### Home > CC3MN > Chapter 3 > Lesson 3.1.3 > Problem3-27

3-27.

1. $(-4)(-2)-6(2-5)$

Circle the terms. $\enclose{circle}[mathcolor="blue"]{\color{black}{\left(-4\right)\left(-2\right)}}-\enclose{circle}[mathcolor="blue"]{\color{black}{6\left(2-5\right)}}$ The negative 4 times negative two is circled. and the 6 times 2 minus 5 is circled.

Simplify within the circled terms. $8-6(-3)$

Circle the terms again. Then simplify and repeat. $\enclose{circle}[mathcolor="blue"]{\color{black}{8}}-\enclose{circle}[mathcolor="blue"]{\color{black}{6\left(-3\right)}}$ Circle the 8 and the 6 times negative 3.

1. $23-(17-3·4)^2+6$

For parts (b), (c), and (d) follow the strategy outlined in part (a).

1. $14(2+3-2·2)\div(4^2-3^2)$

$2$

1. $12.7-18.5+15+6.3-1+28.5$