<span>{(–1, 3), (0, 4), (1, 14), (5, 6), (7, 2)}
</span>
Answer:
20 hours
Step-by-step explanation:
Make the denominators equal for the 2 fractions by multiplying 3 by 10 to make 30. Do the same to the denominator, the answer being 20. So, the 2 fractions will now be 1/30 and 20/30. 1 fits into 20, 20 times.
It 7ft because, 504 divided by 9=56 and then divide it by 8 and you get 7
Isolate the term with
.
![x^2 - 5y^2 + xyz^2 = y - 4 \implies xyz^2 = -x^2 + 5y^2 + y - 4](https://tex.z-dn.net/?f=x%5E2%20-%205y%5E2%20%2B%20xyz%5E2%20%3D%20y%20-%204%20%5Cimplies%20xyz%5E2%20%3D%20-x%5E2%20%2B%205y%5E2%20%2B%20y%20-%204)
Differentiate both sides with respect to
.
![\dfrac{\partial(xyz^2)}{\partial y} = \dfrac{\partial(-x^2 + 5y^2 + y - 4)}{\partial y}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cpartial%28xyz%5E2%29%7D%7B%5Cpartial%20y%7D%20%3D%20%5Cdfrac%7B%5Cpartial%28-x%5E2%20%2B%205y%5E2%20%2B%20y%20-%204%29%7D%7B%5Cpartial%20y%7D)
By the product and chain rules,
![xz^2 + 2xyz \dfrac{\partial z}{\partial y} = 10y + 1](https://tex.z-dn.net/?f=xz%5E2%20%2B%202xyz%20%5Cdfrac%7B%5Cpartial%20z%7D%7B%5Cpartial%20y%7D%20%3D%2010y%20%2B%201)
Solve for the partial derivative, then evaluate at
.
![\dfrac{\partial z}{\partial y} = \dfrac{10y + 1 - xz^2}{2xyz}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cpartial%20z%7D%7B%5Cpartial%20y%7D%20%3D%20%5Cdfrac%7B10y%20%2B%201%20-%20xz%5E2%7D%7B2xyz%7D)
![\dfrac{\partial z}{\partial y} \bigg|_{x=1,y=1,z=1} = \dfrac{10 + 1 - 1}{2} = \boxed{5}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cpartial%20z%7D%7B%5Cpartial%20y%7D%20%5Cbigg%7C_%7Bx%3D1%2Cy%3D1%2Cz%3D1%7D%20%3D%20%5Cdfrac%7B10%20%2B%201%20-%201%7D%7B2%7D%20%3D%20%5Cboxed%7B5%7D)
Answer:
(-2,1) is now at (2,1), (-3,1) is at (3,1), (-4,3) is (4,3), and (-2,4) is at (2,4)
Step-by-step explanation: