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# How many boxes of hex nuts would this be?

Suppose that you were given the job of shipping 25000 hex nuts to a customer. How many boxes of hex nuts would this be? All you have is a hanging scale and a barrel of hex nuts. Describe how you could determine the proper number of pieces without physically counting them out. [info: 8.77 boxes of hex nuts per kg, 114 grams of hex nuts per box, 28.4 grams of hex nuts, and the average mass of one piece of a hex nut is 1.14 grams]

## Research, Knowledge and Information :

### HEX NUT STAINLESS (EACH OR BOX) - Hamilton Marine

NUTS; HEX NUTS; HEX NUT STAINLESS (EACH OR BOX) HEX NUT STAINLESS (EACH OR BOX) no reviews for this product. Login to place a review. HEX NUT STAINLESS (EACH OR BOX)

### Hex Nuts - Nuts - Fasteners - Hardware - The Home Depot

Everbilt 1/2 in.-13 tpi Zinc-Plated Hex Nut (50-Piece per Box) Model# 804710 \$ 8 21 /box. Free shipping with \$45 order. ... The 1/4-20 Zinc-Plated Steel Hex Nuts ...

### HEX NUT BRONZE (EACH OR BOX) - Hamilton Marine

NUTS; HEX NUTS; HEX NUT BRONZE (EACH OR BOX) HEX NUT BRONZE (EACH OR BOX) no reviews for this product. ... HEX NUT BRONZE (EACH OR BOX) Description: Item: HM ...

### Unit 5 – Counting Particles - Objectives - BHS Chem - home

Unit 5 – Counting Particles - Objectives Review Concepts a. ... many boxes of hex nuts would this be? All you have is a hanging scale and a barrel of hex nuts.

### template

If you had 1.00 kg of each kind of hardware, how many boxes of each would you have? Show your work below with units. ... How many boxes of hex nuts would this be?

### Everbilt 1/2 in.-13 Stainless Steel Hex Nut (50 per Box ...

Get Crown Bolt 1/2 in. Coarse Stainless-Steel Hex Nuts (50-Pack) 31930, ... 1/2 in.-13 Stainless Steel Hex Nut (50 per Box) \$ 22 37 /box Choose Your Options.

### Activity: Relative Mass - Mr Montero

Activity: Relative Mass . Purpose . The purpose is to determine the relative mass of different kinds of hardware and ... How many boxes of hex nuts would this be?

### Name Date Pd Lab: Relative Mass - Mr Montero

Name Date Pd Lab: Relative Mass Purpose The purpose is to determine the relative mass of different kinds of hardware and ... How many boxes of hex nuts would this be?

### 1 3/8 - 12, HEX Nuts, Box of 2 - eBay

... 12, HEX Nuts, Box of 2 in Business & Industrial, Fasteners & Hardware, Fastener Nuts, Hex Nuts | eBay. Skip to main ... Details about 1 3/8 - 12, HEX Nuts, Box of 2.

## Suggested Questions And Answer :

### How many boxes of hex nuts would this be?

me hope sumwon spesifi wot SIZE nut em want. If nut dont fit on bolt, it be werthless Basik idea be sumwon tell yu 1 hex nut (av the rite size) wae sumthun like 1 miligram then 25000 nuts=25 grams so skoop out nuts til skale sae 25 grams, then add a fyu more tu plae safe

### Maths mate blue term two sheet five number 32

If this is like a crossword puzzle using numbers instead of letters, you will need to post a picture image to this website with the clues or take a photo of the puzzle and upload it to the website (perhaps by sending it to the server first). If you can't do that, you will need to copy the clues and submit them with details of the puzzle like how many rows and columns; where the black boxes are, using a row-column system. For example, if the puzzle has 10 rows and 10 columns, use letters A to J for the rows and 1 to 10 for the columns. Then you can say where the black boxes are by using the reference: e.g., a black box in the top left hand corner would be A1 black; one in the bottom right-hand corner would be J10 black. Or you could just list the blacks: BLACKS in A1, J10, etc. Then you need to say where the answers go: 1A would mean 1 across; 1D would mean 1 down; so you would write in the ACROSS clues: 1. A2 (5) meaning that 1 across starts in A2 and is 5 digits long; in the DOWN clues: 1. A2 (8) meaning that 1 down starts in A2 and is 8 digits long. It will take you some time and effort to do this, so a picture would be better and you would avoid mistakes. Anyone reading your description would be able to construct the puzzle themselves from the information, and then go about solving the clues.

### 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.

### how to calculate cargo area of a car?

If you mean the volume rather than area you need to fill the space with something that will not damage the interior of the space and will not escape, so water is out of the question for both reasons. You could use shredded paper, rags, newspaper, etc., which would be easy to remove afterwards. Pack as much space as possible. Then empty the filling material into a box of regular dimensions that you can measure. If there's too much material, fill the box to capacity and pack to the same density as you used to pack the cargo space. The volume of the box (length times width times height) gives you the volume of material removed. Empty the box, and then transfer as much of the filling as you can into the box from the cargo space until the box is full again or you've removed all the filling. If the box isn't full you can measure how far the filling rises on the box and so determine the volume. Best to measure the distance from the top layer of material to the top of the box, and check that this distance is the same all round, then subtract this measure from the height of the box. Keep a note of how many times you've filled the box and what the volume of the last quantity of filling was. Multiply the volume of the box by the number of times you filled it and add the final volume of the partially filled box. This should give you an idea of the capacity as long as you packed the box to about the same density as you packed the cargo space. In practice of course, you could never fill the entire cargo space with stuff and a few additional measurements such as maximum height, width and depth may be necessary for ensuring goods will fit into the cargo space.

