c + c + 2c + c + c = 6c
If a=b and b=c then a must = c
(c+4)(c+5)
You want: abc + ab Factor out the common terms which are "a" and "b" ab ( c + 1 )
Depending on your school they will go, Applied Geometry (D average or lower), Geometry ( C and above), Problem Solving A (D in geometry), Algebra 2 (C or better in Geometry), Problem solving B (D or lower in Algebra 2), Calculus AB (C or better in Algebra 2) and Calculus BC (requires AB)
2b + 2c or 2(b + c)
-c + 2 = 5 -c = 3 c = -3
c is a pro-numeral
1. With boolean algebra, 1 + n is always equal to 1, no matter what the value of n is.
c + c + 2c + c + c = 6c
b+b+b+c+c+c+c =3b+4c
c + c + c + c + c = 5 * c.
b + b + b + c + c + c + c = 3b + 4c
4c
If a=b and c=d then (a+c)=(b+d) ? This is proved very simply by the direct application of perhaps the most fundamental statement in all of Algebra: "If equals are added to equals, the sums are equal."
If a=b and b=c then a must = c
(c+4)(c+5)