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+ :))
If AC plus CB equals AB and AC is equal to CB, then point C is the midpoint of segment AB. This means that point C divides the segment AB into two equal parts, making AC equal to CB. Therefore, point C is located exactly halfway between points A and B.
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
AB plus BC equals AC is an example of the Segment Addition Postulate in geometry. This postulate states that if point B lies on line segment AC, then the sum of the lengths of segments AB and BC is equal to the length of segment AC. It illustrates the relationship between points and segments on a line.
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