How do you solve #9- 16\cdot \frac{2^{2} - 1^{3}}{2( - 4)} - 8\div 2#?

Question

Eli Assouline · Accepted Answer

#11#

#"when evaluating expressions with "color(blue)"mixed operations"# #"there is a particular order that must be followed"#

#"follow the order as set out in the acronym PEMDAS"#

[P-parenthesis (brackets), E-exponents (powers), M-multiplication, D-division, A- addition, S-subtraction ]

#"multiplication/division have equal precedence so when"# #"they occur in the same expression evaluate from left"# #"to right. This is also the case with addition/subtraction"#

#9-16xx(2^2-1^3)/(2(-4))-8-:2#

#=9-cancel(16)^-2xx(4-1)/(cancel(-8)^1)-8-:2larrcolor(red)"P/E"#

#=9-(-2xx3)-8-:2larrcolor(red)" tidy up fraction"#

#=9-(-6)-4larrcolor(red)" multiplication/division"#

#=9+6-4larrcolor(red)" evaluate left to right"#

#=11#

Jameson Allen · Answer

undefined To solve the expression $9 - 16 \cdot \frac{2^2 - 1^3}{2(-4)} - \frac{8}{2}$:

First, evaluate the expressions within parentheses and exponents:
$$2^2 = 4$$
$$1^3 = 1$$
$$2^2 - 1^3 = 4 - 1 = 3$$

Now, substitute these values back into the expression:
$$9 - 16 \cdot \frac{3}{2(-4)} - \frac{8}{2}$$

Next, perform the multiplications and divisions:
$$\frac{3}{2(-4)} = \frac{3}{-8} = -\frac{3}{8}$$
$$\frac{8}{2} = 4$$

Now, substitute these values back into the expression:
$$9 - 16 \cdot (-\frac{3}{8}) - 4$$

$$= 9 + \frac{48}{8} - 4$$

$$= 9 + 6 - 4$$

Finally, perform the addition and subtraction:
$$= 15 - 4$$

$$= 11$$

So, the solution to $9 - 16 \cdot \frac{2^2 - 1^3}{2(-4)} - \frac{8}{2}$ is $11$.