HELP PLEASE A tuning fork with a period of 1.95 x 10-3 is struck at the same time as a guitar
string. If 24 beats are heard in 6.0 s, find the possible frequency or frequencies of the guitar string.

The possible approximate frequencies of the guitar string are 516.82 Hz and 508.82 Hz

Step-by-step explanation:

The given period of the tuning fork, T₁ = 1.95 × 10⁻³ s

The number of beats in 6.0 s = 24 beats

Therefore, the beat frequency, = 4 beats per second = 4 Hz

Let f₁ represent the frequency of the tuning fork, and let f₂ represent the frequency of the guitar string, we have;

f₁ = 1/T₁ = 1/(1.95 × 10⁻³) ≈ 512.82 Hz

We have;

Therefore, we get;

f₂ - 512.82 = ± 4

∴ f₂ = 512.82 ± 4

The possible frequencies of the guitar string,  f₂ ≈ 516.82 Hz or f₂ ≈ 508.82 Hz

option 2 is correct.

step-by-step explanation:

we are given two inequality equation

x+y> 5 and -3x+2y< -2

first we draw the graph of both function using t-table method

equation: x+y> 5

for x=0, y=5

for x=-5, y=0

table:

x   :   0     5

y   :   5     0

test point (0,0) , 0> 5

false

equation: -3x+2y< -2

for x=0, y=-1

for x=2, y=4

table:

x   :   0     2

y   : -1       4

test point (0,0) , 0< -2

true

now we draw the graph. see the attachment for graph.

common region would be solution set.

option 2 is correct.