How do you solve #5+3-2(6-:2)^3+|6-7|#?

Question

1485 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-05-09T02:20:19

See a solution process below:

First, execute the operation within the Parenthesis and within the absolute value function:

#5 + 3 - 2(color(red)(6) -: color(red)(2))^ 3 + abs(color(blue)(6) - color(blue)(7)) => 5 + 3 - 2(3)^3 + abs(-1) =>#

#5 + 3 - 2(3)^3 + 1#

Next, execute the operation with the Exponent:

#5 + 3 - 2color(red)((3)^3) + 1 => 5 + 3 - 2 xx 27 + 1#

Then execute the Multiplication operation:

#5 + 3 - color(red)(2 xx 27) + 1 => 5 + 3 - 54 + 1#

Now, execute the addition and subtraction operations from left to right:

#color(red)(5 + 3) - 54 + 1 => 8 - 54 + 1#

#color(red)(8 - 54) + 1 => -46 + 1#

#color(red)(-46 + 1) => -45#