x + 16 = 64 |subtract 16 from both sides
x = 48
The answer is A because the same rule applies
Well, variables are stuff like x and y because they can represent a number. They are variable in an equation. so if you have (3x^3)+(5X^2)+Y+9 then your variables are x X and Y (im assuming the capital X is there and not a typo) if that was a typo and meant to be lowercase, then its just a 2 variable with the x and y
A. (x+3)(4x+5)
(multiply x with 4x+5 then multiply 3 with 4x+5)
=4x^2+5x+12x+15
=4x^2+17x+15
b,c, and d are similar
