The direction vector of the line
L: x=1+t, y=4t, z=2-3t
is <1,4,-3>
which is also the required normal vector of the plane.
Since the plane passes through point (-5,9,10), the required plane is :
Π 1(x-(-5)+4(y-9)-3(z-10)=0
=>
Π x+4y-3z=1
Answer:
Associative and Commutative properties
Step-by-step explanation:
Set each set of parentheses to 0 and solve for x.
X—6 = 0
X = 6
X + 5 = 0
X = -5
X-9 = 0
X = 9
The zeros are 6, -5, 9
Answer: (0,12)
Step-by-step explanation:
There was already a question like this on here so i’m going off the answer already given: 30 hours, i hope it’s right
