Part 6
we have the expression
[tex]\lbrack6\cdot(4-2)+(21\cdot\frac{2}{3}+1)\rbrack+3^2[/tex]
To solve this problem
Apply PEMDAS
P ----> Parentheses first
E -----> Exponents (Powers and Square Roots, etc.)
MD ----> Multiplication and Division (left-to-right)
AS ----> Addition and Subtraction (left-to-right)
that means
step 1
Parentheses first
(4-2)=2
[tex]21\cdot\frac{2}{3}+1[/tex]
solve multiplication
[tex]\begin{gathered} \frac{42}{3}+1 \\ \frac{45}{3} \end{gathered}[/tex]
substitute in the original expression
[tex]\lbrack6\cdot2+\frac{45}{3}\rbrack+3^2[/tex]
Parentheses first again
multiplication
[tex]\begin{gathered} 6\cdot2+\frac{45}{3} \\ 12+\frac{45}{3} \\ \frac{81}{3} \end{gathered}[/tex]
substitute
[tex]\frac{81}{3}+3^2[/tex]
solve exponents first
[tex]\begin{gathered} \frac{81}{3}+9 \\ \frac{108}{3} \\ 36 \end{gathered}[/tex]
Part 2
GCF of 14 and 42
we have
14=2*7
42=2*3*7
so
GFC=2
TRue
Part 3
GCF 16 and 56
16=2*2*2*2
56=2*2*2*7
GCF=8
TRue
Part 4
LCM of 12 and 9
12=2*2*3
9=3*3
LCM=4*9=36
true
Part 5
LCM of 12 and 6
12=2*2*3
6=2*3
LCM=12
false
