How do you solve #2\times ( 14-: 2)^{2} + 5\times 12#?

3 Answers
Sep 21, 2016

Answer:

#158#

Explanation:

Always count the number of terms first.
Simplify each term, following the order of operations - brackets first,
the strongest to weakest :
Powers and roots, then multiply and divide.
Each term will simplify to a single number which are added or subtracted in the last step.

#[2xxcolor(red)((14-: 2))^2]color(blue)( + 5xx 12)" "larr# there are 2 terms

=#[2xxcolor(red)((7)^2)]color(white)(xxxxx)color(blue)( + 60)" "larr# keep them separate

=#" "color(red)(2xx49)color(white)(xxxx.x)color(blue)( + 60)#

=#color(white)(xxx)color(red)(98)color(white)(xxxxxx.x)color(blue)( + 60)" "larr# now add

=#" "158#

Sep 21, 2016

Answer:

158

Explanation:

work:
2 x #(14/2)^2# + 5 x 12
2 x #(7)^2#+ 5 x 12
2 x (49) + 5 x 12
98 + 60
158

Sep 21, 2016

Answer:

158

Explanation:

When dealing with a calculation involving ' mixed' operations we can use the acronym PEMDAS.

#P-"Parenthesis"to("brackets")#

#E-"Exponents"to"powers"#

#M-"Multiplication"#

#D-"Division"#

multiplication and division are of equal precedence.

#A-"Addition"#

#S-"Subtraction"#

Following this order on the given calculation.

#2xx(color(red)(7))^2+5xx12#

#=2xxcolor(blue)(49)+5xx12#

#=(color(magenta)(2xx49))+( color(magenta)(5xx12))#

#=98+60=158#