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if the volume is 357 what is the length, height width

cant find what length height and width

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Length, Width & Height to Volume Calculator - SensorsONE


This is the volume of the rectangular shaped box which corresponds to the dimensions entered for length, width and height. The volume ... Length, Width & Height to ...
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How to Calculate Your Volume - Mendocino County, CA


How to Calculate Your Volume Length X Width X Height = Volume 3 feet X 3 feet X 3 feet = 27 cubic feet 27 cubic feet = 1 cubic yard Example : 5 feet X 8 feet = 40 ...
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Conversion of Volume - Math is Fun


Conversion of Volume. ... Volume is Length by Width by Height : To convert Volume, remember that ... (ie once each for length, width and height)
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If the volume of a box is 357, how do you determine length ...


If the volume of a box is 357, how do you determine length, width and ... length_____ width 14 height 11 surface area_____ volume 2029.72 length*width*height ...
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Edurite.com - How to Find the Height of a Rectangular Prism


To be able to understand how to find the height of a rectangular ... Find the height of a rectangular prism with volume = 70 cm 3 ... Length or Width or Height, ...
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Dimensions - PLATT ELECTRIC SUPPLY


Volume 2—Commercial Distribution CA08100003E ... (717.3) 22.42 (569.5) 14.06 (357.1) FR57P 14 9.34 ... Dimensions Height Width Length FR818N 26 37 ...
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CHAPTER 1 - BASIC TERMS AND CALCULATIONS


CHAPTER 1 - BASIC TERMS AND CALCULATIONS. ... The height (h), base (b), width (w), length (1) ... Volume (V) = length x width x height = 5 m x 10 m x 2 m = 100 m 3.
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Find what are the possible dimensions of a box volume 357?


Find what are the possible dimensions of a box volume 357? ... just do length x width x height You could measure the length, width, and height of the box and multiply ...
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How to find length of object with width - The volume of a ...


... length w is the width and h is the height the volume of the rectangular solid in the illustration is 357 cubic centimeters find the width ... is given by the ...
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Baseline dimensions of the human vagina - Oxford Academic


... age with pelvic flexure width and the height with width at the ... volume (3.0 ml versus 5.0 ... vagina vary along its length. The width of the vagina is ...
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Suggested Questions And Answer :


find the center of mass of the solid of uniform density biunded by the graphs of the equations

x^2+y^2=a^2 is a cylinder with its circular cross-section in the x-y plane; the cylinder is sliced at an angle by the plane z=cy, forming a wedge. y=0 is the x-z plane and z=0 is the x-y plane; these planes and the conditions that y and z are positive impose further constraints on the shape. The cylinder is sliced in half by the x-z plane so the cross-section will be a semicircle and only the upper half is required. The semicircular end rests on the z=0 plane, and doesn't extend beyond the vertex of the wedge. The length of the wedge is ac, because the slope of the angle of the wedge is tan^-1(c) to the y axis and the radius of the wedge is a. Viewed from the side, along the x axis, the wedge is a right-angled triangle with sides a (height), ac (length) and hypotenuse a√(1+c^2). In everyday terms, the wedge looks like the end of a lipstick that has been sliced across. Going back to the side view of the wedge, we can see that the flat end has a height a. As we move along the wedge towards the point the height changes and shrinks to zero when we reach the vertex. The length at a point P(y,z) is a-y. This height is the distance of a chord to the circumference, starting at height a. The first task is to work out the area of a segment. We do this by subtracting the area of a triangle from the area of a sector, given the distance of the chord from the circumference. This distance is a-y, so the distance from the centre of the circle is, conveniently, y. y/a=cosø where ø is half the angle subtended by the arc of the sector. So ø=cos^-1(y/a) and the area of the sector is given by 2ø/2π=ø/π=A/πa^2, or A=a^2ø (ø is measured in radians). The area of the isosceles triangle formed by joining the radial ends of the arc of the sector is B=aysinø=y√(a^2-y^2), and the area of the segment is A-B=a^2cos^-1(y/a)-y√(a^2-y^2). (Note that when y=0, this expression becomes πa^2/2, the area of the semicircle.) The volume of the segment is (a^2cos^-1(y/a)-y√(a^2-y^2))dz where infinitesimal dz is the thickness of the segment. The mass of the segment is directly proportional to its volume because the density is uniform, so the volume is sufficient to represent the mass. The centre of gravity (COG) of the wedge can be found by summing moments about a point G(x,y,z) which is the COG. This sum has to be zero. Since a semicircle is symmetrical in the x-y plane about the y axis, we know the x coord of G must be zero. We need G(0,g,h) where g is the height along the z axis of the COG and h the z position. We need to find the COG of a segment first. The length of the chord of a segment is 2√(a^2-y^2). Earlier we discovered this when we were finding out the areas of the sector and isosceles triangle. Consider just two dimensions. The length of the chord is one dimension and if we create a rectangle from this length and an infinitesimal width dy, the area will be 2√(a^2-y^2)dy. If we think of the rectangle as having mass, its mass is proportional to its area, so area is a sufficient measure for mass. Let's call the COG of the segment point C(x,y)=(0,q), then the moment of the rectangle about the COG=C(0,q) is mass times distance from C=2(q-y)√(a^2-y^2)dy. If we sum the rectangles over the interval y to a we have 2∫((q-y)√(a^2-y^2)dy)=0 for y≤y≤a. The interval looks strange, but it means that y is just a general value, which is determined when we return to the wedge. What we are trying to do is to find q relative to y. We need to evaluate the definite integral, so we split it: 2q∫(√(a^2-y^2)dy)-2∫(y√(a^2-y^2)dy. Call these (a) and (b). To evaluate (a), let y=asinu, then dy=acosudu; √(a^2-y^2)=acosu and the integral becomes 2a^2q∫(cos^2(u)du). Since cos(2u)=2cos^2(u)-1, cos^2(u)=½(cos(2u)+1), so we have a^2q∫((cos(2u)+1)du)=a^2q(sin(2u)/2+u). Since sin(2u)=2sinucosu, this becomes a^2q(sinucosu+u)=a^2q(y/a * √(1-(y/a)^2)+sin^-1(y/a))=qy√(a^2-y^2)+a^2qsin^-1(y/a). Now we apply the limits for y: πa^2q/2-qy√(a^2-y^2)+a^2qsin^-1(y/a). To evaluate (b), let u=a^2-y^2, then du=-2ydy and the integral is +∫√udu = (2/3)u^(3/2)=(2/3)(a^2-y^2)^(3/2). Applying the limits we have: -(2/3)(a^2-y^2)^(3/2). (a)+(b)=πa^2q/2-qy√(a^2-y^2)+a^2qsin^-1(y/a)-(2/3)(a^2-y^2)^(3/2)=0. Multiply through by 6: 3πa^2q-6qy√(a^2-y^2)+6a^2qsin^-1(y/a)-4(a^2-y^2)^(3/2)=0. From this q=4(a^2-y^2)^(3/2)/(3πa^2-6y√(a^2-y^2)+6a^2sin^-1(y/a)). Fearsome though this looks, when y=0 and z=0, q=4a^3/(3πa^2)=4a/(3π), which is in fact the COG of a semicircle. When y=a and z=ac, q=0. Strictly speaking, we should write q(y) instead of just q, showing q to be a dependent variable rather than a constant. To find the COG of the wedge, we need to find the moments of the COGs of the segments. For a segment distance y above and cy along the z axis, we have the above expression for q. So q(y) is the y coord of the COG of the segment and cy is the z coord. The x coord is 0. The mass of the segment is (a^2cos^-1(y/a)-y√(a^2-y^2))dz, as we saw earlier, and this mass can be considered to be concentrated or focussed at the COG of the segment (0,q(y),cy). The COG of the wedge is at G(0,g,h) so the distance of the mass of the segment from G is g-q(y) and h-cy in the y and z directions. The y-moment is (g-4(a^2-y^2)^(3/2)/(3πa^2-6y√(a^2-y^2)+6a^2sin^-1(y/a))(a^2cos^-1(y/a)-y√(a^2-y^2))dz and the z-moment is (h-cy)(a^2cos^-1(y/a)-y√(a^2-y^2))dz. The sums of these individual components are zero. These lead to rather complicated integrals, requiring considerably more space than is available to work out, even if I knew how to do so!
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The volume of rectangular bar is 640 cubic units. The width is 3 units more than the height and the length is 1 unit more than three times the height. Find the dimensions of the bar.

