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

the inverse variation equation shows the relationship between wavelength in meters, x, and frequency, y. y = 3×10^8 / x what is the wavelength for radio waves with frequency 3×10^9? 1×10^(-1) m 3×10^(-1) m 3×10^17 m 9×10^17 m

the inverse variation equation shows the relationship between wavelength in meters, x, and frequency, y. y = 3×10^8 / x what is the wavelength for radio waves with frequency 3×10^9? 1×10^(-1) m 3×10^(-1) m 3×10^17 m 9×10^17 m

Answer

Explanation:

Step1: Identify the given values

We are given the inverse - variation equation $y=\frac{3\times10^{8}}{x}$, where $y$ is the frequency and $x$ is the wavelength. The frequency $y = 3\times10^{9}$.

Step2: Rearrange the equation to solve for $x$

Starting with $y=\frac{3\times10^{8}}{x}$, we can cross - multiply to get $xy = 3\times10^{8}$, then $x=\frac{3\times10^{8}}{y}$.

Step3: Substitute the value of $y$ into the equation for $x$

Substitute $y = 3\times10^{9}$ into $x=\frac{3\times10^{8}}{y}$. So $x=\frac{3\times10^{8}}{3\times10^{9}}$. Using the rule of exponents $\frac{a^{m}}{a^{n}}=a^{m - n}$, we have $x = 3^{1-1}\times10^{8 - 9}=1\times10^{-1}$.

Answer:

$1\times10^{-1}\text{ m}$