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mina [271]
3 years ago
6

The magnitude, M, of an earthquake is defined to be M=log I/S, where I is the intensity of the earthquake (measured by the ampli

tude of the seismograph wave) and S is the intensity of a “standard” earthquake, which is barely detectable. What is the magnitude of an earthquake that is 35 times more intense than a standard earthquake? Use a calculator. Round your answer to the nearest tenth.
–1.5?
–0.5?
1.5?
3.6?
Mathematics
2 answers:
Nitella [24]3 years ago
7 0
The answer is 1.5 have a good day
Brilliant_brown [7]3 years ago
6 0
Given:
M = log I/S

M = magnitude
I = intensity of the earthquake
S = intensity of the standard earthquake

The minimum intensity of a standard earthquake is 10.
Intensity of the earthquake is 35 times the standard earthquake. So,
10 x 35 = 350

M = log 350/10
M = log 35
M = 1.544 or 1.5  Third option
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So, let's substitute 1 for f(x):

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Let's solve for x in each case.

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We have:

1=-\frac{1}{2}x-1

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Let's cancel out the fraction by multiplying both sides by -2. So:

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Flip:

x=-4

So, x is -4.

Case II:

We have:

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Flip:

x=1

This is the solution for our second case.

So, we have:

x_1=-4\text{ or } x_2=1

Now, can check to see if we have to to remove solution(s) that don't work.

Note that x=-4 is the solution to our first equation.

The first equation is defined only if x is less than -2.

-4 <em>is </em>less than -2. So, x=-4 is indeed a solution.

x=1 is the solution to our second equation.

The second equation is defined only if x is greater than or equal to -2.

1 <em>is</em> greater than or equal to -2. So, x=1 is <em>also </em>a solution.

Therefore, our two solutions are:

x_1=-4\text{ or } x_2=1

Out of our answer choices, we can pick D.

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