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Pie
4 years ago
11

The rate StartFraction 165 ounces Over 11 boxes EndFraction describes the relationship between the number of boxes and the weigh

t of the crackers in the
Mathematics
1 answer:
Cloud [144]4 years ago
3 0

Answer:

15 ounces per box

Step-by-step explanation:

The rate StartFraction 165 ounces Over 11 boxes EndFraction describes the relationship between the number of boxes and the weight of the crackers in the boxes. What is the weight, in ounces, of one box?

Total weight of crackers in the boxes = 165 ounces

Total number of boxes = 11 boxes

What is the weight, in ounces, of one box?

Weight per box of crackers =

Total weight of crackers in the boxes / Total number of boxes

= 165 ounces / 11 boxes

= 15 ounces per box

The weight, in ounces, of one box is 15 ounces

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All of the equations to –32 = k + –19
soldi70 [24.7K]

Answer:

k =  - 13

Step-by-step explanation:

- 32 = k - 19 \\  -32 + 19 =  k \\  - 13 = k

8 0
3 years ago
Given the table. find the slope of the line. show all work on separate sheet of paper
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The answer is 1/-0.5 because it is increasing it’s x by 1 and it’s y is decreasing by -0.5
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3 years ago
A 3-lb force acting in the direction of the vector (3,-1) moves an object just over 9 ft from point (0,5) to (6,-2). Find the wo
kodGreya [7K]

Answer: The Work done to move the object is  =23.15 \text { foot pounds }

Step-by-step explanation:

Given;

Force F=3 l b s

Vectors of F= (3,-1)

Moves object to a distance=9 f t s

Let A and B be the displacements Vectors

A= (0,-5)

B= (6,-2)

To Find:

Work done in foot-pounds

Solution:

Work done W=F \times D

Direction of force vector= (3,-1)  (i, j)  

                                        =3i-j

Unit of force vector =\sqrt{3^{2}+\left(-1^{2}\right)}

                                 =\sqrt{9+1}

Force vector=(Force/Unit Vector)Direction of  force vectors

                   F=3 / \sqrt{10} \times(3 i-j)

Direction of motion vector= (B-A)  

                                          = (6,-2)-(0,-5) x i,j)

                                          =(6-0),(-2-5) x (i, j)

                                          =(6,-7) x (i, j)

                                          =6i-7j

Unit of motion vector =\sqrt{6^{2}+\left(-7^{2}\right)}

                                   =\sqrt{36+49}

                                   =\sqrt{85}

Motion Vector= (Distance moved by the object/Unit motion vector) × (Direction of motion vectors)

                        D=9 / \sqrt{85} \times(6 i-7 j)

Workdone W=F \times D

                  = [3/ \sqrt{10} \times (3i-j)] \times [9/ \sqrt{85} \times (6i-7j)]

                  = [3/ \sqrt{10} \times 9/ \sqrt{85}] \times [(3i-j)  \times(6i-7j)]

                  = [3/3.1623 \times 9/9.2195] \times [(3\times6) + ((-1) \times (-7))]

                  = [0.94867 \times 0.97619] \times [18+7]

                  =23.15 \text { foot pounds }

Result:

   Work done to move an object =23.15 \text { foot pounds }

7 0
3 years ago
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Anybody happen to know this?help please
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Answer:

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Step-by-step explanation:

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Llana [10]

Answer:

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Step-by-step explanation:

Hope that this is helpful.

4 0
3 years ago
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