Order of Operations

[fc1123]

Okay I am working on the order of operations the problem is (3-7)^2/2x5

so far I have -4 to the 2nd power divded by 2x5 lost on how to slove this.

 

[mmm4444bot]

lost on how to slove this

Maybe you need to slover more. Should I start frying hamburger? :lol:

Does the given expression look like this?

\(\displaystyle \frac{(3 - 7)^2}{2 \times 5}\)

 

[fc1123]

Re:

lost on how to slove this

Maybe you need to slover more. Should I start frying hamburger? :lol:

You did not type grouping symbols around the denominator, so your typing indicates that the denominator is 2, not 2 times 5.

Please answer my question, and then I will explain.

Does the given expression look like this?

\(\displaystyle \frac{(3 - 7)^2}{2 \times 5}\)

yes it does thanks I wonder how to type it up correctly on here still haven't figured out how to do that on here.

 

[mmm4444bot]

I was trying to delete my statement about your notation, when you posted a reply. I was thinking about something else, so your original post is clear enough.

Okay, you determined that (3 - 7)^2 is (-4)^2.

That is correct.

Do you know how to square -4 ?

The fraction bar is the vertical line. It represents division, but it is also a grouping symbol.

It separates the expression on top (the numerator) from the expression on the bottom (the denominator)

The Order of Operations tells us to do the part on top, then do the part on bottom, then divide the results.

Can you try that?

Cheers ~ Mark

 

[fc1123]

yes is it (-4)^2=-16 so now where does the division come in that it is 2X5=10

 

[mmm4444bot]

yes is it (-4)^2=-16

Oops. No, it's not.

(-4)^2 = (-4)*(-4) = (-1)*(-1)*(4)*(4) = (1)*(4) = 16

where does the division come in that it is 2X5=10

The part on the top is 16, the part on the bottom is 10, and the fraction bar tells us to divide, so we have:

16/10

Reduce this ratio to lowest terms.

 

[lookagain]

Re:

(-1)*(-1)*(4)*(4) = >>> (1)*(4) = 16 <<<

This should be \(\displaystyle (1)(4)^2 = 16\) or
\(\displaystyle (1)(16) = 16.\)

 

[BigGlenntheHeavy]

\(\displaystyle \frac{(3-7)^2}{(2)(5)}, \ note \ the \ vinculum \ (fraction \ bar \ for \ those \ of \ you \ from \ Rio \ Linda)\)

\(\displaystyle is \ drawn \ over \ two \ or \ more \ algebraic \ terms \ to \ show \ that \ they \ are \ to \ be \ considered \ as \ a\)

\(\displaystyle single \ term.\)

\(\displaystyle Hence, \ \frac{(3-7)^2}{(2)(5)} \ = \ \frac{(-4)^2}{10} \ = \ \frac{16}{10} \ = \ \frac{8}{5}.\)

 

[lookagain]

\(\displaystyle \frac{(3-7)^2}{(2)(5)}, \ note \ the \ vinculum \ (fraction \ bar )\)

[/tex]

A fraction bar is a vinculum, but all vinculums are not fraction bars.

http://mathworld.wolfram.com/Vinculum.html