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Nataliya [291]
3 years ago
7

If -xy – 5+ y^2+ x^2= 0 and it is known that dy/dx= y-2x/-x+2y, find all

Mathematics
1 answer:
sleet_krkn [62]3 years ago
3 0

Answer:

There is no point of the form (-1, y) on the curve where the tangent is horizontal

Step-by-step explanation:

Notice that when x = - 1. then dy/dx becomes:

dy/dx= (y+2) / (2y+1)

therefore, to request that the tangent is horizontal we ask for the y values that make dy/dx equal to ZERO:

0 = ( y + 2) / (2 y + 1)

And we obtain y = -2 as the answer.

But if we try the point (-1, -2) in the original equation, we find that it DOESN'T belong to the curve because it doesn't satisfy the equation as shown below:

(-1)^2 + (-2)^2 - (-1)*(-2) - 5 = 1 + 4 + 2 - 5 = 2  (instead of zero)

Then, we conclude that there is no horizontal tangent to the curve for x = -1.

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You want the determinant of the matrix ...

  \left[\begin{array}{ccc}4&3&2\\-3&1&5\\-1&-4&3\end{array}\right]

One way to figure it is as the difference between the sum of products of the down-diagonals and the sum of products of the up-diagonals:

  D = (4)(1)(3) +(3)(5)(-1) +(2)(-3)(-4) -(-1)(1)(2) -(-4)(5)(4) -(3)(-3)(3)

  = 12 -15 +24 +2 +80 +27

  D = 130

The determinant of the coefficient matrix is 130.

_____

Many scientific and graphing calculators and web sites can perform this calculation for you.

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