We are asked to find ratio of two relative intensities. Let's start by rearranging formula for R:

For air-raid siren we have:

For jet engine noise we have:

To find out <span>many times greater is the relative intensity of the air-raid siren than that of the jet engine noise we need to divide these two numbers:
</span>tex] \frac{R_{1}}{R_{2}} = \frac{10^{15}}{10^{12}} =10^{3}=1000[/tex]
The answer is b.
when x decreases, f(x) decreases.
when x increases, f(x) increases.
3/4(2x + 5) = 5(x + 12)
First get rid of any fractions by multiplying my the denominator
4 x 3/4(2x + 5) = 4 x 5(x + 12)
3(2x + 5) = 20(x + 12)
Then get rid of the parentheses
6x + 15 = 20x +240
Then simplify
6x + 15 - 15 = 20x + 240 - 15
6x = 20x + 225
6x - 20x = 20x - 20x + 225
-14x = 225
Now divide on both sides
-14x/-14 = 225/-14
x = -16
Let W = width of package
Let H = height of package
Let L = length of package
The perimeter cab be one of the following:
P = 2(L + W), or
P = 2(L + H)
The perimeter of the cross section cannot exceed 108 in.
When the width is 10 in, then
2(L + 10) <= 108
L + 10 <= 54
L <= 44 in
When the height is 15 in, then
2(L + 15) <= 108
L + 15 <= 54
L <= 39 in
To satisfy both of these conditions requires that L <= 39 in.
Answer: 39 inches
Answer:
The answer is B
Step-by-step explanation:
It is constantly changing every time so it is not linear. (2 x 100.5 is 205, 5 x 80 is 400)