How do you evaluate #(abs(3-3)--4)times5#?
1 Answer
Explanation:
PEMDAS needed:
P: Parenthesis
E: Exponents
M: Multiplication
D: Division
A: Addition
S: Subtraction
First solve what is in the parenthesis:
#( |0| + 4 ) xx 5 larr# two minuses make a plus and the absolute value of#0# is#0#
#=(0 + 4) xx 5#
#=(4) xx 5#
#=20#
