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3 years ago

What is the midpoint of BC

Mathematics

1 answer:

Alex_Xolod [135]3 years ago

5 0

Answer:

(5, 6 )

Step-by-step explanation:

Using the midpoint formula

[ 0.5(x₁ + x₂ ), 0.5(y₁ + y₂ ) ]

with (x₁, y₁ ) = B(2, 1) and (x₂, y₂ ) = C(8, 11)

midpoint = [ 0.5(2 + 8), 0.5(1 + 11) ] = (5, 6 )

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1. Name the x- and y-intercepts for the equation: 2x + y = 6 (0, 3) and (60) (0, -3) and (-6,0) (3,0) and (0,6) (-3,0) and (0, -

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3 years ago

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Step-by-step explanation:

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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

6 0

3 years ago