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3 years ago

PLEASE HELP!!!!! Write the absolute value inequality in the form

Mathematics

2 answers:

Elena-2011 [213]3 years ago

7 0

Answer:

|x - 1 |> 6,

Step-by-step explanation:

First we find the distance between the two points

-5 to 7

7 - -5 =7+5 = 12

We find 1/2 that distance

12/2 = 6

So we are looking for a greater-than inequality because we are looking for the outsides because we want a less than and a greater than

|x - b |> c,

The center is 6 to that is c

|x - b |> 6,

To find the value of b

Let x = 7

7 -b =6

b=1

|x - 1 |> 6,

We can check by using -5

|-5 - 1 |> 6,

Stells [14]3 years ago

7 0

Answer:

|x - 1| > 6

Step-by-step explanation:

x < -5 or x > 7

b is the centre:

(-5+7)/2 = 1

c is the distance from the centre

7 - 1 = 6

1 - (-5) = 6

Since it's an or case, inequality with the modulus would be >

|x - 1| > 6

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Find the Jacobian ∂(x, y, z) ∂(u, v, w) for the indicated change of variables. If x = f(u, v, w), y = g(u, v, w), and z = h(u, v

jeyben [28]

Answer:

The Jacobian ∂(x, y, z) ∂(u, v, w) for the indicated change of variables

= -3072uv

Step-by-step explanation:

Step :-(i)

Given x = 1 6 (u + v) …(i)

Differentiating equation (i) partially with respective to 'u'

$\frac{∂x}{∂u} = 16(1)+16(0)=16$

Differentiating equation (i) partially with respective to 'v'

$\frac{∂x}{∂v} = 16(0)+16(1)=16$

Differentiating equation (i) partially with respective to 'w'

$\frac{∂x}{∂w} = 0$

Given y = 1 6 (u − v) …(ii)

Differentiating equation (ii) partially with respective to 'u'

$\frac{∂y}{∂u} = 16(1) - 16(0)=16$

Differentiating equation (ii) partially with respective to 'v'

$\frac{∂y}{∂v} = 16(0) - 16(1)= - 16$

Differentiating equation (ii) partially with respective to 'w'

$\frac{∂y}{∂w} = 0$

Given z = 6uvw ..(iii)

Differentiating equation (iii) partially with respective to 'u'

$\frac{∂z}{∂u} = 6vw$

Differentiating equation (iii) partially with respective to 'v'

$\frac{∂z}{∂v} =6 u (1)w=6uw$

Differentiating equation (iii) partially with respective to 'w'

$\frac{∂z}{∂w} =6 uv(1)=6uv$

Step :-(ii)

The Jacobian ∂(x, y, z)/ ∂(u, v, w) =

$\left|\begin{array}{ccc}16&16&0\\16&-16&0\\6vw&6uw&6uv\end{array}\right|$

Determinant 16(-16×6uv-0)-16(16×6uv)+0(0) = - 1536uv-1536uv

= -3072uv

Final answer:-

The Jacobian ∂(x, y, z)/ ∂(u, v, w) = -3072uv

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3 years ago

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Answer:

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Step-by-step explanation:

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