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what is the answer to [(x^2)-20x+(23/3)]'-√576+4!=0?

 Aug 19, 2015

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 #3
avatar+118696 
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[(x^2)-20x+(23/3)]'-√576+4!=0

 

(x220x+233)576+4!=0(x220x+233)24+24=0x220x+233=03x260x+23=0x=60±36002766x=60±33246etc

 Aug 19, 2015
 #1
avatar+118696 
0

 [(x^2)-20x+(23/3)]'-√576+4!=0?

 

What is the apostrophy for?

 Aug 19, 2015
 #2
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0

sorry must have put that there by accident...

 Aug 19, 2015
 #3
avatar+118696 
+5
Best Answer

[(x^2)-20x+(23/3)]'-√576+4!=0

 

(x220x+233)576+4!=0(x220x+233)24+24=0x220x+233=03x260x+23=0x=60±36002766x=60±33246etc

Melody Aug 19, 2015

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