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Andreas93 [3]
3 years ago
12

gina sees a rectangular quilt with an area of 3,528 square inches.The length of the quilt is 72 inches. what is the width of the

quilt
Mathematics
2 answers:
yulyashka [42]3 years ago
8 0

Answer:

Step-by-step explanation:

90 x 108 = 10,368 x .015 = 155.52

There is a $35 minimum for all quilting.

Thread will be matched as closely as possible to your quilt. There will be a $5.00 minimum charge on thread. Variegated threads will be a $7.50 charge. We like to use cone threads so please do not bring small spools to be used.

Battings:

We keep the following batts in stock as prepackaged, precut sizes:

Hobbs Poly Down - White and Black, Hobbs Heirloom (80%cotton / 20% poly) Cream, Hobbs Organic Cotton without Scrim (this should not be used on light colored fabrics), Hobbs Wool, Dream Poly—White and Black, Warm and Natural—cotton Cream, Dream Cotton—White and Cream

The following are sizes available:

Crib 46 x 60 Full 92 x 96 Craft 36 x 46

Throw 60 x 60 Queen 92 x 108

Twin 72 x 92 King 121 x 121

Linings:

If you supply the lining, please be sure it measures at least 4” larger than the length and width of the quilt top. We can also supply the lining from our own fabric at the retail cost. If the lining requires piecing, we can do this for a $10 per seam fee.

Bias Bindings can be put on at Quilt ‘N Stitch. We provide the following services. (You must provide the fabric for the binding.)

1. The fee to half bind in $.15 per linear inch. (Half bind is where we attach the binding to the quilt front and you sew down the back.

2. If we do complete binding, finishing on the machine the fee is $.20 per linear inch.

3. If we do complete binding, finishing by hand the fee is $.25 per linear inch..

Yardage required for binding: 60” x45” — 1/2 yard: 90” x 72” — 5/8 yard: 81” x 96” — 2/3 yard: 90” x 108” — 3/4 yard: 120” x 120” — 1 yard.

zloy xaker [14]3 years ago
5 0

Answer:

49 inches is the width

Step-by-step explanation:

3,528÷72=49

49*72=3,528

hope this helps <3

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So, the cost of y nets = y ( cost of 1 net = y ( $23)  = 23 y

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Use undetermined coefficient to determine the solution of:y"-3y'+2y=2x+ex+2xex+4e3x​
Kitty [74]

First check the characteristic solution: the characteristic equation for this DE is

<em>r</em> ² - 3<em>r</em> + 2 = (<em>r</em> - 2) (<em>r</em> - 1) = 0

with roots <em>r</em> = 2 and <em>r</em> = 1, so the characteristic solution is

<em>y</em> (char.) = <em>C₁</em> exp(2<em>x</em>) + <em>C₂</em> exp(<em>x</em>)

For the <em>ansatz</em> particular solution, we might first try

<em>y</em> (part.) = (<em>ax</em> + <em>b</em>) + (<em>cx</em> + <em>d</em>) exp(<em>x</em>) + <em>e</em> exp(3<em>x</em>)

where <em>ax</em> + <em>b</em> corresponds to the 2<em>x</em> term on the right side, (<em>cx</em> + <em>d</em>) exp(<em>x</em>) corresponds to (1 + 2<em>x</em>) exp(<em>x</em>), and <em>e</em> exp(3<em>x</em>) corresponds to 4 exp(3<em>x</em>).

However, exp(<em>x</em>) is already accounted for in the characteristic solution, we multiply the second group by <em>x</em> :

<em>y</em> (part.) = (<em>ax</em> + <em>b</em>) + (<em>cx</em> ² + <em>dx</em>) exp(<em>x</em>) + <em>e</em> exp(3<em>x</em>)

Now take the derivatives of <em>y</em> (part.), substitute them into the DE, and solve for the coefficients.

<em>y'</em> (part.) = <em>a</em> + (2<em>cx</em> + <em>d</em>) exp(<em>x</em>) + (<em>cx</em> ² + <em>dx</em>) exp(<em>x</em>) + 3<em>e</em> exp(3<em>x</em>)

… = <em>a</em> + (<em>cx</em> ² + (2<em>c</em> + <em>d</em>)<em>x</em> + <em>d</em>) exp(<em>x</em>) + 3<em>e</em> exp(3<em>x</em>)

<em>y''</em> (part.) = (2<em>cx</em> + 2<em>c</em> + <em>d</em>) exp(<em>x</em>) + (<em>cx</em> ² + (2<em>c</em> + <em>d</em>)<em>x</em> + <em>d</em>) exp(<em>x</em>) + 9<em>e</em> exp(3<em>x</em>)

… = (<em>cx</em> ² + (4<em>c</em> + <em>d</em>)<em>x</em> + 2<em>c</em> + 2<em>d</em>) exp(<em>x</em>) + 9<em>e</em> exp(3<em>x</em>)

Substituting every relevant expression and simplifying reduces the equation to

(<em>cx</em> ² + (4<em>c</em> + <em>d</em>)<em>x</em> + 2<em>c</em> + 2<em>d</em>) exp(<em>x</em>) + 9<em>e</em> exp(3<em>x</em>)

… - 3 [<em>a</em> + (<em>cx</em> ² + (2<em>c</em> + <em>d</em>)<em>x</em> + <em>d</em>) exp(<em>x</em>) + 3<em>e</em> exp(3<em>x</em>)]

… +2 [(<em>ax</em> + <em>b</em>) + (<em>cx</em> ² + <em>dx</em>) exp(<em>x</em>) + <em>e</em> exp(3<em>x</em>)]

= 2<em>x</em> + (1 + 2<em>x</em>) exp(<em>x</em>) + 4 exp(3<em>x</em>)

… … …

2<em>ax</em> - 3<em>a</em> + 2<em>b</em> + (-2<em>cx</em> + 2<em>c</em> - <em>d</em>) exp(<em>x</em>) + 2<em>e</em> exp(3<em>x</em>)

= 2<em>x</em> + (1 + 2<em>x</em>) exp(<em>x</em>) + 4 exp(3<em>x</em>)

Then, equating coefficients of corresponding terms on both sides, we have the system of equations,

<em>x</em> : 2<em>a</em> = 2

1 : -3<em>a</em> + 2<em>b</em> = 0

exp(<em>x</em>) : 2<em>c</em> - <em>d</em> = 1

<em>x</em> exp(<em>x</em>) : -2<em>c</em> = 2

exp(3<em>x</em>) : 2<em>e</em> = 4

Solving the system gives

<em>a</em> = 1, <em>b</em> = 3/2, <em>c</em> = -1, <em>d</em> = -3, <em>e</em> = 2

Then the general solution to the DE is

<em>y(x)</em> = <em>C₁</em> exp(2<em>x</em>) + <em>C₂</em> exp(<em>x</em>) + <em>x</em> + 3/2 - (<em>x</em> ² + 3<em>x</em>) exp(<em>x</em>) + 2 exp(3<em>x</em>)

4 0
2 years ago
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