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4 years ago

_H2 + _Br2 → ___HBr

Mathematics

1 answer:

Vaselesa [24]4 years ago

3 0

Answer:H2+Br2-->2HBr

Step-by-step explanation:

The amount of atoms on the left side must be equal to the amount of atoms on the right side for each element.

On the left you got 2xH and 2xBr.

This means you must have the same amount on the right.

2HBr =2xH + 2xBr

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Write the linear system:

notsponge [240]

Answer:

(See explanation for further details).

Step-by-step explanation:

C) The linear system is:

$(-1)^{2} \cdot a + (-1)\cdot b + c = 5$

$c = - 4$

$(4)^{2}\cdot a + (4)\cdot b + c = 0$

After some algebraic handling:

$a - b + c = 5$

$c = -4$

$16\cdot a + 4\cdot b + c = 0$

By direct substitution:

$a - b - 4 = 5$

$16\cdot a + 4\cdot b - 4 = 0$

$a - b = 9$

$4\cdot a + b = 1$

$a = 9 + b$

$4\cdot (9+b)+b = 1$

$36 + 5\cdot b = 1$

$5\cdot b = -35$

$b = -7$

$a = 2$

D) The parameter values are: $a = 2$ , $b = -7$ , $c = -4$

E) The quadratic equation is $y = 2\cdot x^{2} - 7\cdot x -4$

8 0

3 years ago

Calculate all four second-order partial derivatives and confirm that the mixed partials are equal. f(x,y)= 2e^xy

nadya68 [22]

ANSWER TO QUESTION 1

The given function is

$f(x,y)=2 {e}^{xy}$

The partial derivative of f with respect to x means we are treating y as a constant. The first derivative is

$f_{x} = 2y {e}^{xy}$

and the second derivative with respect to x is,

$f_{xx} = 2 {y}^{2} {e}^{xy}$

ANSWER TO QUESTION 2

The given function is

$f(x,y)=2 {e}^{xy}$

The partial derivative of f with respect to y means we are treating x as a constant. The first derivative is

$f_{y} = 2x{e}^{xy}$

and the second derivative with respect to y is

$f_{yy} = 2 {x}^{2} {e}^{xy}$

ANSWER TO QUESTION 3

Our first mixed partial is

$f_{xy}$

We need to differentiate
$f_{x} = 2y {e}^{xy}$
again. But this time with respect to y.

Since this is a product of two functions of y, we apply the product rule of differentiation to obtain,

$f_{xy} = 2y( {e}^{xy})' + ({e}^{xy})(2y)'$

$f_{xy} = 2xy {e}^{xy} + 2{e}^{xy}$

ANSWER TO QUESTION 4

The second mixed partial is

$f_{yx}$

We need to differentiate
$f_{y} = 2x{e}^{xy}$

again. But this time with respect to x.

Since this is a product of two functions of x, we apply the product rule of differentiation to obtain,

$f_{yx} = 2x({e}^{xy})' + ({e}^{xy})(2x)'$

$f_{yx} = 2xy {e}^{xy} + 2{e}^{xy}$

Hence,

$f_{xy} = 2xy {e}^{xy} + 2{e}^{xy} =f_{yx}$