Look at the expression below. Use the order of operations to simplify the expression. Show all work. (10-6)[(5 + 2) + 3]
@Hero @ganeshie8 @Mertsj
@zepdrix @iambatman
solve operations in brackets and parenthesis first remember pemdas
\[\Large\rm (10-6)[\color{orangered}{(5 + 2)} + 3]\]Innermost brackets first :)
ok
Lots of brackets in this one :o can get a little tricky.
5+3 is 8
Well, we would like to do the orange part first, since that's how they applied the brackets.
yes
5 + 2 is 7, yes? :)\[\Large\rm (10-6)[\color{orangered}{(5 + 2)} + 3]\]\[\Large\rm (10-6)[\color{orangered}{(7)} + 3]\]So we dealt with that first set of brackets, we'll drop them.\[\Large\rm (10-6)[7+3]\]
Keep in mind, there is NO DIFFERENCE between round and square brackets. We just use different shapes sometimes.
ok
\[\Large\rm \color{orangered}{(10-6)}[7+3]\]Then how bout this set of brackets? c:
10-6 is 4
\[\Large\rm \color{orangered}{(4)}[7+3]\]Ok good!\[\Large\rm (4)\color{orangered}{[7+3]}\]How bout these brackets?
7+3 is 10
\[\Large\rm (4)\color{orangered}{[10]}\]Good good good c:\[\Large\rm (4)[10]\]This last operation is multiplication. That's what it means when you have brackets touching, or touching a number.
4*10 is 40
Yayyyy good job \c:/
can you answer more
I can try c: I have a lil spare time
3 ∙ (5 + 9) ÷ 7 – 6
3 * (5 + 9) / 7 – 6
\[\Large\rm 3 \times (5 + 9) \div 7-6\]Hmm so what operation should we prioritize? Which is most important? `multiply`, `brackets`, `divide`, `subtract` those are the operations we see in this problem. Which one first? :)
brackets
Ok good, that gives you uhhh, 14, yes?
yes
then we multiply
|dw:1415583460134:dw|
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