All categories

3 years ago

PLEASE HELP MEEEEEEEEEEE

Mathematics

1 answer:

Lesechka [4]3 years ago

3 0

Answer: This hard behind question

Step-by-step explanation:

You might be interested in

Which scatter plot represents the data shown in the table?

svetoff [14.1K]

The answer would be A :)

3 0

2 years ago

How many lines of symmetry does it have

Anarel [89]

Answer:

Step-by-step explanation:

3 diangonalss joined togeheter is call a symertry

4 0

2 years ago

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?

zloy xaker [14]

Answer: 15%

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