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timofeeve [1]
3 years ago
7

Help sorry nothing was attached last time

Mathematics
1 answer:
kicyunya [14]3 years ago
8 0

Given:

The area of a rectangle is 99 square yd.

Length of the rectangle = 7 yd more than twice the width.

To find:

The dimensions of the rectangle.

Solution:

Let x be the width of the rectangle. Then, length of the rectangle is:

Length=2x+7

Area of a rectangle is:

A=length\times width

A=(2x+7)\times x

A=2x^2+7x

The area of a rectangle is 99 square yd.

2x^2+7x=99

2x^2+7x-99=0

Splitting the middle term, we get

2x^2+18x-11x-99=0

2x(x+9)-11(x-9)=0

(x+9)(2x-11)=0

Using zero product property, we get

(x+9)=0 and (2x-11)=0

x=-9 and x=\dfrac{11}{2}

x=-9 and x=5.5

Width of the rectangle cannot be negative. So, x=5.5 yd.

Now, the length of the rectangle is:

Length=2x+7

Length=2(5.5)+7

Length=11+7

Length=18

Therefore, the length of the rectangle is 18 yd and the width of the rectangle is 5.5 yd.

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Andreyy89

Answer:

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3 0
3 years ago
If you create a matrix, C, to show the inventory at the end of July, the value of the entry represented by C22 is ?
Inga [223]

Answer:

C_{22} = 389

Max(A_{31}) =376

Step-by-step explanation:

Given

See attachment for complete question

Solving (a): The entry C22

First, matrix C represents the inventory at the end of July.

The entry of C is calculated as:

C = Inventory - Sales

i.e.

C = \left[\begin{array}{ccc}543&356&643\\364&476&419\\376&903&409\end{array}\right]  -  \left[\begin{array}{ccc}102&78&97\\98&87&59\\54&89&79\end{array}\right]

C = \left[\begin{array}{ccc}543-102&356-78&643-97\\364-98&476-87&419-59\\376-54&903-89&409-79\end{array}\right]

C = \left[\begin{array}{ccc}441&278&546\\266&389&360\\322&814&330\end{array}\right]

Item C22 means the entry at the second row and the second column.

From the matrix

C_{22} = 389

Solving (b): The maximum A31 possible.

From the given data, we have:

Inventory = \left[\begin{array}{ccc}543&356&643\\364&476&419\\376&903&409\end{array}\right]

Unit\ Sales =   \left[\begin{array}{ccc}102&78&97\\98&87&59\\54&89&79\end{array}\right]

From the matrices above.

A31 means entry at the 3rd row and 1st column.

So, the possible values of A31 are:

A_{31} = 376

and

A_{31} = 54

By comparison, 376 > 54

So:

Max(A_{31}) =376  

8 0
2 years ago
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