Is y=1 a linear equation
1 answer:
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Answer:
The Jacobian ∂(x, y, z) ∂(u, v, w) for the indicated change of variables
= -3072uv
Step-by-step explanation:
<u>Step :-(i)</u>
Given x = 1 6 (u + v) …(i)
Differentiating equation (i) partially with respective to 'u'
Differentiating equation (i) partially with respective to 'v'
Differentiating equation (i) partially with respective to 'w'
Given y = 1 6 (u − v) …(ii)
Differentiating equation (ii) partially with respective to 'u'
Differentiating equation (ii) partially with respective to 'v'
Differentiating equation (ii) partially with respective to 'w'
Given z = 6uvw ..(iii)
Differentiating equation (iii) partially with respective to 'u'
Differentiating equation (iii) partially with respective to 'v'
Differentiating equation (iii) partially with respective to 'w'
<u>Step :-(ii)</u>
The Jacobian ∂(x, y, z)/ ∂(u, v, w) =
Determinant 16(-16×6uv-0)-16(16×6uv)+0(0) = - 1536uv-1536uv
= -3072uv
<u>Final answer</u>:-
The Jacobian ∂(x, y, z)/ ∂(u, v, w) = -3072uv
Answer:
A)
Step-by-step explanation:
We are trying to find how long it will take Carlos to have $500 TOTAL, so we need to subtract what he already has ($150) to get the value we want to divide by $25 to figure out how many months it will take him to save up the rest. The given answer above is just a simplified way of writing , if that makes more sense.