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Mekhanik [1.2K]
2 years ago
13

What is the common difference, d, in the arithmetic sequence defined by the formula below? an=2n+1

Mathematics
1 answer:
marishachu [46]2 years ago
5 0
<h3><u>Answer:</u></h3>

  • d = 2

<h3><u>Solution:</u></h3>

We are given that the arithmetic progression is defined by :

➝ 2n + 1

<em>Therefore, </em>

  • <u>For </u><u>first </u><u>term</u>

➙ n = 1

➝ 2 × 1 + 1

➝ 2 + 1

➝ 3

  • <u>For </u><u>second </u><u>term</u>

➙ n = 2

➝ 2 × 2 + 1

➝ 4 + 1

➝ 5

  • <u>Common </u><u>difference</u>

➙ 2nd term - 1st term

➝ 5 - 3

➝ 2

<h3><u>More </u><u>information</u><u>:</u></h3>

  • The difference between the successive term and the preceding term is the difference of an arithmetic progression. It is always same for the same arithmetic progression.

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14y + 6 + 3y

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The sum of the digits of a two-digit number is 13. Twice the first digit is 1 less than the second digit. What is the two digit
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6 and 7

Step-by-step explanation:

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3 years ago
Bob makes three different silverware - spoons, forks and knives. Each unit (dozen) of spoon gives him a profit of $9, forks of $
Gala2k [10]

Answer:

Solution is:

z  (max) = 687.5

x₁   =  0

x₂   = 16

x₃   = 21

Step-by-step explanation:

From the problem statement we have:

Resources:      machine time ( 183 h)      labor (250  h)   steel  185 (pounds)

Unit

Spoons                          4                                     4                          3

forks                               4                                     9                          2

knives5                          5                                     5                          4

Profit $                           9                                    20                       17.5

Objective function z =  9*x₁  +  20*x₂  + 17.5 *x₃   to maximize

Subject to:

Availability of machine time  :  183 h

4*x₁  +  4*x₂ +   5*x₃ ≤  183

Availability of labor  :  250 h

4*x₁  +  9*x₂  +  5*x₃  ≤  250

Availability of steel  :  185 pounds

3*x₁  +  2*x₂  +  4*x₃  ≤  185

Requirement:

x₂ ≥ 16

General constraints:

x₁  ≥ 0             x₃      ≥  0 all integers

After 6 iteration the solution using AtomZmath on-line solver

z  (max) = 687.5

x₁   =  0

x₂   = 16

x₃   = 21

Resources used:

Machine time: 16* 4 + 21*5  =  64  +  105  = 169

remains   183 - 169  = 14 h

Labor:  16*9  +  21* 5  =  144  +  105  =  249

remains   250 -  249  =  1 h

Steel :  16*2  +  21*4  =  32  + 84  =  116

remains   185  -  116  = 69 pounds.

If it is decided that 20 units of forks are to be made then

we will need   4*4 = 16   h of machine time

                         9*4 =  36 h  of labor

                         2*4 = 8 pounds of steel

We can get that from abandom to make one unit of x₃ ??

No because as we said we need 36 hours  of  labor ( we still have 1 we need 35 more hours ) if we make 20 x₃ insted of 21 we get only 5 hours.

z we got is maximum

5 0
3 years ago
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