+85 Arrange the following numbers in increasing order:
Latex version:
\(\begin{align*} A &= \frac{2^{1/2}}{4^{1/6}}\\ B &= \sqrt[12]{128}\vphantom{dfrac{2}{2}}\\ C &= \left( \frac{1}{8^{1/5}} \right)^2\\ D &= \sqrt{\frac{4^{-1}}{2^{-1} \cdot 8^{-1}}}\\ E &= \sqrt[3]{2^{1/2} \cdot 4^{-1/4}}\vphantom{dfrac{2}{2}} \end{align*}\)
In plain text:
A = (2^(1/2))/(4^(1/6))
B = 128^(1/12)
C = (1/8^(1/5))^2
D = (4^-1/((2^-1)(8^-1)))^(1/2)
E = (2^(1/2)(4^(-1/4)))^(1/3)
I've gotten the answer C,E,A,D,B because C = 0.43527528164, E = 1, A = 1.5, D = 4, and B = 10 2/3, but it's incorrect. How do I solve this?
.
Thanks for helping!
Here is the increasing order:
C = 0.43527528164, E = 1, A=1.122462048, B=1.498307077, D = 2
+118687
\begin{align*} A &= \frac{2^{1/2}}{4^{1/6}}\\ B &= \sqrt[12]{128}\vphantom{dfrac{2}{2}}\\ C &= \left( \frac{1}{8^{1/5}} \right)^2\\ D &= \sqrt{\frac{4^{-1}}{2^{-1} \cdot 8^{-1}}}\\ E &= \sqrt[3]{2^{1/2} \cdot 4^{-1/4}}\vphantom{dfrac{2}{2}} \end{align*}
\(\begin{align*} A &= \frac{2^{1/2}}{4^{1/6}}=2^{0.5-0.\dot3}=2^{1/6}=2^{2/12}\\ B &= \sqrt[12]{128}\vphantom{dfrac{2}{2}}=\sqrt[12]{2^7}\vphantom{dfrac{2}{2}}=2^{7/12} \end{align*} etc\)
Change them all to powers of 2.
And stop accusing other people of cheating (on other threads)
It is irritating.
