the inverse variation equation shows the relationship between wavelength in meters, x, and frequency, y.\ny…

the inverse variation equation shows the relationship between wavelength in meters, x, and frequency, y.\ny = \frac{3\times10^{8}}{x}\nwhat is the wavelength for radio waves with frequency (3\times10^{9})?\n(1\times10^{-1}\text{ m})\n(3\times10^{-1}\text{ m})\n(3\times10^{17}\text{ m})\n(9\times10^{17}\text{ m})

the inverse variation equation shows the relationship between wavelength in meters, x, and frequency, y.\ny = \frac{3\times10^{8}}{x}\nwhat is the wavelength for radio waves with frequency (3\times10^{9})?\n(1\times10^{-1}\text{ m})\n(3\times10^{-1}\text{ m})\n(3\times10^{17}\text{ m})\n(9\times10^{17}\text{ m})

Answer

Explanation:

Step1: Substitute given frequency into equation

Given $y = \frac{3\times10^{8}}{x}$ and $y = 3\times 10^{9}$. Substitute $y$ into the equation: $3\times 10^{9}=\frac{3\times10^{8}}{x}$.

Step2: Solve for $x$

Cross - multiply to get $3\times 10^{9}\times x=3\times10^{8}$. Then $x=\frac{3\times10^{8}}{3\times 10^{9}}$. Using the rule $\frac{a^{m}}{a^{n}}=a^{m - n}$, we have $x = 10^{8-9}=10^{-1}$. And $\frac{3}{3}=1$, so $x = 1\times10^{-1}$ m.

Answer:

$1\times 10^{-1}$ m