Will give brainlist or whateve.

1. a note has a wavelength of 0.58399 m. if the speed of sound is 343.00 m/s, what pitch is this note?

2. a note has a wavelength of 0.69450 m. if the speed of sound is 343.00 m/s, what pitch is this note?

3. a note has a wavelength of 0.49108 m. if the speed of sound is 343.00 m/s, what pitch is this note?

4. an a note has a frequency of 440.00 hz. calculate its wavelength in meters.

5. an e note has a frequency of 659.26 hz. calculate its wavelength in meters.

6. a c note has a frequency of 523.25 hz. calculate its wavelength in meters.

7. a note has a wavelength of 0.43750 m. if the speed of sound is 343.00 m/s, what pitch is this note?

8. a note has a wavelength of 0.77954 m. if the speed of sound is 343.00 m/s, what pitch is this note?

9. an f note has a frequency of 698.46 hz. calculate its wavelength in meters.

10. a b note has a frequency of 493.88 hz. calculate its wavelength in meters.

1:A

2:E

3:C

4:0.5840

5:0.6945

6:0.4911

7:D

8:G

9:0.4375

10:0.5203

#1) 0.77955f = 343

Divide both sides by 0.77955:

0.77955f/0.77955 = 343/0.77955

f = 439.997 ≈ 440. This is for A.

#2) 0.52028f = 343

Divide both sides by 0.52028, and you get f = 659.260.  This is for E.

#3) 0.65552f = 343

Divide both sides by 0.65552, and you get f = 523.25.   This is for C.

#4) 587.33λ = 343

Divide both sides by 587.33 and you get λ = 0.583999 ≈ 0.5840.

#5) 493.88λ = 343

Divide both sides by 493.88, and we get λ = 0.6945.

#6) 698.46λ = 343

Divide both sides by 698.46 and you get λ = 0.49108 ≈ 0.4911.

#7) 0.5840f = 343

Divide both sides by 0.5840 and you get f = 587.3288 ≈ 587.33.  This is for D.

#8) 0.4375f = 343

Divide both sides by 0.4375 and you get f = 784. which is G.

#9) 783.99λ = 343

Divide both sides by 783.99 and you get λ = 0.4375.

#10) 659.26λ = 343

Divide both sides by 659.26 and you get λ = 0.52028 ≈ 0.5203.

hope this helped <3

i don't see a text, but i am gonna assume this *orbit* the curved path, usually elliptical, described by a planet, satellite, spaceship, etc., around a celestial body, as the sun.

c is having 4 valence electrons and is bonded to three hydrogens and one sulphur atom, therefore, the number of bond pair is equal to four. as all the valence electrons are used in forming the bonds so, there is no lone pair left on carbon.

s is having 6 valence electrons and is bonded to one other sulfur atom and one carbon atom, therefore, the number of bond pair is equal to 2. as only two valence electrons are used in forming the bonds so, four electrons are left. therefore, two lone pairs are present on s.