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Mathematics 13 Online
OpenStudy (anonymous):

i dont undersatnd my homework its order of operations (A)

OpenStudy (amistre64):

1. groupings 2. divide or multiply left to right 3. add or subtract left to right

OpenStudy (amistre64):

groupings tend to be easiest when done from the inside-out

OpenStudy (anonymous):

but it has two parentesis

OpenStudy (amistre64):

im sure it does, care to share it?

OpenStudy (amistre64):

if your trying to use that equation button; it might be easier just to type it outright

OpenStudy (anonymous):

its i cant type it

OpenStudy (amistre64):

i tried reading your mind, but all I got was boy clutter :)

OpenStudy (anonymous):

its difficult

OpenStudy (anonymous):

ill try and type it

OpenStudy (amistre64):

[(1+2(3-2))+14]^2 ??

OpenStudy (anonymous):

ok my paper is on this page just go to math-drills.com order of operations

OpenStudy (amistre64):

then what?

OpenStudy (anonymous):

yea its that page now go to positie and negative intigers three steps (A)

OpenStudy (anonymous):

i meant positive

OpenStudy (amistre64):

and which one are you stuck on?

OpenStudy (anonymous):

number 1

OpenStudy (anonymous):

send me an attachment to see if your in the right page

OpenStudy (amistre64):

[ (10 div (-10) )^7 ]^2

OpenStudy (anonymous):

yes im stuck in that one

OpenStudy (amistre64):

its just another way to write: 1

OpenStudy (amistre64):

but lets work thru it; what do youthink we should do first?

OpenStudy (anonymous):

the parenteses

OpenStudy (amistre64):

which set :) (10 div (-10)) (...)^7 0r (...)^2

OpenStudy (anonymous):

the (...)^7

OpenStudy (amistre64):

ok, but youlll prolly understand it better if we do the: (10 div (-10)) first; either way is fine tho

OpenStudy (anonymous):

ok

OpenStudy (amistre64):

\[(\frac{10}{-10})^7=\frac{10^7}{-10^7}=\frac{10,000,000}{-10,000,000}\]

OpenStudy (anonymous):

im sorry but i dont get it

OpenStudy (amistre64):

i didnt think you would for that one :) lets me guide you to an easier way

OpenStudy (anonymous):

ok i think that (-10) goes first

OpenStudy (amistre64):

the inner most grouping is: (10 div(-10)) what can you tell me about this value?

OpenStudy (amistre64):

have you done fractions yet? \[(10\div (-10))=\frac{10}{-10}\]

OpenStudy (anonymous):

ummm....[(10div(-10) the -10 turns into positive

OpenStudy (amistre64):

the -10 does not turn positive; that is not in any rule of math :)

OpenStudy (anonymous):

ohh sorry im confused

OpenStudy (amistre64):

10 times ___ = -10 can you fill in the blank?

OpenStudy (anonymous):

_10

OpenStudy (anonymous):

imean -1

OpenStudy (amistre64):

-1 is correct; which tells us that: \(10\div(-10)= -1\)

OpenStudy (anonymous):

do i write that down in my paper

OpenStudy (amistre64):

that is part of the process yes, there are 2 more groupings to go

OpenStudy (amistre64):

lets go to the next in line, that ^7

OpenStudy (anonymous):

ok

OpenStudy (amistre64):

is this correct so far? \[(10\div(-10))^7=(-1)^7\]

OpenStudy (anonymous):

are the ones at the top 7 or?

OpenStudy (anonymous):

because i wrot 7 is it ok

OpenStudy (amistre64):

we have: \[\left((numbers)^7\right)^2\]

OpenStudy (amistre64):

we figured out that the "numbers" = -1 so lets move outwards to the \(()^7\) part right?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

imma show you what i wrote

OpenStudy (anonymous):

(10div(-10))^7=(-1)

OpenStudy (amistre64):

\((-1)^7\) means: -1 x -1 x -1 x -1 x -1 x -1 x -1 (-1 x -1) x (-1 x -1) x (-1 x -1) x (-1) (1 x 1 x 1) x -1 1 x -1 -1 thats fine as long as you know that: \((-1)^7=(-1)\)

OpenStudy (anonymous):

ok do i have my work correct

OpenStudy (amistre64):

so far so good, just one last grouping to take care of

OpenStudy (anonymous):

ok

OpenStudy (amistre64):

\((-1)^2=?\)

OpenStudy (anonymous):

is it 2

OpenStudy (amistre64):

when we work it thru the groupings it looks like this: \[\left((10\div(-10))^7\right)^2\] \[\left((-1)^7\right)^2\] \[(-1)^2\] \[1\]

OpenStudy (amistre64):

and no, -1 x -1 = 1, not 2 :)

OpenStudy (anonymous):

is that the answer

OpenStudy (amistre64):

well, it is the problem i have been working with you on hasnt it?

OpenStudy (anonymous):

yes

OpenStudy (amistre64):

then i believe that that is the answer :)

OpenStudy (anonymous):

can you help me on #2 or are you too busy

OpenStudy (amistre64):

my time is up, i gotta get ready for tomorrow, good luck tho ;)

OpenStudy (anonymous):

what do i do if i need some more help im new here

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