8w=40
You get area by doing length x width. In this case, the “length” is the height. So you would need to figure out what you multiply 8 by to get 40.
In standard form the two equations are
7x +y = 5
7x +y = -5
The equations describe lines that are parallel.
first one is 0 second one is 2 and then the last one is 0
Let
. The tangent plane to the surface at (0, 0, 8) is
![\nabla f(0,0,8)\cdot(x,y,z-8)=0](https://tex.z-dn.net/?f=%5Cnabla%20f%280%2C0%2C8%29%5Ccdot%28x%2Cy%2Cz-8%29%3D0)
The gradient is
![\nabla f(x,y,z)=\left(1-8yze^{xyz},1-8xze^{xyz},1-8xye^{xyz}\right)](https://tex.z-dn.net/?f=%5Cnabla%20f%28x%2Cy%2Cz%29%3D%5Cleft%281-8yze%5E%7Bxyz%7D%2C1-8xze%5E%7Bxyz%7D%2C1-8xye%5E%7Bxyz%7D%5Cright%29)
so the tangent plane's equation is
![(1,1,1)\cdot(x,y,z-8)=0\implies x+y+(z-8)=0\implies x+y+z=8](https://tex.z-dn.net/?f=%281%2C1%2C1%29%5Ccdot%28x%2Cy%2Cz-8%29%3D0%5Cimplies%20x%2By%2B%28z-8%29%3D0%5Cimplies%20x%2By%2Bz%3D8)
The normal vector to the plane at (0, 0, 8) is the same as the gradient of the surface at this point, (1, 1, 1). We can get all points along the line containing this vector by scaling the vector by
, then ensure it passes through (0, 0, 8) by translating the line so that it does. Then the line has parametric equation
![(1,1,1)t+(0,0,8)=(t,t,t+8)](https://tex.z-dn.net/?f=%281%2C1%2C1%29t%2B%280%2C0%2C8%29%3D%28t%2Ct%2Ct%2B8%29)
or
,
, and
.
(See the attached plot; the given surface is orange, (0, 0, 8) is the black point, the tangent plane is blue, and the red line is the normal at this point)
Answer:
hi lets be friends
Step-by-step explanation: