Solve the following equations:
To solve the first equation, \(100c^2 = 9\), we begin by isolating the variable \(c\). Dividing both sides by 100 gives \(c^2 = \frac{9}{100}\). Taking the square root of both sides, we find that \(c = \pm\sqrt{\frac{9}{100}}\), which simplifies to \(c = \pm\frac{3}{10}\), or \(c = \pm0.3\).
To solve the second equation, \(64 = \frac{v^2}{5} + 19\), we first isolate the term with the variable \(v\). Subtracting 19 from both sides, we get \(64 - 19 = \frac{v^2}{5}\), which simplifies to \(45 = \frac{v^2}{5}\). Multiplying both sides by 5, we find \(v^2 = 45 \times 5 = 225\). Taking the square root of both sides, we obtain \(v = \pm\sqrt{225}\), so \(v = \pm15\).
For the third expression, \(s \cdot s \cdot 5 - 19\), we perform algebraic simplification. Since \(s \cdot s = s^2\), the expression can be written in standard form as \(5s^2 - 19\).
These types of algebraic equations appear frequently in physics and geometry. For instance, calculating the side length \(s\) of a square with a known area involves solving an equation like \(A = s^2\). Similarly, determining the velocity \(v\) of an object from kinetic energy formulas often requires rearranging terms and calculating roots, just as we did to solve for \(c\) and \(v\) above. Knowing how to manipulate and solve these equations is a fundamental skill for modeling real-world behavior.