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2 years ago

100 10 100 100 10 100 )))))))))

Mathematics

2 answers:

maw [93]2 years ago

8 0

Answer:

Step-by-step explanation:

Thanks

VashaNatasha [74]2 years ago

3 0

Thanks for the points.

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Help me with these 3 questions, please

Zina [86]

Answer:

1. x = -4y ---> y = (-1/4)x

slope = -1/4. y-intercept = (0,0)

2. y = -2x + 4

3. y = (1/3)x - 1

Step-by-step explanation:

1. Re-write your equation so that x is on the right and y is on the left:

x = -4y ---> y = (-1/4)x

slope = -1/4. y-intercept = (0,0)

2. y-intercept = (0,4) ----> P1

x-intercrpt = (2,0) ----> P2

slope m = (y2 - y1) / (x2 - x1)

= (0 - 4)/(2 - 0)

= -2

therefore, y - y1 = mx - x1 ---> y - 4 = -2x

or y = -2x + 4

3. y-intercept = (0,-1)

x-intercept = (3,0)

m = (0 - (-1)) / (3 -0) = 1/3

y - (-1) = (1/3)x - 0 ---> y = (1/3)x - 1

5 0

3 years ago

PLEASE HELP!!!! Will give brainliest

pashok25 [27]

Answer:

Trapezium:

perimeter= 232

area= 2052

Triangle:

perimeter= 221

area= 4088

Step-by-step explanation:

6 0

3 years ago

jeka94

Answer:

g(x)

because it is a quadratic equation it is mirrored the other one isn’t even a function

8 0

3 years ago

Evaluate the diagram below

tigry1 [53]

Answer:

Only option (A) is incorrect.

So , true statements are (B), (C) &(D).

#$# THANK YOU #$#

4 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