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Zink96
Group Title
Find the value of 2 times the square root of 3, all over the sqaure root of 10 in simplest form.
 one year ago
 one year ago
Zink96 Group Title
Find the value of 2 times the square root of 3, all over the sqaure root of 10 in simplest form.
 one year ago
 one year ago

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philo1234 Group TitleBest ResponseYou've already chosen the best response.1
\[\frac{ 2\sqrt{3} }{ \sqrt{10} }\] is this the question?
 one year ago

philo1234 Group TitleBest ResponseYou've already chosen the best response.1
Rationalize the denominator, by multiplying the numerator and denominator by the radical in the denominator: \[\frac{ 2\sqrt{3} *\sqrt{10}}{ \sqrt{10}*\sqrt{10} }\]
 one year ago

philo1234 Group TitleBest ResponseYou've already chosen the best response.1
This simplifies to: \[\frac{ 2\sqrt{30} }{ \sqrt{100} }\]
 one year ago

philo1234 Group TitleBest ResponseYou've already chosen the best response.1
What is the square root of 100?
 one year ago

Zink96 Group TitleBest ResponseYou've already chosen the best response.0
50? Im not quite sure...
 one year ago

philo1234 Group TitleBest ResponseYou've already chosen the best response.1
What number can you square to give you 100? or what number can u multiply by itself to give you 100
 one year ago

philo1234 Group TitleBest ResponseYou've already chosen the best response.1
good so now we have:\[\frac{ 2\sqrt{30} }{ 10 }\] Can you simplify this and tell me what you got?
 one year ago

Zink96 Group TitleBest ResponseYou've already chosen the best response.0
2 sqrt 3 over 5?
 one year ago

philo1234 Group TitleBest ResponseYou've already chosen the best response.1
Nope you can only divide a radical by a radical. Therefore it is incorrect to divide 30 by 10. But you can simplify the numbers that are not radicals, by dividing 2 by 2 in the numerator and 10 by 2 in the denominator: You should get as your answer  \[\frac{ \sqrt{30} }{ 5 }\]
 one year ago

philo1234 Group TitleBest ResponseYou've already chosen the best response.1
Do you understand how I got this?
 one year ago

Zink96 Group TitleBest ResponseYou've already chosen the best response.0
Oh Okay... yeah i got it now, thanks!
 one year ago
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