Answer:
48
Step-by-step explanation:
I'm assuming all the points are on one line..
In that case you'd subtract the big segment (67) by the little segment (19) to get the other little segment
N³ - 2n² - 15n = 0
"n"
n (n² + 2n - 15) = 0
n = 0
n² + 2n - 15 = 0
Δ = (2)² - 4(1)(-15)
Δ = 4 + 60 = 64
n' = (-2+8) / 2 = 6/2 =3
n'' = (-2-8) / 2 = -10/2 = -5
Solution:
S {-5 , 0 , 3 }
Answer with Step-by-step explanation:
Let F be a field .Suppose 
and 
We have to prove that a has unique multiplicative inverse.
Suppose a has two inverses b and c
Then, 
where 1 =Multiplicative identity
(cancel a on both sides)
Hence, a has unique multiplicative inverse.
Can someone please help me with this problem
Answer:
the relation is not one-to-one.
Step-by-step explanation:
it can't because every number is in the 4th quadrant.
