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Mandarinka [93]
2 years ago
8

2.

Mathematics
1 answer:
9966 [12]2 years ago
6 0

Answer:

(b) -m, m + 3

Step-by-step explanation:

x² − 3x − m(m + 3) = 0

x² − 3x = m(m + 3)

x² − 3x + 9/4 = m(m + 3) + 9/4

(x − 3/2)² = m(m + 3) + 9/4

(x − 3/2)² = m² + 3m + 9/4

(x − 3/2)² = (m + 3/2)²

x − 3/2 = ±(m + 3/2)

x − 3/2 = m + 3/2, -m − 3/2

x = m + 3, -m

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anzhelika [568]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The solution is  \frac{\delta  (x,y)}{\delta (u, v)} | = 10e^{-4u}

Step-by-step explanation:

From the question we are told that

        x =  e^{-2a} cos (5v)

and  y  =  e^{-2a} sin(5v)

Generally the absolute value of the determinant of the Jacobian for this change of coordinates is mathematically evaluated as

     | \frac{\delta  (x,y)}{\delta (u, v)} | =  | \ det \left[\begin{array}{ccc}{\frac{\delta x}{\delta u} }&{\frac{\delta x}{\delta v} }\\\frac{\delta y}{\delta u}&\frac{\delta y}{\delta v}\end{array}\right] |

        = |\ det\ \left[\begin{array}{ccc}{-2e^{-2u} cos(5v)}&{-5e^{-2u} sin(5v)}\\{-2e^{-2u} sin(5v)}&{-2e^{-2u} cos(5v)}\end{array}\right]  |

Let \   a =  -2e^{-2u} cos(5v),  \\ b=-2e^{-2u} sin(5v),\\c =-2e^{-2u} sin(5v),\\d=-2e^{-2u} cos(5v)

So

     \frac{\delta  (x,y)}{\delta (u, v)} | = |det  \left[\begin{array}{ccc}a&b\\c&d\\\end{array}\right] |

=>    \frac{\delta  (x,y)}{\delta (u, v)} | = | a *  b  - c* d |

substituting for a, b, c,d

=>    \frac{\delta  (x,y)}{\delta (u, v)} | =  | -10 (e^{-2u})^2 cos^2 (5v) - 10 e^{-4u} sin^2(5v)|

=>   \frac{\delta  (x,y)}{\delta (u, v)} | =  | -10 e^{-4u} (cos^2 (5v)   + sin^2 (5v))|

=>  \frac{\delta  (x,y)}{\delta (u, v)} | = 10e^{-4u}

7 0
3 years ago
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My name is Ann [436]

Answer:

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Step-by-step explanation:

If you use stoichometry you can convert your units

9.83 seconds

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Answer:

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