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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I believe $30 but i am not completely sure :)
Answer: Inequality
Step-by-step explanation:
From the question, if we can describe 15×-10 as an expression, then we would describe 6×-2< 35 as an inequality. An
inequality is used to compare two values, and shows if one is less than, or greater than, or maybe not equal to the other value.
For example, a ≠ b means that a is not equal to b and a < b means a is less than b while a > b means a is greater than b. From the question, 6×-2< 35 means that 6x - 2 is less than 35.
We have that
<span>f(x) = 3x + 2
</span><span>g(x) = x - 3
[</span>g(x) - f(x)]=(x-3)-(3x+2)--------> (x-3-3x-2)-----> -2x-5
the domain is all real numbers
the answer is
<span>–2x – 5; all real numbers</span>