A Piece Of String 25 1/2 Inches Long Will Be Cut Into 3/4 Inch Pieces. How Many Pieces Will There

A piece Of string 25 1/2 inches long will be cut into 3/4 inch pieces. How many pieces will there be? PLZzzzz Help reply Asap have a test...?

Number of pieces = Total length of string / length of each piece
= (25 1/2) / (3/4)
= (51/2) / (3/4)
= (51/2) x (4/3)
= 204/6
= 34 pieces

A piece of string that is 63 inches long is cut into 3 parts such that the length of the parts of the string are in the ratio 5 to 6 to 10.?

Let x = length of first string
Let y = length of second string
Let z = length of third string

The three strings came from a 63" in string before getting cut, so the sum of the three must be 63:

x + y + z = 63

and we know they are in a ratio of 5:6:10, so we can set up two additional equations from that:

x/y = 5/6
x/z = 5/10 = 1/2

Now that we have three unknowns and three equations, we can solve this. I'll start with solving the second and the third equation to be x in terms of either y or z.

x/y = 5/6
x = 5y/6

x/z = 1/2
x = z/2

Since both are equal to x, both are equal to each other:

5y/6 = z/2

and we can now solve this for z:

z = 5y/3

Now we have x in terms of z and z in terms of y , so we can substitute everything into the original equation and solve for y:

x + y + z = 63
z/2 + y + z = 63
z + 2y + 2z = 126
2y + 3z = 126
2y + 3(5y/3) = 126
6y + 15y = 378
21y = 378
y = 18

Now that we have y, we can work back to solve for x and z:

z = 5y/3
z = 5(18)/3
z = 5(6)
z = 30

x = z/2
x = 30/2
x = 15

Your lengths are:

15", 18", and 30"

A 28 inch long string is to be cut into two pieces, one piece to form a square and the other to form a cirlce?

length of wire, L = 28 inches

let the radius of circle = r in

let the side of square = s in

28 = circumference of circle + perimeter of square

28 = 2π r + 4s

4s = 28 - 2π r

s = 7 - (1/2)π r

now total area, A = area of circle + area of square

A = π r^2 + s^2

substituting s value

A = π r^2 + [ 7 - (1/2)π r]^2

since r ≥ 0 and [ 7 - (1/2)π r ] ≥ 0, the domain is 0 ≤ r ≤ 14/π

So the end points are 0 and 14/ π

now differentiate A with respect to r

dA/dr = 2π r + 2[7 - (1/2)π r](-(1/2)π

dA/dr = 2π r - 7π + π^2 r /2

dA/dr = π [ 2 r + π r/2 - 7]

equating dA/dr to zero

π [ 2 r + π r/2 - 7] = 0

2 r + π r/2 - 7 = 0

r (2 + π/2) = 7

r = 7 / ( π/2 +2) = 14 / (π + 4)

find out A at end points 0, 14 /π and critical point 14 / ( π +4)

A = π r^2 + [ 7 - (1/2)π r]^2

A(0) = 49 in^2

A(14/π) = π (14 /π)^2 + [ 7 - (1/2)π 14/π]^2

= 196 /π + (49 +49 - 98) = 196 /π

A(14/π) = 62.4 in^2

A(14 /(π+4) ) = π ( 14 /π+4)^2 + [ 7 - (1/2)π 14 /(π+4) ]^2

= 196 π / (π+4)^2 + 49 + 49 π^2/ (π+4)^2 - 98π / (π+4)

= 196 / (π +4)

= 27.5 in^2

so maximum area is 62.4 in^2 when r = 14/π and s = 7 - (1/2)π 14/π = 0

to maximize area wire should not cut, whole wire is to made into circle.

Minimum area, 27.5 in^2 occurs at r = 14/( π+4)= 1.96 in

s = 7 - (1/2)π(1.96) = 3.92 in

so wire should be cut at 2π(1.96) = 12.3 in to make circle and remaining piece of 15.7 in piece should be made into square of side 3.92 in

I inherited a 500 pound ball of string. Should I cut it into four inch pieces and market them as Stringettes?

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What would be the best way to cut a solid piece of wood into a perfect circle with a 20 inch diameter?

If you really want a perfect circle and a clean cut, you'll probably need to use a router with a narrow cutter, and make up (or buy) a trammel.

The trammel can be just a piece of wood, a bit more than 20" long that's bolted to the router baseplate at one end, and has a bolt at the other. A bolt of about 5mm diameter would be about right.

Drill a 5mm hole at the centre of where you want the 20 inch hole to be. Bolt the free end of the trammel through this hole - but not so tightly that it won't swing. Your router will now swing around the centre hole in a perfect circle. You may need to adjst where along the trammel's length the router sites to get the cutter exactly 20" from the centre point.

When it's all set up right, start your router and cut the hole. If your wood's thicker than a few millimetres, you'll probably need to cut a shallow groove first, then deepen it on a second pass, and repeat until you've cut all the way through.

You can buy adjustable trammels to fit your router if you don't want to make your own.

You can see some pics and description of a simple home-made trammel in use here: http://www.woodworkingtips.com/etips/etip030201wb.html

If a rope four yards long is cut into three equal pieces, each piece will be?

x = shortest piece x+2 = next piece x+4 = next piece x+6 = 4th piece upload all of them at the same time and they = 152, then remedy for x x + x+2 + x+4 + x+6 = 152 ; combine words 4x + 12 = 152 ; subtract 12 from both area 4x = one hundred forty ; divide both area by 4 x = 35 ft or A.