Answer:
C. π ft ²
Step-by-step explanation:
:)
The cross product of the normal vectors of two planes result in a vector parallel to the line of intersection of the two planes.
Corresponding normal vectors of the planes are
<5,-1,-6> and <1,1,1>
We calculate the cross product as a determinant of (i,j,k) and the normal products
i j k
5 -1 -6
1 1 1
=(-1*1-(-6)*1)i -(5*1-(-6)1)j+(5*1-(-1*1))k
=5i-11j+6k
=<5,-11,6>
Check orthogonality with normal vectors using scalar products
(should equal zero if orthogonal)
<5,-11,6>.<5,-1,-6>=25+11-36=0
<5,-11,6>.<1,1,1>=5-11+6=0
Therefore <5,-11,6> is a vector parallel to the line of intersection of the two given planes.
A) $702,000 is the answer mate
Answer:
![\sqrt[]{\frac{x+8}{4}}-3](https://tex.z-dn.net/?f=%5Csqrt%5B%5D%7B%5Cfrac%7Bx%2B8%7D%7B4%7D%7D-3)
Step-by-step explanation:

First rewrite
as y

Now swap y and x

Add 8 on both sides.


Divide by 4.


Extract the square root on both sides.
![\sqrt[]{\frac{x+8}{4}}=\sqrt[]{(y+3)^2}](https://tex.z-dn.net/?f=%5Csqrt%5B%5D%7B%5Cfrac%7Bx%2B8%7D%7B4%7D%7D%3D%5Csqrt%5B%5D%7B%28y%2B3%29%5E2%7D)
![\sqrt[]{\frac{x+8}{4}}=y+3](https://tex.z-dn.net/?f=%5Csqrt%5B%5D%7B%5Cfrac%7Bx%2B8%7D%7B4%7D%7D%3Dy%2B3)
Subtract 3 on both sides.
![\sqrt[]{\frac{x+8}{4}}-3=y+3-3](https://tex.z-dn.net/?f=%5Csqrt%5B%5D%7B%5Cfrac%7Bx%2B8%7D%7B4%7D%7D-3%3Dy%2B3-3)
![\sqrt[]{\frac{x+8}{4}}-3=y](https://tex.z-dn.net/?f=%5Csqrt%5B%5D%7B%5Cfrac%7Bx%2B8%7D%7B4%7D%7D-3%3Dy)