Let the night be H, then width=H+3 and length=3H+1, so volume=640=H(H+3)(3H+1). H(3H^2+10H+3)=640; 3H^3+10H^2+3H=640, or 3H^3+10H^2+3H-640=0. By a bit of trial and error, we can find that H=5 satisfies the equation, so we can use synthetic division to reduce the cubic to a quadratic: 5 | 3 10.....3 -640 .....3 15 125  640 .....3 25 128 | 0 ⇒ 3H^2+25H+128=0, which doesn't factorise, so H=5 and width=8 and length=16 units. So the dimensions are length=16, width=8 and height=5 units.
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Using the numerical technique, calculate the volume under the plane z=ax^2+by and over the rectangle R={(x,y):0≤x≤3 and 1≤y≤2}.

The usual way to solve this is by using double integrals. The volume is given by the summation of zdxdy, which is the volume of a tiny cuboid with length z, width dx and height dy. The limits are those of the rectangular width and height. For the purposes of illustration S will be used as the integration sign and square brackets will be used for the limits: [lower,upper]. S[A,B](...)dv represents the integral between A and B of the expression in the round brackets with respect to variable v. When integration takes place the variable which is not the one implied in the integration is treated as a constant, so ax^2 is treated as a constant when integrating with respect to y. z=ax^2+by; volume=S[0,3](S[1,2]zdy)dx=S[0,3](S[1,2](ax^2+by)dy)dx= S[0,3](ax^2y+by^2/2)[1,2])dx=S[0,3](2ax^2-ax^2+2b-b/2)dx=S[0,3](ax^2+3b/2)dx= (ax^3/3+3bx/2)[0,3]=9a+9b/2=volume. The same answer is generated if x is integrated first before y.
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A box in the shape of a rectangular prism has a volume of 2500cm^3.

The volume is length times width times height=x(x+5)x/4=2500. Multiply through by 4: x^2(x+5)=10000 x^3+5x^2-10000=0=(x-20)(x^2+25x+500). The quadratic doesn't factorise, so x=20. Width=20cm, length=25cm, depth=5cm.
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what is the maximum size of the volume of a 8.5inches x 11inches paper

The total surface area of a cube is 6a^2, where a is the length of its side. A cube produces the maximum volume. If the area of the paper is completely used to form a cube 6a^2=11*8.5 and a=3.9476". The maximum volume is a^3=61.52 cu in approx. If, however, the area of the paper could be used to cover a sphere of radius r, then 4πr^2=11*8.5, and r=√11*8.5/4π=2.7277" producing a sphere with volume=4πr^3/3=85.01 cu in approx. A more practical solution consists of folding the paper to form an open box. If we use L, W and H to represent the length, width and height of the box we can write: volume V=LWH; W+2H=8.5, L+2H=11, so we can express L and W in terms of H: W=8.5-2H, L=11-2H. V=H(11-2H)(8.5-2H)=93.5H-39H^2+4H^3. dV/dH=93.5-78H+12H^2=0 at a maximum or minimum. The next derivative gives us -78+24H. First, solve the quadratic: H=(78±√(78^2-4*12*93.5))/24=(39±√399)/24=4.9146" or 1.5854". By substituting these values into the second derivative, we can see that H=1.5854" is a maximum, while H=4.9146" is a minimum, and results in negative W and V. From H we can find L=11-2H=7.8292" and W=5.3292", V=66.1482 cu in. The paper has 4 squares side 1.5854" cut out from the corners. The paper then folds up to make an open box with the volume V. On the graph the red curve is the volume function V(H). The maximum is clear to see at 66.1482 cu in. The sloping lines show the width and length (W(H) in blue and L(H) in green), both of which must be positive to form a box with H. So the volume only has meaning when V, H, W and L are all positive (all in quadrant 1). When V is at a maximum, H, W and L are all positive. H is positive by being to the right of the origin and V, W and L are positive because they're all above the H-axis.
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A box is to be made from a 12 by rectangular piece of aluminum by cutting 3x inches square from each corner. Express the volume of the box as a function of X,and find the domain

