How do you evaluate (1+ 3^ { 2} ) \div 2\cdot 3^ { 2}(1+32)÷232?

2 Answers
Dec 13, 2017

45

Explanation:

You would want to follow the order of operations BEDMAS (Bracket, exponent, division, multiplication, addition, subtraction.)

Let's start by solving what's in the brackets, so

(1+3^2)÷2*3^2(1+32)÷232

= (1+9)÷2*3^2=(1+9)÷232

(10)÷2*3^2(10)÷232

Next, we can solve for the exponents

(10)÷2*3^2(10)÷232

= 10÷2*9=10÷29

Since we see the division prior to the multiplication, we have to solve that portion first.

10÷2*910÷29

= 5*9=59

Now all we have left is to multiply.

5*959

=45=45

Dec 13, 2017

5/959

Explanation:

= (1+3^2)/(2*3^2)=1+32232

= 1/(2*3^2) + 3^2/(2*3^2)=1232+32232

= 1/(2*3^2) + 1/2=1232+12

= (1/2) *(1/3^2 + 1)=(12)(132+1)

= (1/2) *(1/9 + 1)=(12)(19+1)

= (1/2) *((1+9)/9)=(12)(1+99)

= 10/(2*9)=1029

= 5/9=59