anonymous
  • anonymous
11* 3 sqrt 6+8* 4 sqrt 6 +2* 3sqrt 6-3* 4sqrt 6
Mathematics
chestercat
  • chestercat
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hba
  • hba
\[\huge \color{red}{\text{Welcome to Open Study}}\ddot\smile\]
anonymous
  • anonymous
hi
hba
  • hba
\[11*3\sqrt{6}+8*4\sqrt{6}+2*3\sqrt{6}-3*4\sqrt{6},Right ?\]

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anonymous
  • anonymous
A) 18 sqrt 12 B) 18 sqrt 6 C) 5 4 sqrt 12+13 3 sqrt 12 D
anonymous
  • anonymous
D) 5 4 sqrt 6+133 sqrt 6
hba
  • hba
Can you simplify it ?
anonymous
  • anonymous
im not sure what im doing with this problem it is asking me to simplify
hba
  • hba
We just have to simply add them up but before we do this we have to multiply the terms Example 2*3sqrt{6}=6sqrt{6}
anonymous
  • anonymous
when i simplified it i got 59 sqrt 6 but thats not one of the options
hba
  • hba
Show me you working :)
anonymous
  • anonymous
let me try it your way so 11 3 sqrt 6 is ....33 sqrt 6 i think
hba
  • hba
Yeah good :)
hba
  • hba
Now do the others and then add,subtract them :)
anonymous
  • anonymous
ok
anonymous
  • anonymous
33 sqrt 6+32 sqrt 6+6 sqrt 6-12 sqrt 6
anonymous
  • anonymous
it still comes out to 59 sqrt 6
hba
  • hba
Is your question correct my friend ?
hba
  • hba
lol :P
anonymous
  • anonymous
ya i think so its just im taking this class on flvs and they always have the wierdest answer ... like you solve the problem and then its wrong and apparently their answer is right
hba
  • hba
Haha lol :P
anonymous
  • anonymous
hold on the question itself is a picture ill attach it
anonymous
  • anonymous
1 Attachment
hba
  • hba
lol,the question is diffrent lemme show you this.
hba
  • hba
You will add these \[\huge\ a \sqrt[3]{6}+b \sqrt[3]{6}=(a+b)\sqrt[3]{6}\] You add this and also add the other.
anonymous
  • anonymous
i dont understand that
hba
  • hba
@starrlight \[\huge\ 11\sqrt[3]{6}+8\sqrt[4]{6}+2\sqrt[3]{6}-3\sqrt[4]{6}\] Add them straight up :)
hba
  • hba
@starrlight Um,getting it ?
anonymous
  • anonymous
no
hba
  • hba
\[\huge\ 11\sqrt[3]{6}+2\sqrt[3]{6}+8\sqrt[4]{6}-3\sqrt[4]{6}\]
anonymous
  • anonymous
oh ok

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