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Archy [21]
2 years ago
15

HelP pls i’ll mark as brainliest if correcttt

Mathematics
1 answer:
sveta [45]2 years ago
3 0

Answer:

answer is 42 in²

:-)))

....

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(a) Write 7.97 x 10 ^-6 as an ordinary number.
Vaselesa [24]

Answer:

0.00000797

Step-by-step explanation:

3 0
1 year ago
Find the pattern, then write the next two letters.<br> J, F, M, A, M,_, _
RoseWind [281]

Answer:

its answer J,F,M,A,M,J,J

Step-by-step explanation:

<h3>its a name of months next 5 words is A,S,O,N,and D</h3>
5 0
3 years ago
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A factory produce bicycles at a rate of 110+0.5t^2-0.9t bicycles per week (t in weeks)
klasskru [66]

Answer:

221

Step-by-step explanation:

Day 8 is the first day of the second week.

Day 21 is the last day of week 3.

We need to know the n umber of bicycles made from t = 1 to t = 3

The function is b(t) = 110 + 0.5t^2 - 0.9t, where t is in weeks.

We need to integrate the function with the limits of 1 to 3.

\int_{1}^{3} (110 + 0.5t^2 - 0.9t) dt

\int_{1}^{3} (110 + \dfrac{t^2}{2} - \dfrac{9t}{10}) dt

= 110t + \dfrac{t^3}{6} - \dfrac{9t^2}{20} \Biggr|_{1}^{3}

= 330.45 - 109.72

= 220.7333

Answer: 221

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
What is the slope? 3 or 1/3 or -1/3 or -3
maria [59]
I think it is 1/3 because rise over run and so you go up 1 and over 3
6 0
3 years ago
Read 2 more answers
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