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4 years ago

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Suppose that x ∗ is an optimal basic feasible solution, and consider an optimal basis associated with x ∗ . Let B and N be the s

et of basic and nonbasic indices, respectively. Let I be the set of nonbasic indices i for which the corresponding reduced costs are zero.

1 answer:

ch4aika [34]4 years ago

6 0

If I’m honest I’d love to help but I’m pretty bad at math so

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The range of the following relation {(3,-5), (1,2), (-1,-4), (-1,2)} is

charle [14.2K]

Answer:

Range: {-5, 2, -4, 2)

Step-by-step explanation:

1. By definition, the range of the relation is the output value or the value of y.

2.The value of y is the second number in each ordered pair.

3. Therefore, keeping the information above on mind, you can conclude that the range of the relation (3,-5), (1,2), (-1,-4), (-1,2)} is the shown below:

range: {-5, 2, -4, 2)

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Answer:

Cos(A) = 12 / 13

Step-by-step explanation:

Cos(A) = Adjacent / Hypotenuse

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3 years ago

The exterior angle of an equilateral triangle is obtuse

kifflom [539]

Is this a true or false question?

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3 years ago

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Find the solution of the problem (1 3. (2 cos x - y sin x)dx + (cos x + sin y)dy=0.

lakkis [162]

Answer:

$2*sin(x)+y*cos(x)-cos(y)=C_1$

Step-by-step explanation:

Let:

$P(x,y)=2*cos(x)-y*sin(x)$

$Q(x,y)=cos(x)+sin(y)$

This is an exact differential equation because:

$\frac{\partial P(x,y)}{\partial y} =-sin(x)$

$\frac{\partial Q(x,y)}{\partial x}=-sin(x)$

With this in mind let's define f(x,y) such that:

$\frac{\partial f(x,y)}{\partial x}=P(x,y)$

and

$\frac{\partial f(x,y)}{\partial y}=Q(x,y)$

So, the solution will be given by f(x,y)=C1, C1=arbitrary constant

Now, integrate $\frac{\partial f(x,y)}{\partial x}$ with respect to x in order to find f(x,y)

$f(x,y)=\int\ 2*cos(x)-y*sin(x)\, dx =2*sin(x)+y*cos(x)+g(y)$

where g(y) is an arbitrary function of y

Let's differentiate f(x,y) with respect to y in order to find g(y):

$\frac{\partial f(x,y)}{\partial y}=\frac{\partial }{\partial y} (2*sin(x)+y*cos(x)+g(y))=cos(x)+\frac{dg(y)}{dy}$

Now, let's replace the previous result into $\frac{\partial f(x,y)}{\partial y}=Q(x,y)$ :

$cos(x)+\frac{dg(y)}{dy}=cos(x)+sin(y)$

Solving for $\frac{dg(y)}{dy}$

$\frac{dg(y)}{dy}=sin(y)$

Integrating both sides with respect to y:

$g(y)=\int\ sin(y) \, dy =-cos(y)$

Replacing this result into f(x,y)

$f(x,y)=2*sin(x)+y*cos(x)-cos(y)$

Finally the solution is f(x,y)=C1 :

$2*sin(x)+y*cos(x)-cos(y)=C_1$

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4 years ago

It says this is wrong, so what’s the right answer?

victus00 [196]

Answer:

it is right are you able to show the multiple choices

Step-by-step explanation:

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