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/Step-by-step Explanation:
Given:
a = no. of DVDs
b = no. of CDs
Total cost ($) for DVDs and CDs = 
a. The cost for 1 DVD = 2.5(a) = 2.5(1) = $2.5
The cost for 1 CD = 2(b) = 2(1) = $2
b. Total cost for 4 DVDs and 4 CDs:
Substitute a = 4, and b = 4 into the equation.


Total cost for 4 DVDs and 4 CDs = $18
c. To find out if $20 would be enough to buy 6 DVDs and 3 CDs, substitute a = 6, and b = 3 into the equation. Solve for the total cost to see if it equals $20 or is less than $20. If it's greater than $20, then it won't be enough.


6 DVDs and 3 CDs cost more than $20, therefore, $20 is not enough.
C = 0.25r
this is becasue quarters have a value of 0.25 and c is the total cents
hope this helps and have a great day! :)
Answer:
The amount of change she received was
dollar.
Step-by-step explanation:
Price of each carton of milk is
cents and she bought 3 cartons of milk.
We know that, <u>1 dollar = 100 cents</u>
So, 1 cent =
dollar.
and
cents
dollar.
Thus, <u>the total price of 3 cartons of milk</u>
dollar.
She gave the cashier $10.
So, the amount of change she received
dollar.