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Lynna [10]
2 years ago
13

A dual value (shadow price) for a constraint in a maximization problem means: a) as the right hand side value decreases, the obj

ective value function will increase b) as the right hand side value increases, the objective value function will increase c) as the right hand side value increases, the objective value function will decrease d) as the right hand side value increases, the objective value function value will stay the same
Mathematics
1 answer:
grin007 [14]2 years ago
3 0

Answer:B. As the right hand side value increases, the objective value function will increase

Step-by-step explanation: The dual value also known as the shadow price of a commodity that is not commonly priced in the market, the price of the commodity is not known. The dual value of a constraint can also be described as the rate of change of the objective function to the constant value which is placed at the right hand side of that constraint. The shadow price for a constraint in a maximization problem is Described as the Increase experienced at the Objective function when the right hand side value increases.

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Damon is buying meat by the pound at the deli. For every 5 pounds of meat he buys, he pays $8. What is the cost, per pound, for
Irina-Kira [14]

Answer:

$1.60 per pound

Step-by-step explanation:

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2 years ago
The volume of a cylinder is 9080 in.³ the same height of the cylinder is 20 inches what is the radius of the cylinder express yo
insens350 [35]

Answer:

12.02 inches

Step-by-step explanation:

recall, volume of a cylinder is given by

V = πr²h

where r is the radius of the cylinder and h = the height

hence

9080 = π x r² x 20

r² = 9080 ÷ 20π = 144.513

r = √144.513 = 12.02 inches

7 0
2 years ago
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Distribute and simplify these radicals. 12 x(-1+ √5)
Serga [27]
Hello there!

Here is the answer worked out for you!
12x[-1+sqrt(5)]
-12x+12xsqrt(5)

Remember that you can only add like terms so you can't add -12x to 12xsqrt(5). The radical makes it an unlike term.

I hope this helped and best wishes!

5 0
3 years ago
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the seven houses on Kent lane are numbered from 1 to 7 .the local newspaper choose two of the houses at random to receive a free
USPshnik [31]
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8 0
3 years ago
Solve the following equations: (a) x^11=13 mod 35 (b) x^5=3 mod 64
tino4ka555 [31]

a.

x^{11}=13\pmod{35}\implies\begin{cases}x^{11}\equiv13\equiv3\pmod5\\x^{11}\equiv13\equiv6\pmod7\end{cases}

By Fermat's little theorem, we have

x^{11}\equiv (x^5)^2x\equiv x^3\equiv3\pmod5

x^{11}\equiv x^7x^4\equiv x^5\equiv6\pmod 7

5 and 7 are both prime, so \varphi(5)=4 and \varphi(7)=6. By Euler's theorem, we get

x^4\equiv1\pmod5\implies x\equiv3^{-1}\equiv2\pmod5

x^6\equiv1\pmod7\impleis x\equiv6^{-1}\equiv6\pmod7

Now we can use the Chinese remainder theorem to solve for x. Start with

x=2\cdot7+5\cdot6

  • Taken mod 5, the second term vanishes and 14\equiv4\pmod5. Multiply by the inverse of 4 mod 5 (4), then by 2.

x=2\cdot7\cdot4\cdot2+5\cdot6

  • Taken mod 7, the first term vanishes and 30\equiv2\pmod7. Multiply by the inverse of 2 mod 7 (4), then by 6.

x=2\cdot7\cdot4\cdot2+5\cdot6\cdot4\cdot6

\implies x\equiv832\pmod{5\cdot7}\implies\boxed{x\equiv27\pmod{35}}

b.

x^5\equiv3\pmod{64}

We have \varphi(64)=32, so by Euler's theorem,

x^{32}\equiv1\pmod{64}

Now, raising both sides of the original congruence to the power of 6 gives

x^{30}\equiv3^6\equiv729\equiv25\pmod{64}

Then multiplying both sides by x^2 gives

x^{32}\equiv25x^2\equiv1\pmod{64}

so that x^2 is the inverse of 25 mod 64. To find this inverse, solve for y in 25y\equiv1\pmod{64}. Using the Euclidean algorithm, we have

64 = 2*25 + 14

25 = 1*14 + 11

14 = 1*11 + 3

11 = 3*3 + 2

3 = 1*2 + 1

=> 1 = 9*64 - 23*25

so that (-23)\cdot25\equiv1\pmod{64}\implies y=25^{-1}\equiv-23\equiv41\pmod{64}.

So we know

25x^2\equiv1\pmod{64}\implies x^2\equiv41\pmod{64}

Squaring both sides of this gives

x^4\equiv1681\equiv17\pmod{64}

and multiplying both sides by x tells us

x^5\equiv17x\equiv3\pmod{64}

Use the Euclidean algorithm to solve for x.

64 = 3*17 + 13

17 = 1*13 + 4

13 = 3*4 + 1

=> 1 = 4*64 - 15*17

so that (-15)\cdot17\equiv1\pmod{64}\implies17^{-1}\equiv-15\equiv49\pmod{64}, and so x\equiv147\pmod{64}\implies\boxed{x\equiv19\pmod{64}}

5 0
2 years ago
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