Mathematics
9 Online
OpenStudy (anonymous):
Solve the open sentence. 6 < 3b OR 3b + 12 < 9
A.b < –1 OR b > –6
B.b < –1 OR b < 3
C.b < 2 OR b > –3
D.b > 2 OR b < –1
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OpenStudy (tapha):
hmm
OpenStudy (tapha):
ok
OpenStudy (anonymous):
can you help me
OpenStudy (tapha):
sure
pooja195 (pooja195):
solve them serperatly
6 < 3b OR 3b + 12 < 9
for 6<3b just divide both sides by 3
3b+12<9
subtract 12 from both sides and divide by 3
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OpenStudy (tapha):
c i think
OpenStudy (tapha):
try it
OpenStudy (anonymous):
ok ill try it
OpenStudy (tapha):
pooja not intrested
pooja195 (pooja195):
@tapha I messaged you not to give direct answers if you wanna get in trouble keep giving out answers
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OpenStudy (tapha):
ok
pooja195 (pooja195):
@WildQueen did you try?
OpenStudy (tapha):
i really gtg
OpenStudy (anonymous):
yes i still dont get it
pooja195 (pooja195):
What dont you get?
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OpenStudy (anonymous):
how i divide 3 and subtract 12
pooja195 (pooja195):
Lets do this one first :)
6 < 3b
in this you divide by 3
what are you left with?
OpenStudy (anonymous):
you do the other side
pooja195 (pooja195):
Right but one you divide by 3 what are you left with?
pooja195 (pooja195):
\[\huge~\rm~6 /3< 3b/3\]
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OpenStudy (anonymous):
after u divide it u subtract 12
OpenStudy (ddos_dragon):
When you do the inverse operation of something, that is the same integer, it... vanishes.
OpenStudy (anonymous):
oh ok
pooja195 (pooja195):
@WildQueen we do not subtract by 12 in this one
pooja195 (pooja195):
That is for the after the OR part
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pooja195 (pooja195):
What are you left with once both sides are divide by 3?
OpenStudy (anonymous):
then u subtract 12 and divide 3 again
OpenStudy (ddos_dragon):
No..no..no..no
pooja195 (pooja195):
No ignore that part. Just do the part i told you.
OpenStudy (ddos_dragon):
The division first.
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pooja195 (pooja195):
Focus only on this
\[\huge~\rm~6 /3< 3b/3\]
OpenStudy (ddos_dragon):
Remember that inverse trick.
pooja195 (pooja195):
What will the inequality look like if we divide both sides by 3?
OpenStudy (anonymous):
ok you will get 3 or -3
OpenStudy (ddos_dragon):
Erm... Not exactly.
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pooja195 (pooja195):
You are dividing a positive by a positive there are no negatives.
OpenStudy (ddos_dragon):
We know one side has nothing but the B, because of the inverse trick. What about the other?
OpenStudy (anonymous):
oh ok ok so u get 3
OpenStudy (ddos_dragon):
6/3 = ?
OpenStudy (anonymous):
3
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OpenStudy (ddos_dragon):
No.
OpenStudy (ddos_dragon):
How many times does 3 fit into 6?
pooja195 (pooja195):
Im gonna let @Ddos_Dragon finish helping u.
OpenStudy (anonymous):
ok
OpenStudy (ddos_dragon):
So, how many times does 3 go into 6?
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OpenStudy (anonymous):
2
OpenStudy (ddos_dragon):
YES!
OpenStudy (ddos_dragon):
2 < B
OpenStudy (ddos_dragon):
There is your first inequality.
OpenStudy (anonymous):
ok i got you
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OpenStudy (ddos_dragon):
Now for, 3b + 12 < 9
OpenStudy (anonymous):
ok ready
OpenStudy (ddos_dragon):
First, divide 3 on both sides.
OpenStudy (anonymous):
how will that look like
OpenStudy (ddos_dragon):
9/3
3/3b (Inverse trick removes both threes.)
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OpenStudy (ddos_dragon):
So all you need to do is figure out how many times 3 goes into 9.
OpenStudy (anonymous):
3 times
OpenStudy (ddos_dragon):
Perfect!
OpenStudy (ddos_dragon):
Leaving us with, b + 12 < 3
OpenStudy (anonymous):
so now u subtract
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OpenStudy (ddos_dragon):
b + 12 -12 (Inverse trick!!!!!)
3- 12
OpenStudy (anonymous):
is it 9
OpenStudy (ddos_dragon):
Ermm. No.
OpenStudy (anonymous):
oh
OpenStudy (anonymous):
then how do u solve it
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OpenStudy (ddos_dragon):
3 - 12 = ?
If the bigger number is on the right it is a negative.
OpenStudy (ddos_dragon):
BRB
OpenStudy (anonymous):
ok
OpenStudy (ddos_dragon):
Back. And its closed XD