Answer:
B.
![s = \sqrt{ (R - 3)/(t) }](https://img.qammunity.org/2023/formulas/advanced-placement-ap/high-school/lde03wnv3pw2advv41c5bp037k6tzej43x.png)
Explanation:
Solve for s in the equation:
![R = t {s}^(2) + 3](https://img.qammunity.org/2023/formulas/advanced-placement-ap/high-school/7n56d28h8erhdbjn4868hkx5n24p798mvk.png)
Make s the subject of formula by collecting like terms.
![t {s}^(2) = R - 3](https://img.qammunity.org/2023/formulas/advanced-placement-ap/high-school/zxr4jqs9gdtgrdylfoqev0dat73dcv0ux0.png)
Divide both sides by the coefficient of s²
![\frac{t {s}^(2) }{t} = (R - 3)/(t)](https://img.qammunity.org/2023/formulas/advanced-placement-ap/high-school/caykt74ekp2fa068ydhnv4g3r985v1au5j.png)
![{s}^(2) = (R - 3)/(t)](https://img.qammunity.org/2023/formulas/advanced-placement-ap/high-school/6nq9z7wh2xayjmow0cxevtuhws1ik6lav2.png)
Square root both sides because s is squared.
![\sqrt{ {s}^(2) } = \sqrt{ (R - 3)/(t) }](https://img.qammunity.org/2023/formulas/advanced-placement-ap/high-school/8r5g6dt1v0xhx1ixlfo3v2w8snf0sjj89c.png)
Therefore:
is the final answer.
Option A is almost correct but the square root is not for only
but for
![(R - 3)/(t)](https://img.qammunity.org/2023/formulas/advanced-placement-ap/high-school/2609efs9u176d9ykjl2k7hse2jw4btj57q.png)
I hope this helps