Mathematics
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OpenStudy (rootbeer003):
Which pair of numbers is not a factor pair of 40?
4 and 10
–4 and –10
–8 and 5
20 and 2
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OpenStudy (rootbeer003):
AYEEEEEE
OpenStudy (anonymous):
oh no!!!
OpenStudy (rootbeer003):
WELCOME
OpenStudy (anonymous):
what is \(-8\times 5\)?
OpenStudy (rootbeer003):
-45
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OpenStudy (anonymous):
hint, it is not \(40\)
OpenStudy (anonymous):
:D well for the third one, what does -8 x 5 make?
OpenStudy (rootbeer003):
40*
OpenStudy (frozen15):
@satellite73 I couldn't message you but are you good at g.um? English sorry for intrupting the question
OpenStudy (anonymous):
no not 40
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OpenStudy (anonymous):
you got one minus sign right? where did it go?
OpenStudy (rootbeer003):
OHHHH IM DUMBB
OpenStudy (anonymous):
Yup
OpenStudy (rootbeer003):
WOOT
OpenStudy (rootbeer003):
um no satelite get back here
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OpenStudy (anonymous):
what??
OpenStudy (anonymous):
How is that wrong???
OpenStudy (rootbeer003):
Which of the following numbers is prime?
111
121
131
141
OpenStudy (anonymous):
\[-8\times 5=-40\neq 40\]
OpenStudy (anonymous):
oh another question
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OpenStudy (rootbeer003):
im not done
OpenStudy (anonymous):
111 is divisible by 3 since \(1+1+1=3\)
OpenStudy (anonymous):
\[121=11^2\] not prime
OpenStudy (rootbeer003):
but it asks which is prime
OpenStudy (anonymous):
141 is also divisible by 3 since \(1+4+1=6\)
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OpenStudy (anonymous):
as usual, it is C 131
OpenStudy (anonymous):
you now how to tell if a number is divisible by 3?
OpenStudy (rootbeer003):
Determine the prime factorization of 99.
34
34• 2 • 5
34• 11
32• 11
OpenStudy (anonymous):
??
OpenStudy (rootbeer003):
HOW
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OpenStudy (anonymous):
\[99=9\times 11\] right ?
OpenStudy (rootbeer003):
ye
OpenStudy (anonymous):
and since \(9=3\times 3\) that means \[99=3\times 3\times 11=3^2\times 11\] which does not see bot be one of your choices
OpenStudy (rootbeer003):
mhm
OpenStudy (anonymous):
unless your last choice is \(3^2\times 11\) and not \(32\times 11\) which is what you wrote
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OpenStudy (rootbeer003):
mhmm
OpenStudy (rootbeer003):
so d
OpenStudy (anonymous):
yeah D
OpenStudy (rootbeer003):
alright
OpenStudy (rootbeer003):
Determine the prime factorization of 476.
22 • 32 • 13
22 • 7 • 19
22• 7 • 17
2 • 35
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OpenStudy (anonymous):
donkey work
OpenStudy (rootbeer003):
hay im tryin
OpenStudy (anonymous):
want to do it the long boring way ?
OpenStudy (rootbeer003):
no
OpenStudy (rootbeer003):
wanna do the ' i learn ' it way
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OpenStudy (rootbeer003):
>:)
OpenStudy (anonymous):
ok then it is clear that \(476\) is divisble by 2 right, because it is even
OpenStudy (rootbeer003):
yass
OpenStudy (anonymous):
ok so step one is find half of \(476\) i.e. divide it by 2
what do you get?
OpenStudy (rootbeer003):
238
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OpenStudy (anonymous):
ok so step one \[476=2\times 238\] but \(238\) is also even, what is half of 238?
OpenStudy (rootbeer003):
119
OpenStudy (anonymous):
ok so we are at \[476=2^2\times 119\]
119 is not even, s it is not divisible by 2
OpenStudy (anonymous):
\[1+1+9=11\] so it is not divisible by 3
OpenStudy (anonymous):
it is obviously not divisible by 5
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OpenStudy (anonymous):
try 7
OpenStudy (rootbeer003):
k
OpenStudy (rootbeer003):
17!
OpenStudy (rootbeer003):
took me awhile to process that
OpenStudy (anonymous):
ok so \[476=2^2\times 7\times 17\]
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OpenStudy (rootbeer003):
cccccc
OpenStudy (rootbeer003):
hello?
OpenStudy (anonymous):
hi!
OpenStudy (rootbeer003):
Find the greatest common factor of the pair of numbers.
104 and 260
4
13
20
52
OpenStudy (anonymous):
yeah C
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OpenStudy (anonymous):
\[104=2\times 52\] right? does \(52\) go in to 260 evenly?
OpenStudy (rootbeer003):
ya
OpenStudy (anonymous):
then pick that one
OpenStudy (rootbeer003):
dddddddddddddd
OpenStudy (rootbeer003):
?
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OpenStudy (anonymous):
do i have to do everything?
yes D
OpenStudy (rootbeer003):
lagg
OpenStudy (rootbeer003):
Find the greatest common factor of the pair of numbers.
348 and 426
12
6
9
3
OpenStudy (rootbeer003):
ill look into this one
OpenStudy (anonymous):
it has be be even right, because both numbers are even
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OpenStudy (anonymous):
see if it is 12, if not, it is 6
OpenStudy (rootbeer003):
well i dont want to get it wrong
OpenStudy (rootbeer003):
um 6
OpenStudy (rootbeer003):
cause its even
OpenStudy (anonymous):
lage real bad hold on
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OpenStudy (anonymous):
you are right, 6
OpenStudy (rootbeer003):
ready 4 some more?
OpenStudy (anonymous):
if i can get past this lage
OpenStudy (rootbeer003):
simplify
OpenStudy (rootbeer003):
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OpenStudy (anonymous):
\[\frac{a^9}{a^4}=a^{9-4}\]
OpenStudy (rootbeer003):
c
OpenStudy (rootbeer003):
waitt
OpenStudy (rootbeer003):
a
OpenStudy (anonymous):
A
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OpenStudy (rootbeer003):
iz a?
OpenStudy (anonymous):
yup
OpenStudy (rootbeer003):
OpenStudy (anonymous):
i have never heard of this but the only answer that makes sense is "power of products"
OpenStudy (rootbeer003):
huh k
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OpenStudy (rootbeer003):