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blagie [28]
2 years ago
7

I'll mark you as brainliestgive serious answers​

Mathematics
1 answer:
AlexFokin [52]2 years ago
5 0

Answer:

1) 2/4

= 1/2

2) 6/9

= 2/3

3) 10/12

= 5/6

4) 8/12

= 2/3

5) 12/32

= 3/8

6) 27/45

= 3/5

7) 35/63

= 5/9

8) 64/96

= 2/3

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Consider the following.x = t − 2 sin(t), y = 1 − 2 cos(t), 0 ≤ t ≤ 8πSet up an integral that represents the length of the curve.
sweet [91]

The length of the parametric curve (call it <em>C </em>) is given by

\displaystyle\int_C\mathrm ds=\int_0^{8\pi}\sqrt{\left(\frac{\mathrm dx}{\mathrm dt}\right)^2+\left(\frac{\mathrm dy}{\mathrm dt}\right)^2}\,\mathrm dt

We have

x=t-2\sin t\implies\dfrac{\mathrm dx}{\mathrm dt}=1-2\cos t

y=1-2\cos t\implies\dfrac{\mathrm dy}{\mathrm dt}=2\sin t

Now,

\left(\frac{\mathrm dx}{\mathrm dt}\right)^2+\left(\frac{\mathrm dy}{\mathrm dt}\right)^2=5-4\cos t

so that the arc length integral reduces to

\displaystyle\int_0^{8\pi}\sqrt{5-4\cos t}\,\mathrm dt

which has an approximate value of 53.4596.

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A product can be made in sizes huge, average and tiny which yield a net unit profit of $14, $10, and$5, respectively. Three cent
navik [9.2K]

Answer:

The model is:

z = 14* X₁₁ + 14*X₁₂ + 5*X₁₃ + 10*X₂₁ + 10*X₂₂ + 10*X₂₃ + 5*X₃₁ + 5*X₃₂ + 5*X₃₃    to maximize

Subject to:

First center               X₁₁  +  X₂₁  + X₃₁  ≤  550

Second center         X₁₂  +  X₂₂  + X₃₂  ≤ 750

Third center               X₁₃  + X₂₃  + X₃₃  ≤ 275                  

22* X₁₁  + 16* X₂₁  + 9*X₃₁     ≤   11000

22* X₁₂  + 16* X₂₂  + 9*X₃₂   ≤   2700

22*X₁₃  + 16* X₂₃  +  9*X₃₃  ≤  3400

X₁₁  +  X₁₂  + X₁₃  ≤  710

X₂₁  + X₂₂ + X₂₃  ≤  900

X₃₁ + X₃₂ + X₃₃  ≤  350

2700*(X₁₁  +  X ₂₁  + X₃₁)  -  11000*(X₁₂ + X₂₂ + X₃₂) = 0

3400*(X₁₁  +  X ₂₁  + X₃₁) - 11000*( ( X₁₃ + X₂₃ + X₃₃) = 0

Xij >= 0

Step-by-step explanation:

Let´s call Xij   product size i produced in center j

According to this, we get the following set of variable

X₁₁    product size huge produced in center 1

X₁₂    product size huge produced in center 2

X₁₃   product size huge produced in center 3

X₂₁   product size average produced in center 1

X₂₂   product size average produced in center 2

X₂₃   product size average produced in center 3

X₃₁  product size-tiny produced in center 1

X₃₂ product size-tiny produced in center 2

X₃₃ product size-tiny produced in center 3

Then Objective function is

z = 14* X₁₁ + 14*X₁₂ + 5*X₁₃ + 10*X₂₁ + 10*X₂₂ + 10*X₂₃ + 5*X₃₁ + 5*X₃₂ + 5*X₃₃

Constrains

Center capacity

1.-   First center               X₁₁  +  X₂₁  + X₃₁  ≤  550

2.-   Second center         X₁₂  +  X₂₂  + X₃₂  ≤ 750

3.- Third center               X₁₃  + X₂₃  + X₃₃  ≤ 275

Water available

1.-  22* X₁₁  + 16* X₂₁  + 9*X₃₁     ≤   11000

2.-  22* X₁₂  + 16* X₂₂  + 9*X₃₂   ≤   2700

3.-   22*X₁₃  + 16* X₂₃  +  9*X₃₃  ≤  3400

Demand constrain

Product huge

X₁₁  +  X₁₂  + X₁₃  ≤  710

Product average

X₂₁  + X₂₂ + X₂₃  ≤  900

Product tiny

X₃₁ + X₃₂ + X₃₃  ≤  350

Fraction SP/CC must be the same

First and second centers  fraction SP/CC  

(X₁₁  +  X ₂₁  + X₃₁)/ 11000   =  (X₁₂ + X₂₂ + X₃₂)/ 2700

2700*(X₁₁  +  X ₂₁  + X₃₁)  -  11000*(X₁₂ + X₂₂ + X₃₂) = 0

First and third centers  fraction SP/CC  

(X₁₁  +  X ₂₁  + X₃₁)/ 11000   = ( X₁₃ + X₂₃ + X₃₃)/ 3400

3400*(X₁₁  +  X ₂₁  + X₃₁) - 11000*( ( X₁₃ + X₂₃ + X₃₃) = 0

The model is:

z = 14* X₁₁ + 14*X₁₂ + 5*X₁₃ + 10*X₂₁ + 10*X₂₂ + 10*X₂₃ + 5*X₃₁ + 5*X₃₂ + 5*X₃₃

Subject to:

First center               X₁₁  +  X₂₁  + X₃₁  ≤  550

Second center         X₁₂  +  X₂₂  + X₃₂  ≤ 750

Third center               X₁₃  + X₂₃  + X₃₃  ≤ 275                  

22* X₁₁  + 16* X₂₁  + 9*X₃₁     ≤   11000

22* X₁₂  + 16* X₂₂  + 9*X₃₂   ≤   2700

22*X₁₃  + 16* X₂₃  +  9*X₃₃  ≤  3400

X₁₁  +  X₁₂  + X₁₃  ≤  710

X₂₁  + X₂₂ + X₂₃  ≤  900

X₃₁ + X₃₂ + X₃₃  ≤  350

2700*(X₁₁  +  X ₂₁  + X₃₁)  -  11000*(X₁₂ + X₂₂ + X₃₂) = 0

3400*(X₁₁  +  X ₂₁  + X₃₁) - 11000*( ( X₁₃ + X₂₃ + X₃₃) = 0

Xij >= 0

6 0
2 years ago
A square piece of paper has an area of 36 square inches what is the perimeter of the paper​
Gekata [30.6K]

Lets represent the side of the square as s , since we know the area is 36. write s

{s}^{2}  = 36

s = \sqrt{36}

the square root of 36 is 6. So, we know the side is 6. The formula for perimeter is P= s+s+s+s , but since it is a square and all the sides are equal, for a perimeter of a square you can do p= 4s , now substitute 6 for s to get 4(6)=24 so the perimeter of a square would be 24 inches

6 0
3 years ago
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