Giving 40 points to whoever answer this correctly RSTU is a parallelogram. If m∠TSV = 31° and m∠SVT = 126°, explain how you can find the measure of ∠URV. Show all steps of your work, and refer to any properties of triangles, parallelograms, or triangle congruency theorems as necessary to justify your response.

we are given

i=3.2\times 10^{-6}

i_0=1\times 10^{-12}

now, we can plug values

\beta =10log(\frac{3.2\times 10^{-6}}{1\times 10^{-12}} )

\beta =10\log _{10}\left(3200000\right)

\beta =65.05150db[/</p><p>(b)</p><p>we are given decibel level goes up by 80db</p><p>so, </p><p>[tex]\beta _2-\beta_1=80

we are given

i_1=3.2\times 10^{-6}

now, we can use formula

\beta =10log(\frac{i}{i_0} )

now, we can find beta 1 and beta 2

\beta_1 =10log(\frac{i_1}{i_0} )

\beta_2 =10log(\frac{i_2}{i_0} )

\beta _2-\beta_1=10log(\frac{i_2}{i_0} )-10log(\frac{i_1}{i_0} )

\beta _2-\beta_1=10log(\frac{i_2\times i_0}{i_0\times i_1} )

\beta _2-\beta_1=10log(\frac{i_2}{i_1} )

now, we can plug values

80=10log(\frac{i_2}{i_1} )

8=log(\frac{i_2}{i_1} )

\frac{i_2}{i_1} =10^8

i_2 =10^8\times i_1

now, we can plug i1

i_2 =10^8\times (3.2\times 10^{-6})

[tex]i_2 =320watts/m^2[/

we are given

i=3.2\times 10^{-6}

i_0=1\times 10^{-12}

now, we can plug values

\beta =10log(\frac{3.2\times 10^{-6}}{1\times 10^{-12}} )

\beta =10\log _{10}\left(3200000\right)

\beta =65.05150db[/</p><p>(b)</p><p>we are given decibel level goes up by 80db</p><p>so, </p><p>[tex]\beta _2-\beta_1=80

we are given

i_1=3.2\times 10^{-6}

now, we can use formula

\beta =10log(\frac{i}{i_0} )

now, we can find beta 1 and beta 2

\beta_1 =10log(\frac{i_1}{i_0} )

\beta_2 =10log(\frac{i_2}{i_0} )

\beta _2-\beta_1=10log(\frac{i_2}{i_0} )-10log(\frac{i_1}{i_0} )

\beta _2-\beta_1=10log(\frac{i_2\times i_0}{i_0\times i_1} )

\beta _2-\beta_1=10log(\frac{i_2}{i_1} )

now, we can plug values

80=10log(\frac{i_2}{i_1} )

8=log(\frac{i_2}{i_1} )

\frac{i_2}{i_1} =10^8

i_2 =10^8\times i_1

now, we can plug i1

i_2 =10^8\times (3.2\times 10^{-6})

[tex]i_2 =320watts/m^2[/

john earns $14.00 for two hours of work.