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zubka84 [21]
3 years ago
5

-10 + square root of 2x+1 = -5

Mathematics
1 answer:
Citrus2011 [14]3 years ago
5 0

Answer:

Step-by-step explanation:

-10+sqrt2x+1 = -5

sqrt2x+1 = -5+10

sqrt2x+1 = 5

2x+1 = sqrt5

2x = sqrt5-1

x = (sqrt5-1)/2

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I'm lazy but have a lot of points
ohaa [14]

Answer: 37

Step-by-step explanation:

Well DCB is a right angle, and there is already half which is 53 degrees, subtract 90 by 53 and you will get your answer.

4 0
3 years ago
Suppose a certain manufacturing company produces connecting rods for 4- and 6-cylinder automobile engines using the same product
gregori [183]

Answer:

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

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

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Answer:

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Step-by-step explanation:

Placing a - sign before the X will make it flip across the y-axis. After the X, and it flips across the X-axis.

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kvv77 [185]
THE ANSWER IS 27.5,BECAUSE WHEN I ADDED IT 25+2.5 I GOT 27.5.THIS IS HOW I GOT MY ANSWER.
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A square tile has a side 4 cm long. use the appropriate formula model to find the area. what is the area of the square? include
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S^2 = A
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