3^2-4[9-(-3+6)4]
2006-08-31 14:51:51 · 7 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
3^2-4[9-(-3+6)4]
first do the ones in Parenthesis
3^2 - 4 [9 - (3) 4]
then you multiply 3 by 4
3^2 - 4 [9 - 12]
then subtract the ones inside the parenthesis
3^2 - 4[-3]
do the one with exponents
9 - 4[-3]
do the multiplication
9 + 12
then add
21
All you have to do is follow the rule
PEMDAS or
Parenthises - Exponent - Multiplication - Division - Addition - Subtraction
2006-08-31 15:09:23 · answer #1 · answered by Hi-kun 2 · 1⤊ 0⤋
21
2006-09-01 02:03:58 · answer #2 · answered by xavierbondoc_15 1 · 0⤊ 0⤋
3^2 - 4[9-(-3+6)4]
I presume that it is (-3+6)^4 and solving
9 - 4[9-(3)^4]
9-4[9-81]
9-4[-72]
9+288 = 297
If it is (-3+6)*4
then 9-4[9-(3)*4]
9-4[9-12]
9-4[-3]
9+12
21
2006-08-31 15:12:32 · answer #3 · answered by Peter T 2 · 1⤊ 0⤋
3^2 -4[9-(3)4]
3^2 -4[9-12]
3^2 -4[-3]
3^2 +12
9+12
21
2006-08-31 15:15:54 · answer #4 · answered by daeylcq 2 · 1⤊ 0⤋
Assuming it's 3 squared minus the rest the answer is 21.
2006-08-31 15:16:48 · answer #5 · answered by TheDude 3 · 0⤊ 1⤋
This is not a polynomial, it is a numerical expression and 21 is the correct answer
2006-08-31 16:18:18 · answer #6 · answered by MollyMAM 6 · 0⤊ 1⤋
no
2006-09-04 14:16:52 · answer #7 · answered by chris m 5 · 0⤊ 0⤋