The answer is C 28 I hope this helped :)
well, this is just a matter of simple unit conversion, so let's recall that one revolution on a circle is just one-go-around, radians wise that'll be 2π, and we also know that 1 minute has 60 seconds, let's use those values for our product.
![\cfrac{300~~\begin{matrix} r \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~ }{~~\begin{matrix} min \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~ }\cdot \cfrac{2\pi ~rad}{~~\begin{matrix} r \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~ }\cdot \cfrac{~~\begin{matrix} min \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~ }{60secs}\implies \cfrac{(300)(2\pi )rad}{60secs}\implies 10\pi ~\frac{rad}{secs}\approx 31.42~\frac{rad}{secs}](https://tex.z-dn.net/?f=%5Ccfrac%7B300~~%5Cbegin%7Bmatrix%7D%20r%20%5C%5C%5B-0.7em%5D%5Ccline%7B1-1%7D%5C%5C%5B-5pt%5D%5Cend%7Bmatrix%7D~~%20%7D%7B~~%5Cbegin%7Bmatrix%7D%20min%20%5C%5C%5B-0.7em%5D%5Ccline%7B1-1%7D%5C%5C%5B-5pt%5D%5Cend%7Bmatrix%7D~~%20%7D%5Ccdot%20%5Ccfrac%7B2%5Cpi%20~rad%7D%7B~~%5Cbegin%7Bmatrix%7D%20r%20%5C%5C%5B-0.7em%5D%5Ccline%7B1-1%7D%5C%5C%5B-5pt%5D%5Cend%7Bmatrix%7D~~%20%7D%5Ccdot%20%5Ccfrac%7B~~%5Cbegin%7Bmatrix%7D%20min%20%5C%5C%5B-0.7em%5D%5Ccline%7B1-1%7D%5C%5C%5B-5pt%5D%5Cend%7Bmatrix%7D~~%20%7D%7B60secs%7D%5Cimplies%20%5Ccfrac%7B%28300%29%282%5Cpi%20%29rad%7D%7B60secs%7D%5Cimplies%2010%5Cpi%20~%5Cfrac%7Brad%7D%7Bsecs%7D%5Capprox%2031.42~%5Cfrac%7Brad%7D%7Bsecs%7D)
Remember that the radicand (the area under the root sign) must be positive or zero for a radical with an even index (like the square root or fourth root, for example). This is because two numbers squared or to the fourth power, etc. cannot be negative, so there are no real solutions when the radicand is negative. We must restrict the domain of the square-root function.
If the domain has already been restricted to

, we can work backwards to add 11 to both sides. We see that

must be under the radicand, so the answer is
A.
1, ___, ____, _____, _____, _____, ____, 29
29-1 = 28
28/7 = 4
1, 5, 9, 13, 17, 21, 25, 29