Answer :
We start with the equation
[tex]$$9e^2f = 49.$$[/tex]
Our goal is to solve for [tex]$e$[/tex].
Step 1. Divide both sides of the equation by [tex]$9f$[/tex] (assuming [tex]$f>0$[/tex] so that division is valid):
[tex]$$e^2 = \frac{49}{9f}.$$[/tex]
Step 2. Take the square root of both sides. Remembering that taking the square root introduces a [tex]$\pm$[/tex] sign, we have
[tex]$$e = \pm \sqrt{\frac{49}{9f}}.$$[/tex]
Step 3. Simplify the square root in the numerator and the denominator separately:
[tex]$$\sqrt{49} = 7 \quad \text{and} \quad \sqrt{9f} = 3\sqrt{f}.$$[/tex]
Thus,
[tex]$$e = \pm \frac{7}{3\sqrt{f}}.$$[/tex]
This expression shows that the solutions for [tex]$e$[/tex] are
[tex]$$e = \pm \frac{7}{3\sqrt{f}}.$$[/tex]
Notice that one of the answer choices was written as
[tex]$$e = \pm \frac{7\sqrt{f}}{3f}.$$[/tex]
Since
[tex]$$\frac{\sqrt{f}}{f} = \frac{1}{\sqrt{f}},$$[/tex]
this option is equivalent to our obtained result.
Therefore, the final answer is
[tex]$$e = \pm \frac{7}{3\sqrt{f}}.$$[/tex]
[tex]$$9e^2f = 49.$$[/tex]
Our goal is to solve for [tex]$e$[/tex].
Step 1. Divide both sides of the equation by [tex]$9f$[/tex] (assuming [tex]$f>0$[/tex] so that division is valid):
[tex]$$e^2 = \frac{49}{9f}.$$[/tex]
Step 2. Take the square root of both sides. Remembering that taking the square root introduces a [tex]$\pm$[/tex] sign, we have
[tex]$$e = \pm \sqrt{\frac{49}{9f}}.$$[/tex]
Step 3. Simplify the square root in the numerator and the denominator separately:
[tex]$$\sqrt{49} = 7 \quad \text{and} \quad \sqrt{9f} = 3\sqrt{f}.$$[/tex]
Thus,
[tex]$$e = \pm \frac{7}{3\sqrt{f}}.$$[/tex]
This expression shows that the solutions for [tex]$e$[/tex] are
[tex]$$e = \pm \frac{7}{3\sqrt{f}}.$$[/tex]
Notice that one of the answer choices was written as
[tex]$$e = \pm \frac{7\sqrt{f}}{3f}.$$[/tex]
Since
[tex]$$\frac{\sqrt{f}}{f} = \frac{1}{\sqrt{f}},$$[/tex]
this option is equivalent to our obtained result.
Therefore, the final answer is
[tex]$$e = \pm \frac{7}{3\sqrt{f}}.$$[/tex]
