Hello :
the normal vector of the plane is : d' = [4,-1,5]...(vector perpendicular to the plane)<span>
d </span><span>⊥ d' because : (4)(2)+(-1)(3)+(5)(1)=0
</span>the line is perpendicular to the plane .
Answer:
Step-by-step explanation:
subtract llol
first 1
x . y
1 . 2
3 . 4 rewrite =y/x
2-4/1-3
-2/2=-1
do for the rest bye
Answer:
19683
Step-by-step explanation:
As the formula of the geometric sequence:
![a_{n} =a_{1} *r^{n-1}](https://tex.z-dn.net/?f=a_%7Bn%7D%20%3Da_%7B1%7D%20%2Ar%5E%7Bn-1%7D)
In the sequence of 1, 3, 9..., you have a1 = 1 and r = 3. Therefore:
![a_{10} =1 *3^{10-1} = 3^{9} =19683](https://tex.z-dn.net/?f=a_%7B10%7D%20%3D1%20%2A3%5E%7B10-1%7D%20%20%3D%203%5E%7B9%7D%20%3D19683)
Hope this helps!