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:
(16 x^8 - x^3 + 6)/(2 x^3)
Step-by-step explanation:
Simplify the following:
(64 x^8 - 4 x^3 + 24)/(8 x^3)
Factor 4 out of 64 x^8 - 4 x^3 + 24:
(4 (16 x^8 - x^3 + 6))/(8 x^3)
4/8 = 4/(4×2) = 1/2:
Answer: (16 x^8 - x^3 + 6)/(2 x^3)
Answer:
Change in the cost of each book printed = $25
Cost to get started = $1150
Step-by-step explanation:
Equation representing the relation between number of copies of the books (x) and total cost of publishing a book (y) is,
y = 1150 + 25x
This equation is in the form of a linear equation,
y = mx + b
Where m represents the change in cost of printing each book and y-intercept of the line 'b' represents the cost before any book is printed.
By comparing these equations,
m = 25
Therefore, change in the cost of each book printed = $25
b = 1150
Which represents the cost to get started = $1150
