Help with Operations with Radicals. Will Give Medals! Is the simplified form of 2square root of 3 • 2square root of 6 rational? Yes No Is the simplified form of 2square root of 3 + 3square root of 3 rational? Yes No
@Andras Help me please?
\[2\sqrt{3}*2\sqrt{6}=2*3*\sqrt{3*6}=6\sqrt{18}=6\sqrt{2*3^2}=6*3\sqrt{2}=18\sqrt{2}\]
So the first one is irrational
Oh sorry man I already solved them I need help with two important questions so I can finish my module :-((((((
\[2\sqrt{3}+3\sqrt{3}=(2+3)\sqrt{3}=5\sqrt{3}\]
Good
@Andras Can you help me?
Maybe, I need to see the problem first
Explain how the Quotient of Powers was used to simplify this expression. 2 to the fifth power, over 8 = 22 By finding the quotient of the bases to be one fourth and cancelling common factors By finding the quotient of the bases to be one fourth and simplifying the expression By simplifying 8 to 23 to make both powers base two and subtracting the exponents By simplifying 8 to 23 to make both powers base two and adding the exponents
2 to the fifth power, over 8 = 22 that is not true 2 to the fifth power, over 8 = 2^2
\[\frac{ 2^5 }{ 8 }=\frac{ 2^5 }{ 2^3 }=2^(5-3)=2^2=4\]
So what is it?
By simplifying 8 to 23 to make both powers base two and subtracting the exponents
* to 2^3 and I had 2^(5-3)
Clear?
Nope
\[\frac{ 2^5 }{ 8 }=\frac{ 2^5 }{ 2^3 }=2^{\left( 5-3 \right)}=2^2=4\]
So the ssecond one?
No the third one: By simplifying 8 to 2^3 to make both powers base two and subtracting the exponents
This is the last one man
Rewrite the radical as a rational exponent. the cube root of 2 to the seventh power 2 to the 3 over 7 power 2 to the 7 over 3 power 2^21 24
\[\sqrt[3]{2^7}=2^{\frac{ 7 }{ 3 }}\]
I think it's the first or third
:-) It is the second, wrong guess
Oops :)
Thanks though, heres a medal :-)
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