Question

1. Determine the distance from the bridge for three consecutive frets. Round each answer to 3 decimal place. For example, you can calculate the distance from the bridge for the 4th, 5th and 6th fret.
2. Create the ratios of the highest fret to the previous fret and simplify to a decimal. Create the ratio of the next highest fret to the previous fret and simplify to a decimal. What are the values? For example, if your calculated the distance to the 4th, 5th and 6th fret, you would create the two ratios: 6th fret distance/5th fret distance, and 5th fret distance/4th fret distance. How are these ratios two related? Does this remind you of something we studied earlier in the course?

Answer 1

The distance of each subsequent  fret from the bridge is constant multiple

of the distance of the previous fret from the bridge.

• Distance from the bridge for the 4th fret is approximately 55.185 cm

• Distance from the bridge for the 5th fret is approximately 52.087 cm

• Distance from the bridge for the 6th fret  is approximately 49.164 cm

Reason:

1. The equation for the distance of the x-th fret from the bridge in

centimeters is given as follows;

• $d = 21.9 \times \left(2\right)^{\frac{20 - x}{12} }$

Therefore, for three consecutive frets, we have;

$At \ the \ 4th \ fret; \ d = 21.9 \times \left(2\right)^{\frac{20 - 4}{12} } \approx 55.185$

The distance from the bridge for the 4th fret = 55.185 cm

$At \ the \ 5th \ fret; \ d = 21.9 \times \left(2\right)^{\frac{20 - 5}{12} } \approx 52.087$

The distance from the bridge for the 5th fret = 52.087 cm$At \ the \ 6th \ fret; \ d = 21.9 \times \left(2\right)^{\frac{20 - 6}{12} } \approx 49.164$

The distance from the bridge for the 6th fret = 49.164 cm

2. The ratio are;

• $\dfrac{Distance \ to \ the \ 6th \ fret }{Distance \ to \ the \ 5th \ fret} = \dfrac{49.164}{52.087} = \dfrac{21.9 \times \left(2\right)^{\frac{14}{12} }}{21.9 \times \left(2\right)^{\frac{15}{12} }} =e^{\dfrac{1}{\dfrac{Ln(2)}{12} } }$

• $\dfrac{Distance \ to \ the \ 5th \ fret }{Distance \ to \ the \ 4th \ fret} = \dfrac{52.087}{55.185} = \dfrac{21.9 \times \left(2\right)^{\frac{15}{12} }} {21.9 \times \left(2\right)^{\frac{16}{12} }} =e^{\dfrac{1}{\dfrac{Ln(2)}{12} } }$

Therefore, the ratio of the distances of each fret to the distance of the

previous fret is a constant, which is similar to a geometric progression.

Therefore;

The distances of the frets from the bridge form a geometric

progression or sequence.

Learn more here:

brainly.com/question/25244113