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.

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Answer 1 · 2017-01-29T07:18:52

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

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#