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Strike441 [17]
2 years ago
15

The first term of an arithmetic sequence is 4 and the tenth term is 67.

Mathematics
1 answer:
Dmitry [639]2 years ago
6 0

Answer:

Step-by-step explanation:

Formula

tn = a1 + (n-1)* d

Givens

tn = 67 (it is the last term or the nth term

n = 10

a1 = 4

find d

Solution

67 = 4 + (10 - 1)*d

67 = 4 + 9*d               Subtract 4 from both sides

67 - 4 = 9d            

63 = 9d                      Divide by 9

63/9 = 9d/9              

7  = d

Remark: This is a fundamental application of the arithmetic sequence. You pretty much have to know how to juggle it's parts.

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Answer:

x  =  7

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Step-by-step explanation:

Let call  

x numbers of church goup and

y numbers of Union Local

Then  

First contraint

2*x + 2*y ≤ 20

Second one

1*x + 3*y ≤ 16

Objective Function

z = 150*x + 200*y

Then the system is

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We will solve by using the Simplex method

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First Table

z           x          y          s₁           s₂          Cte

1        -150    -200       0            0     =      0

0          2         2           1            0     =    20

0          1          3          0            1      =     16

First iteration:

Column pivot   ( y column ) row pivot (third row) pivot 3

Second table

z           x                y          s₁           s₂                Cte

1       -250/3           0          0         200/3    =    3200/3

0      - 4/3               0          -1           2/3       =    -20/3

0         1/3               1            0           1/3       =    -20/3

Second iteration:

Column pivot  ( x column ) row pivot  (second row)  pivot  -4/3

Third table

z           x                y          s₁           s₂                Cte

1            0               0      750/12    700/6    =   4950/3

0            1                0        3/4          -1/2     =    7

0            0                1         -1/4          1/2     =  9/3

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