A symphony orchestra is arranged in a semicircle around the conductor. The rows are about 3 feet apart, with the first row about 6 feet from the conductor. A mid-sized orchestra has about 6 rows of chairs. What is the approximate delay between the instant when a play at the 6th row start a note and when the conductor hears the note?

19 ms

Step-by-step explanation:

We need to find the time it takes for the sound from the 6th row gets to the conductor.

Since there are 6 rows and each row is 3 feet apart, since the orchestra is arranged in a semi-circular manner around the conductor, the distance from the 6 th row to the 1st row is (6 - 1) × 3 = 5 × 3 = 15 feet.

Now, since the conductor is 6 feet from the first row, his distance from the 6th row is 15 feet + 6 feet = 21 feet.

So, time approximate delay between the instant when a play at the 6th row start a note and when the conductor hears the note, t = distance between 6th row and conductor/speed of sound.

Speed of sound = 343 m/s

distance between 6th row and conductor = 21 feet. In meters, it is 21 × 0.3048 m = 6.4008 m

t = 6.4008 m ÷ 343 m/s

= 0.019 s

= 19 ms

answer: the answer is 40.82

step-by-step explanation: 3.14 x 13(or half of 26) to the power of 2

(f - g)(x) = 5x^3 - 3

step-by-step explanation:

rearrange these functions vertically, with like terms in separate columns:

f(x) = 5x^3 - 2

g(x) =         +1

now subtract, column by column:

(f - g)(x) = 5x^3 - 3