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Dmitry_Shevchenko [17]
3 years ago
15

PLEASE HELP MEEEEEEEEEEE

Mathematics
1 answer:
Lesechka [4]3 years ago
3 0

Answer: This hard behind question

Step-by-step explanation:

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The answer would be A :)
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How many lines of symmetry does it have
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14

Step-by-step explanation:

3 diangonalss joined togeheter is call a symertry

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Its pretty easy if you are good at math! Please help!
aleksley [76]

Step-by-step explanation:

Explanation:

The trick is to know about the basic idea of sequences and series and also knowing how i cycles.

The powers of i will result in either: i, −1, −i, or 1.

We can regroup i+i2+i3+⋯+i258+i259 into these categories.

We know that i=i5=i9 and so on. The same goes for the other powers of i.

So:

i+i2+i3+⋯+i258+i259

=(i+i5+⋯+i257)+(i2+i6+⋯+i258)+(i3+i7+⋯+i259)+(i4+i8+⋯+i256)

We know that within each of these groups, every term is the same, so we are just counting how much of these are repeating.

=65(i)+65(i2)+65(i3)+64(i4)

From here on out, it's pretty simple. You just evaluate the expression:

=65(i)+65(−1)+65(−i)+64(1)

=65i−65−65i+64

=−65+64

=−1

So,

i+i2+i3+⋯+i258+i259=-1

8 0
3 years ago
a laptop was originally sold for $975. The laptop is now on sale for $828.75. what is the percent markdown?
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Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
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

3 0
3 years ago
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