<h2>Answer</h2>
2
<h2>Explanation</h2>
First, we are going to use the law of fractional exponents: ![a^{\frac{1}{n} =\sqrt[n]{a}](https://tex.z-dn.net/?f=a%5E%7B%5Cfrac%7B1%7D%7Bn%7D%20%3D%5Csqrt%5Bn%5D%7Ba%7D)
We can infer form our expression that 
and 
, so let's replace the values:
![16^{\frac{1}{4} }=\sqrt[4]{16}](https://tex.z-dn.net/?f=16%5E%7B%5Cfrac%7B1%7D%7B4%7D%20%7D%3D%5Csqrt%5B4%5D%7B16%7D)
Notice that we can also decompose 16 into prime factors to get 
, so we can rewrite our expression as follows:
![\sqrt[4]{16}=\sqrt[4]{2^4}](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B16%7D%3D%5Csqrt%5B4%5D%7B2%5E4%7D)
Finally, we can use the rule of radicals: ![\sqrt[n]{a^n} =a](https://tex.z-dn.net/?f=%5Csqrt%5Bn%5D%7Ba%5En%7D%20%3Da)
, so:
![\sqrt[4]{2^4}=2](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B2%5E4%7D%3D2)
Answer:1/3
Step-by-step explanation:
can you give branliest??
Period =2r /b
assume frequency factor meaning angular frequency "b"
then
2r= 2r/b => b=1
A. sine function whose period 2pi radians
Answer:
k = 3 1/5
Step-by-step explanation:
We want to isolate k, or move all numbers to the other side.
1/2 (4 - k) = 2/5
(4 - k) = 2/5 * 2
-k = 4/5 - 4
k = 4 - 4/5
k = 3 1/5
