Solution :
Let
be the unit vector in the direction parallel to the plane and let
be the component of F in the direction of
and
be the component normal to
.
Since, 


Therefore, 
From figure,

We know that the direction of
is opposite of the direction of
, so we have



The unit vector in the direction normal to the plane,
has components :


Therefore, 
From figure,

∴ 

Therefore,


Answer:
20+20+20+20+20+20= 120
Step-by-step explanation:
2^6 = 120 hope this helps! plz mark brainliest
Answer:
from those are similar, the ratio of side will be 40/24=x/20 so x=33.3333 is equal 33 and one-third.
Answer:
1 and 2/4 a cup
Step-by-step explanation:
Ummm hi :) Ann’s Kama abs