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Likurg_2 [28]
3 years ago
8

Solve for y. 3X+2y=12

Mathematics
1 answer:
Montano1993 [528]3 years ago
3 0
3x + 2y = 12

2y = -3x + 12
  y = -1.5x + 6
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8x^2+18x-5=0 find the discriminant
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The formula for the discriminant is the square root of b^2 - 4ac

8x^2+18x-5= 0

a = 8, b=18, c= -5

Square root of 324 +160

Square root of 484 = 22                The discriminant is 22


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3 years ago
Convert 501% to a decimal.<br> How would I do this?
jonny [76]
Puedo patearla tan fuerte como pueda.

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Read 2 more answers
Please help me with this sum I will mark you as brainest​
Murljashka [212]

Answer:

Step-by-step explanation:

3/6 x 8

= ( 3 x 8 ) / 6

= 24 / 6

4/7 x 2

= ( 4 x 2 ) / 7

= 8 / 7

9/2 x 3

= ( 9 x 3 ) / 2

= 27 / 2

8/5 x 7

= ( 8 x 7 ) / 5

= 56 / 5

3/9 x 5

= ( 3 x 5 ) / 9

= 15/9

7/4 x 4

= ( 7 x 4 ) / 4

= 28 / 4

6/2 x 6

= ( 6 x 6 ) / 2

= 36 / 2

8/3 x 9

= ( 8 x 9 ) / 3

= 72 / 3

9/4 x 8

= ( 9 x 8 ) / 4

= 72 / 4

5/3 x 6

= ( 5 x 6 ) / 3

= 30 / 3

2/7 x 3

= ( 2 x 3 ) / 7

= 6 / 7

8/4 x 7

= ( 8 x 7 ) / 4

= 72 / 4

6/4 x 2

= ( 6 x 2 ) / 4

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3 0
3 years ago
Better Products, Inc., manufactures three products on two machines. In a typical week, 40 hours are available on each machine. T
Kaylis [27]

Answer:

z (max)  =  1250 $

x₁  = 25    x₂  =  0   x₃  =  25

Step-by-step explanation:

                                Profit $    mach. 1      mach. 2

Product 1     ( x₁ )       30             0.5              1

Product 2    ( x₂ )       50             2                  1

Product 3    ( x₃ )       20             0.75             0.5

Machinne 1 require  2 operators

Machine   2 require  1  operator

Amaximum of  100 hours of labor available

Then Objective Function:

z  =  30*x₁  +  50*x₂  +  20*x₃      to maximize

Constraints:

1.-Machine 1 hours available  40

In machine 1    L-H  we will need

0.5*x₁  +  2*x₂  + 0.75*x₃  ≤  40

2.-Machine 2   hours available  40

1*x₁  +  1*x₂   + 0.5*x₃   ≤  40

3.-Labor-hours available   100

Machine 1     2*( 0.5*x₁ +  2*x₂  +  0.75*x₃ )

Machine  2       x₁   +   x₂   +  0.5*x₃  

Total labor-hours   :  

2*x₁  +  5*x₂  +  2*x₃  ≤  100

4.- Production requirement:

x₁  ≤  0.5 *( x₁ +  x₂  +  x₃ )     or   0.5*x₁  -  0.5*x₂  -  0.5*x₃  ≤ 0

5.-Production requirement:

x₃  ≥  0,2 * ( x₁  +  x₂   +  x₃ )  or    -0.2*x₁  - 0.2*x₂ + 0.8*x₃   ≥  0

General constraints:

x₁  ≥   0       x₂    ≥   0       x₃     ≥   0           all integers

The model is:

z  =  30*x₁  +  50*x₂  +  20*x₃      to maximize

Subject to:

0.5*x₁  +  2*x₂  + 0.75*x₃  ≤  40

1*x₁  +  1*x₂   + 0.5*x₃       ≤  40

2*x₁  +  5*x₂  +  2*x₃        ≤  100

0.5*x₁  -  0.5*x₂  -  0.5*x₃  ≤ 0

-0.2*x₁  - 0.2*x₂ + 0.8*x₃   ≥  0

x₁  ≥   0       x₂    ≥   0       x₃     ≥   0           all integers

After 6 iterations with the help of the on-line solver AtomZmaths we find

z (max)  =  1250 $

x₁  = 25    x₂  =  0   x₃  =  25

6 0
3 years ago
The formula for the resistance of a conductor with voltage V and current I is r= v/1. Solve for V.
VikaD [51]
V=i x r is the formula 
8 0
3 years ago
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