Answer:
[x+6y+2z][x²+(6y)²+(2z)²-6xy-12yz-2xz]
Step-by-step explanation:
x³+216y³+8z³-36xyz
x³+(6y)³+(2z)³-3×6×2×xyz
As we know
a³+b³+c³-3abc=(a+b+c)(a²+b²+c²-ab-bc-ca)
Let a=x
b=6y
c=2z
Now.
[x+6y+2z][(x²+(6y)²+(2z)²-x×6y-6y×2z-x×2z]
[x+6y+2z][x²+(6y)²+(2z)²-6xy-12yz-2xz]
Find answer is the given attachment
1
-
8 that is the same as 1 over 8
<span>if v belongs to V, then we can find scalars a1,a2,...,an, such that
v=a1*v1+a2*v2+...+an*vn,
L1(v)=L1(a1*v1+a2*v2+...+an*vn)
=a1*L1(v1)+a2*L1(v2)+...+an*L1(vn)
=a1*L2(v1)+a2*L2(v2)+...+an*L2(vn)
=L2(a1*v1+a2*v2+...+an*vn)
=L2(v)</span>
Answer:
D. A'(1,-7), B'(2,-1), C'8,-2) and D'(7,-5)
Step-by-step explanation:
Probably too late, but I took the quiz
