-14/15+-2v/3+v/3 that is the answer
Answer:
Distance to the xy-plane = |z|
Distance to the yz-plane = |x|
Distance to the xz-plane = |y|
Step-by-step explanation:
The distance from P(x,y,z) to the xy-plane is by definition the magnitude of the vector that goes from the perpendicular projection of P over the xy-plane to the point P, which is exactly the magnitude of the vector (0,0,z) = |z| the absolute value of z
Similarly, the distance from P to the yz-plane is |x| and the distance from P to the xz-plane is |y|
Distance to the xy-plane = |z|
Distance to the yz-plane = |x|
Distance to the xz-plane = |y|
The answer is x-18
Do need the complete answer of your question??
.6k+.9=1.6
.6k=.7
k=7/6= 1 1/6
k= 1 1/6
Hope this helps.
