<span>The best answer is C.add 2 to both sides of the equation.
</span>5=x-2
5+2=x-2+2
7=x-0
x=7
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
#SPJ1
Step-by-step explanation:
x + y = 9. => 2x + 2y = 18.
2x + 2y = 18
- (2x - 3y = -12)
=> 5y = 30, y = 6.
Therefore x + (6) = 9, x = 3.
The solution is x = 3 and y = 6.
Answer:
12a+60 i think lol
Step-by-step explanation:
this is how i did it though:
12(a+b)
12a+60
-5 1/3, -2 1/2, 3.75, 4.2 ( this is the order from least to greatest)