### digital picture

(a) So the number of pixels in the border is: (x + 2)(x + 2) - (x - 2)(x - 2) = x^2 + 4x + 4 - (x^2 - 4x + 4) = 8x (b) So when x = 5, (a) = 8 x 5 = 40 (c) It would be best to do this on some kind of graph paper: Draw a square which is 7 cm long and 7 cm wide (obviously it doesn't matter what size the squares are, I’m just using cm for convenience), then inside this box draw another box that is 3 cm long and 3 cm wide, then count how many square cm there are in the border area (the area between the big box and the small box). This will come to 40 pixels, just as in part (b) (d) Simply place the value x = 500 into the original equation that you worked out, i.e. no. of pixels = 8x = 8 x 500 = 4000

### 13.5' square feet equals how many inches

Let's see if we can give you a better understanding of feet, square feet, inches and square inches. You can easily change one foot into inches by just multiplyng by 12. You can't convert a square foot into inches, but you can convert it to square inches. Here's how. You have a box of 1" tiles, so each tile is a square of side 1". Picture a square with side 1'. How many 1" tiles will you need to fill it? You need 12 along the top, but you also need 12 down the side, so you need 12 rows of 1" tiles, which is 12 times 12=144. That's how you convert 1 sq ft to sq inches. You multiply by 144. So 3.5 sq ft=3.5*144=432+72=504 sq in. You would need 504 1" tiles to fill a 3.5 ft square.

### how many gallon,s in a 5'long 2'6" wide 3'10" tank

is it shaped like a box...90deg angels at all korners????? then volyume=leng*wide*deep 5ft*(2ft 6in)*(3ft 10 inch) 2ft 6=2.5 ft 3 ft 10=3 & 10/12or 3.8333333 gt 5*2.5*3.8333333=47.916666666 kubik ft 1 kubik ft=7.480681 gallon volyume=358.449 gallon

### How can one can connect nine dots, configured into a box, with three (3) straight lines?

To connect them continuously there are many ways. Number the dots in rows: row1: 1 2 3; row2: 4 5 6; row3: 7 8 9. Rectangular connection: 1-2-3-6-9-8-7-4-5(-1); 3-2-1-4-7-8-9-6-5(-3); etc. Zigzag or Snake: 1-4-7-8-5-2-3-6-9(-1); 1-2-3-6-5-4-7-8-9(-5-1); etc. Triangular connection: 4-7-8-9-5-1-2-3-6(-5-4) The brackets just show how you join the last dot to the first. You can rotate all these patterns or start at different points to make the same patterns with different numbered dots, Continuous connection with 4 straight lines: Although we're going to connect only 9 dots, to illustrate what to do I've started with a square of 4 by 4 dots and numbered them 1 to 16 in rows: 1-4, 5-8, 9-12, 13-16. Join the dots in the following order: 8 12 (16) 11 6 (1) 2 3 4 7 10. The 9 dots you want to join are in 3 rows of 3 inside the 4X4 square: row 1: 2 3 4; row 2: 6 7 8; row 3: 10 11 12. The numbers in the brackets don't belong to the 3X3 square, because they're just to show which way you draw. You can see that the numbered dots in this smaller square all appear just once and you have used just 4 lines to join them without having to lift your pen off the paper!

### a box has drill bits.sizes increase to 5/8 cm in steps of 1/16.how many drill bits are in box?

Assuming that the lowest is 1/16cm, the next would be 1/8cm, then 3/16cm. The largest is 5/8cm which is 10/16cm. So the drill bits range in size from 1 to 10 sixteenth of a cm. Therefore there must be 10 drill bits, assuming that no more than one in each size is needed.

### How many pounds of Cashews that cost \$4.25 per pound should be mixed with 4 pounds of m&m's that cost \$1.24 per pound to obtain a mixture that cost \$3.30 per pound?

How many pounds of Cashews that cost \$4.25 per pound should be mixed with 4 pounds of m&m's that cost \$1.24 per pound to obtain a mixture that cost \$3.30 per pound? Let there be C pounds of Cashew nuts Let there be M=4 pounds of M&M's Total weight of mixture is W = C + M = C + 4 Total cost of mixture is P = cost of cashews + cost of M*M's P = C*4.25 + M*1.24 = 4.25C + 4*1.24 = 4.25C + 4.96 And total cost of mixture is W*3.30 P = 3.30W = 4.25C + 4.96 3.30(C + 4) = 4.25C + 4.96 3.30C + 13.20 = 4.25C + 4.96 8.24 = 0.95C C = 8.24/0.95 = 8.673 Answer: weight of Cashews to be mixed with 4 lbs M&M's is: 8.673 lbs