The sound wave with a <u>frequency of 20</u> waves/sec is 800 longer than the wavelength of a sound wave with a <u>frequency of 16,000</u> waves/sec
<h3>Calculating wavelength </h3>
From the question, we are to determine how many times longer is the first sound wave compared to the second sound water
Using the formula,
v = fλ
∴ λ = v/f
Where v is the velocity
f is the frequency
and λ is the wavelength
For the first wave
f = 20 waves/sec
Then,
λ₁ = v/20
For the second wave
f = 16,000 waves/sec
λ₂ = v/16000
Then,
The factor by which the first sound wave is longer than the second sound wave is
λ₁/ λ₂ = (v/20) ÷( v/16000)
= (v/20) × 16000/v)
= 16000/20
= 800
Hence, the sound wave with a <u>frequency of 20</u> waves/sec is 800 longer than the wavelength of a sound wave with a <u>frequency of 16,000</u> waves/sec
Learn more on Calculating wavelength here: brainly.com/question/16396485
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Answer:
X =30/7
Step-by-step explanation:
The first way would be something like:
-7x + 10 = -20. The first thing we would do is to subtract 10 from both sides.
Therefore,
-7x+10-10=-20-10
Simplified, we get
-7x=-30
Remove the negative symbols
7x =30
x=30/7
Second example:
10=7x - 20
Add 20 to both sides.
10+20=7x-20+20
Simplified,
30=7x
x = 30/7
5+6.50+7.21=18.71 30-18.71=$11.29 left
Answer:
93
Step-by-step explanation:
The average is calculated as
average = 
On 3 quizzes
= 73 ( multiply both sides by 3 )
sum = 219
On 4 quizzes, letting x be the score on fourth test
= 78 ( multiply both sides by 4 )
sum + x = 312 , that is
219 + x = 312 ( subtract 219 from both sides )
x = 93
She must score 93 on the next quiz
