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Vlad [161]
2 years ago
12

A model car travels 20 meters in 15 seconds how many will it travel in 45 seconds ​

Mathematics
2 answers:
Monica [59]2 years ago
8 0

Answer: <em>60 meters</em>

Step-by-step explanation:

<em>if the car travels 20 meters in 15 seconds</em>

<em>you want to take 45 and divide it by 15, giving you 3</em>

<em>now you want to multiply 20x3</em>

<em>which will get you </em><em>60 meters</em>

dlinn [17]2 years ago
5 0

Answer: 60 meters

Step-by-step explanation: you can divide 20 by 15 to get 3/4 and 3/4 times 45 is 60

oof sorry if its wrong or the explanation is too lazy

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Generally the constraint that sets next week are shown below

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Generally the constrain that sets next week maximum production of connecting rod for 6 cylinder  to  W_6 or  0  is  

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Generally the constrain that limits the production of connecting rods  for both 4 cylinder and 6 cylinders  is

     x_4 \le W_4 *  s_6

=>   x_4 \le 5000 *  s_6

     x_4 \le W_6 *  s_4

=>    x_4 \le 8000 *  s_4

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The minimum cost of production for next week is  

   U  =  M_4 *  x_4 + M_6 * x_6 + C_4 * s_4 + C_6 * s_6

=>  U  =  13x_4 + 16x_6 + 2000 s_4 + 3500 s_6

Step-by-step explanation:

The cost for the four cylinder production line is  C_4 =  \$2,100

The cost for the six cylinder production line is  C_6 = \$3,500

The manufacturing cost for each four cylinder is  M_4= \$13

 The manufacturing cost for each six cylinder is M_6= \$16

  The weekly production capacity for 4 cylinder connecting rod is W_4 = 5,000

   The weekly production capacity for 6 cylinder connecting rod is W_6 = 8,000

Generally the constraint that sets next week are shown below

Generally the constrain that sets next week maximum production of connecting rod for 4 cylinder  to  W_4 or  0 is  

     x_4 \le W_4 *  s_4

    x_4 \le 5000 *  s_4

Generally the constrain that sets next week maximum production of connecting rod for 6 cylinder  to  W_6 or  0  is  

     x_6 \le W_6 *  s_6

     x_6 \le 8,000 *  s_6

Generally the constrain that limits the production of connecting rods  for both 4 cylinder and 6 cylinders  is

     x_4 \le W_4 *  s_6

=>   x_4 \le 5000 *  s_6

     x_4 \le W_6 *  s_4

=>    x_4 \le 8000 *  s_4

     s_4 + s_6 = 1

The minimum cost of production for next week is  

   U  =  M_4 *  x_4 + M_6 * x_6 + C_4 * s_4 + C_6 * s_6

=>  U  =  13x_4 + 16x_6 + 2000 s_4 + 3500 s_6

3 0
3 years ago
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