters) of each side?

Let a=length of cube side, a^3=216 so a=6 (6*6*6=216) metres.

### How do I solve width cubed equals Eight meters cubed minuse one meter?

I think it's easier to try some numbers, 2 x 4 x1 =8 W=2m, L=4m, H = 1m, V=8m^3

### The owner of rectangular lot wants to put a fence around the area using 120 meter barb wire. The length of the rectangular lot is x. Express the area in square meters of the field as a function of x.

The wire represents the perimeter of the lot=2(length+width)=120, or length+width=60, so, if length=x, x+width=60 and width=60-x metres. The area is length times width=x(60-x)=60x-x^2. We can write f(x)=60x-x^2 where area=f(x) in square metres.