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.

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.

Cameron Arnold · Accepted Answer

#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#

Zoe Adams · Answer

undefined To solve the expression $3^0 (5^0 - 6^{-1} \times 3)/ 2^{-1}$ correctly, you should follow the order of operations, also known as PEMDAS/BODMAS, which stands for Parentheses/Brackets, Exponents/Orders (i.e., powers and square roots, etc.), Multiplication and Division (from left to right), and Addition and Subtraction (from left to right).

1. First, evaluate the expressions within parentheses and brackets. In this case, there are none.

2. Next, deal with exponents. Here, you have exponents involving negative numbers, $3^0$, $5^0$, and $6^{-1}$. $3^0$ and $5^0$ are both equal to 1.

3. After handling exponents, perform any multiplications and divisions. Since multiplication and division have equal precedence, you should proceed from left to right. Therefore, calculate $6^{-1} \times 3$ first.

4. Then, you'll have the expression $3^0 (5^0 - \frac{3}{6})/ 2^{-1}$. Next, calculate $2^{-1}$, which is equal to $\frac{1}{2}$.

5. Now, you have $3^0 (5^0 - \frac{3}{6})/ \frac{1}{2}$. Next, evaluate $3^0$, $5^0$, and $\frac{3}{6}$.

6. Finally, perform the division, followed by any remaining multiplication.

Following this order, you'll eventually arrive at the solution, which is 4.

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.