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gregori [183]
3 years ago
12

How many 5-millimeter-long spiders would it take to make a line of spiders 1 meter long?

Mathematics
1 answer:
Bogdan [553]3 years ago
3 0

Answer:

1000

Step-by-step explanation:

10milimeter=1 centimeter

100 centimeters=1 meter

10x100=1000

Let me know if im right :)

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Read 2 more answers
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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