Question

Answer is 4 which order to solve? `3^0 (5^0 - 6^-1 * 3)/ 2^-1` I don't know if I should bring the `6^-1` down first or multiply by 3 first.

Answer 1

`3^0(((5^0-6^(-1)*3))/2^(-1))=1`

Explanation:

Following the order of operations, with the parentheses made explicit:

First, perform any operations within parentheses.

`3^0color(red)((((5^0-6^(-1)*3))/2^(-1)))`

Within those parentheses, we treat numerators and denominators as having parentheses around them, and so perform operations within those first.

`=3^0((color(red)((5^0-6^(-1)*3)))/2^(-1))`

Evaluate any exponents. Recall that if `x!=0`, then `x^0 = 1` and that `x^-a = 1/x^a`.

`=3^0(((color(red)(5^0)-6^(-1)*3))/2^(-1))`

`=3^0(((1-color(red)(6^(-1))*3))/2^(-1))`

`=3^0(((1-color(red)(1/6^1)*3))/2^(-1))`

`=3^0(((1-1/6*3))/2^(-1))`

Perform any multiplication or division, going left to right.

`=3^0(((1-color(red)(1/6*3)))/2^(-1))`

`=3^0(((1-color(red)((3)/6)))/2^(-1))`

`=3^0(((1-1/2))/2^(-1))`

Perform any addition or subtraction, going left to right.

`=3^0(((color(red)(1-1/2)))/2^(-1))`

`=3^0((1/2)/2^(-1))`

All operations in the numerator have been completed. Moving to the denominator, we have an exponent to evaluate.

`=3^0((1/2)/color(red)(2^(-1)))`

`=3^0((1/2)/color(red)(1/2^1))`

`=3^0((1/2)/(1/2))`

We now perform the remaining division. Recall that any nonzero number divided by itself is `1`.

`=3^0(color(red)((1/2)/(1/2)))`

`=3^0(1)`

All operations within parentheses have been evaluated. Going back, we now evaluate the remaining exponents.

`=color(red)(3^0)*1`

`=1*1`

And finally, we perform the remaining multiplication.

`=1`