pre algeba ????? [1+3(9+12)]-4 to third power

[TylerSteen]

I am unsure how to solving this [1+3(9+12)]-4 to third power

 

[mmm4444bot]

Hi Tyler:

The order of operations tells us to start inside the innermost set of grouping symbols.

For this exercise, the innermost set of grouping symbols contains (9 + 12).

Next, do the multiplication by 3 to get rid of these innermost parentheses.

Then do the addition of 1. This finishes the work inside the square brackets.

Next, do the subtraction of 4 to get rid of the square brackets.

Finish by raising your last result to the third power.

If you need more help, then PLEASE show your work and try to explain why you're stuck.

Cheers,

~ Mark

 

[Deleted member 4993]

I am unsure how to solving this [1+3(9+12)]-4 to third power

Is your problem

$\displaystyle [1 \, + \, 3 \, (9 \, + \, 12)] \, - 4^3$

or

$\displaystyle [{1 \, + \, 3 \, (9 \, + \, 12)} \, - 4]^3$

For either case,

Have you learned Order of operations - PEMDAS?

Go to:

http://www.purplemath.com/modules/orderops.htm

for a quick review with worked examples.

If you are still stuch - write back showing your work/thoughts, so that we know where to begin to help you.

 

[TylerSteen]

Hey thanks for all your help. I do appreciate it.


The order of operations tells us to start inside the innermost set of grouping symbols.

For this exercise, the innermost set of grouping symbols contains (9 + 12).

Next, do the multiplication by 3 to get rid of these innermost parentheses.

Then do the addition of 1. This finishes the work inside the square brackets.

Next, do the subtraction of 4 to get rid of the square brackets.

Finish by raising your last result to the third power.

If you need more help, then PLEASE show your work and try to explain why you're stuck.

Cheers,

~ Mark [/size][/quote]

 

[Denis]

I am unsure how to solving this [1+3(9+12)]-4 to third power

Tyler, try and be a bit clearer next time: [1 + 3(9 + 12)] - 4^3
64 - 4^3 = 64 - 64 = 0

 

[mmm4444bot]

Is your problem

$\displaystyle [1 \, + \, 3 \, (9 \, + \, 12)] \, - 4^3$

or

$\displaystyle [{1 \, + \, 3 \, (9 \, + \, 12)} \, - 4]^3$

Hi Tyler:

My post above assumes that your exercise is the second expression listed by Subhotosh.

~ Mark