I think it’s D or a I think so
28/80 and then you simplify it by dividing the numerator and denominator by 4 and you get 7/20.
B. 7/20 is the correct answer
![\displaystyle\lim_{n\to\infty}\sqrt[n]{\left|\left(\frac{5n+15}{2n-1}\right)^n\right|}=\lim_{n\to\infty}\frac{5n+15}{2n-1}=\dfrac52](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Clim_%7Bn%5Cto%5Cinfty%7D%5Csqrt%5Bn%5D%7B%5Cleft%7C%5Cleft%28%5Cfrac%7B5n%2B15%7D%7B2n-1%7D%5Cright%29%5En%5Cright%7C%7D%3D%5Clim_%7Bn%5Cto%5Cinfty%7D%5Cfrac%7B5n%2B15%7D%7B2n-1%7D%3D%5Cdfrac52)
Since this limit exceeds 1, the series diverges.
<u>Solution </u><u>1</u><u> </u><u>:</u><u>-</u>
Given expression ,
We know that, ![a^{\frac{m}{n}} = \sqrt[n]{a^m}](https://tex.z-dn.net/?f=%20a%5E%7B%5Cfrac%7Bm%7D%7Bn%7D%7D%20%3D%20%5Csqrt%5Bn%5D%7Ba%5Em%7D%20)
Therefore , this can be written as ,
![\implies \sqrt[4]{13^3}](https://tex.z-dn.net/?f=%5Cimplies%20%5Csqrt%5B4%5D%7B13%5E3%7D%20)
<u>Solution</u><u> </u><u>2</u><u>:</u><u>-</u>
Given expression ,
![\implies \sqrt[3]{24^{10}}](https://tex.z-dn.net/?f=%5Cimplies%20%5Csqrt%5B3%5D%7B24%5E%7B10%7D%7D%20)
We know that,
Therefore, this can be written as ,
Answer: pi/8
Step-by-step explanation:
adding or multiplying an irrational by a rational will result in another irrational. So we could take something like pi and divide it until it’s between 2/7 and 3/7. I did that and apparently pi/8 works, so boom
