It is possible, depending on what on earth AC and BC are!
(a - 5b)(c - 5d)
You cannot prove it since it is not true for a general quadrilateral.
In this case you want to group the terms so they have at least two terms in common. First step group and rewrite it: abc + a'bc + a'b'c' + a'b'c + ab'c' + abc' = Use the rule Identities x(y+z)=xy+xz: bc(a+a') + a'b'(c'+c) + ac'(b'+b) = Use the rule Identities x+x'=1: bc (1) + a'b'(1) + ac'(1) = Use the rule Identities x(1) = x: bc+a'b'+ac'
6(b - ac + b2 - bc)
15 units
never A+ :))
5
The distributive property states that a(b + c) = ab + ac. This only works in your case if you meant to write 15(x + 20). That would equal 15x + 300.
yes because ab plus bc is ac
x(x - 1)(4x - 15)
If line BE is the bisector of segment AC, it means that BE divides AC into two equal segments. Therefore, if AB is 7, then AC must be twice that length, making AC equal to 14.
15 ac + 20 bc + 6 ad + 8 bd =3a (5c + 2d) + 4b (5c + 2d) =(3a + 4b) (5c + 2d)
You could conclude that B lies between A and C.
Alternator and ac generator is equal
associative? single replacement
An offset AC wave. It will be offset by the magnitude of the DC applied.
A+BC+AC+B=A+BC+AC+B unless any of these variables has an assigned value.