The cutouts reduce the length and width by 6x each: 12-6x and W-6x where W=width (not given). The height of the box is 3x when the sides and ends are folded up. The volume, V=3x(W-6x)(12-6x)=3x(12W-6xW-72x+36x^2)=36xW-18Wx^2-216x^2+108x^3=108x^3-18x^2(W+12)+36xW=18x(6x^2-(W+12)x+2W) cubic inches. The domain of x must be so as to keep the sides with non-zero length. Therefore 12-6x>0, 12>6x or x<2. And W-6x>0 so x Read More: ...

volume of a solid bounded by the gragh of the equation √(x/a)+√(y/b)+√(y/c) =1 and the coordinate planes

I read the equation as √(x/a)+√(y/b)+√(z/c)=1 √(z/c)=1-√(y/b)-√(x/a) and z/c=(1-√(y/b)-√(x/a))^2. If we consider a cuboid (rectangular prism) of height z and length and width dx and dy, then the volume of the cuboid is zdxdy. The double integral will be ∫∫zdxdxy, representing the sum of an infinite number of cuboids with an infinitesimal cross-section, is the volume of the required solid. In this double integral we replace z by c((1-√(y/b)-√(x/a))^2). Now we need to work out the limits. The positive square roots are implied. Put z=0 into the equation. What we get is the x-y plane and a view of the end face of the solid. The curve meets the y-axis at (0,b) and the y-axis at (a,0). These are the y and x intercepts. Therefore we have the x and y limits for the double integral. For z>0 (up to z=c) we have laminae building up from the x-y plane. The area of one infinitesimally thin lamina is the area under the curve bounded by the intercepts. The sum of the laminae is the volume of the required solid. When z=c we reach the limit of the solid. The question does not ask for evaluation of the double integral, but the process is to integrate c((1-√(y/b)-√(x/a))^2) wrt x treating y as a constant, between the limits x=0 and x=a; then the result is integrated wrt y between the limits y=0 and y=b. The limit of z=c isn't used because the equation defining z has been used as the integrand. The answer is c∫∫((1-√(y/b)-√(x/a))^2)dxdxy with 0≤x≤a and 0≤y≤b. The picture below shows in "net" form the general shape of the solid as folded-down axes to produce a 2-dimensional impression. By folding up the axes so that they form a 3-dimensional set of axes it's possible to imagine what the solid looks like: With Z-X as the base plane fold up Y-Z and X-Y so that the markers -2 and -1 in Y-Z are "stitched" to the markers 2 and 1 in X-Y  making the X, Y and Z axes orthogonal like the corner of a cube. The result is a sort of cobweb strung across the corner, where the curves are the strands of the cobweb fastened to the "walls". Now mentally fill in the body of the cobweb, which is the required solid.
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Use a right hand riemann sum to approximate the integr based off the values in the table. Integral from 0 to 14 of f(x) dx Table: (0,-1), (3,-2), (5,-1), (9,0), (13,-1), (14,0)

The Riemann sum is based on a simple concept: approximating the area under a graph using rectangles. The area of a rectangle is length times width, so width is the difference between two close values of x while length, or height,  is either the value of the function for the lower value of x or the higher. The rectangle that sits on the right above the graph will be an overestimate of the area while the rectangle sitting on the left under the graph will be an underestimate. There are 6 points so there will be 5 rectangles. If we take the first rectangle we have a right hand value of x of 3 and a right-hand value of f(x) of -2. So the width of the rectangle along the x axis is 3 and the height below the x axis is 2 making the area 2*3=6. Rectangle 1 has area 6. Rectangle 2 has width 5-3=2 and height=1 making the area=2. Rectangle 3 has width 9-5=4 and height=0 making area=0. Rectangle 4 has width 13-9=4 and height=1 making area=4. Rectangle 5 has width 14-13=1 and height=0 making area=0. Add these areas together: 6+2+0+4+0=12 square units.
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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!   
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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.  
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