QuestionFind the frequency of note F# which is 3 half steps below A3 (220 Hz) using $F(x)=F_{0}(1.059463)^{x}$ . Round to the nearest whole number.

Studdy Solution

STEP 1

Assumptions1. The reference frequency $_{0}$ is the frequency of the note $A_{3}$ , which is $220 \mathrm{~Hz}$ . . The note F\# is3 half steps below $A_{3}$ , so $x=-3$ .
3. The frequency of a note is modeled by the function $(x)=_{0}(1.059463)^{x}$ .

STEP 2

We need to find the frequency of the note F\#, which is half steps below $A_{}$ . We can do this by substituting the values of $_{0}$ and $x$ into the function.
$(x)=_{0}(1.059463)^{x}$

STEP 3

Substitute the values of $_{0}$ and $x$ into the function.
$(-3)=220(1.059463)^{-3}$

STEP 4

Calculate the value of $(1.059463)^{-3}$ .
$1.059463^{-3} \approx0.841395$

STEP 5

Substitute the value of $(1.059463)^{-3}$ back into the function.
$(-3)=220 \times0.841395$

STEP 6

Calculate the frequency of the note F\#.
$(-3)=220 \times0.841395 \approx185$ The frequency of the note F\# is approximately185 Hz, rounded to the nearest whole number.

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QuestionFind the frequency of note F# which is 3 half steps below A3 (220 Hz) using $F(x)=F_{0}(1.059463)^{x}$ . Round to the nearest whole number.

Studdy Solution

STEP 1

Assumptions1. The reference frequency $_{0}$ is the frequency of the note $A_{3}$ , which is $220 \mathrm{~Hz}$ . . The note F\# is3 half steps below $A_{3}$ , so $x=-3$ .
3. The frequency of a note is modeled by the function $(x)=_{0}(1.059463)^{x}$ .

STEP 2

We need to find the frequency of the note F\#, which is half steps below $A_{}$ . We can do this by substituting the values of $_{0}$ and $x$ into the function.
$(x)=_{0}(1.059463)^{x}$

STEP 3

Substitute the values of $_{0}$ and $x$ into the function.
$(-3)=220(1.059463)^{-3}$

STEP 4

Calculate the value of $(1.059463)^{-3}$ .
$1.059463^{-3} \approx0.841395$

STEP 5

Substitute the value of $(1.059463)^{-3}$ back into the function.
$(-3)=220 \times0.841395$

STEP 6

Calculate the frequency of the note F\#.
$(-3)=220 \times0.841395 \approx185$ The frequency of the note F\# is approximately185 Hz, rounded to the nearest whole number.

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QuestionFind the frequency of note F# which is 3 half steps below A3 (220 Hz) using F(x)=F0(1.059463)xF(x)=F_{0}(1.059463)^{x}F(x)=F0​(1.059463)x. Round to the nearest whole number.

Studdy Solution

STEP 1

STEP 2

We need to find the frequency of the note F\#, which is half steps below AA_{}A​. We can do this by substituting the values of 0_{0}0​ and xxx into the function.(x)=0(1.059463)x(x)=_{0}(1.059463)^{x}(x)=0​(1.059463)x

STEP 3

Substitute the values of 0_{0}0​ and xxx into the function.(−3)=220(1.059463)−3(-3)=220(1.059463)^{-3}(−3)=220(1.059463)−3

STEP 4

Calculate the value of (1.059463)−3(1.059463)^{-3}(1.059463)−3.1.059463−3≈0.8413951.059463^{-3} \approx0.8413951.059463−3≈0.841395

STEP 5

Substitute the value of (1.059463)−3(1.059463)^{-3}(1.059463)−3 back into the function.(−3)=220×0.841395(-3)=220 \times0.841395(−3)=220×0.841395

STEP 6

Calculate the frequency of the note F\#.(−3)=220×0.841395≈185(-3)=220 \times0.841395 \approx185(−3)=220×0.841395≈185The frequency of the note F\# is approximately185 Hz, rounded to the nearest whole number.

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QuestionFind the frequency of note F# which is 3 half steps below A3 (220 Hz) using $F(x)=F_{0}(1.059463)^{x}$ . Round to the nearest whole number.

We need to find the frequency of the note F\#, which is half steps below $A_{}$ . We can do this by substituting the values of $_{0}$ and $x$ into the function.
$(x)=_{0}(1.059463)^{x}$

Substitute the values of $_{0}$ and $x$ into the function.
$(-3)=220(1.059463)^{-3}$

Calculate the value of $(1.059463)^{-3}$ .
$1.059463^{-3} \approx0.841395$

Substitute the value of $(1.059463)^{-3}$ back into the function.
$(-3)=220 \times0.841395$

Calculate the frequency of the note F\#.
$(-3)=220 \times0.841395 \approx185$ The frequency of the note F\# is approximately185 Hz, rounded to the nearest whole number.