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3 years ago

9

Helpppppppppppppppppppppppppppppp

1 answer:

n200080 [17]3 years ago

4 0

For this case we first write the equation of which we will use:
I (db) = 10log (l / l)
We substitute the value of l.
We have then:
l = 10 ^ 8lo
Substituting in the given equation:
I (db) = 10log ((10 ^ 8lo) / lo)
Rewriting:
I (db) = 10 * log (10 ^ 8)
I (db) = 80
Answer:
I (db) = 80
option 4

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3 years ago

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Step-by-step explanation:

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3 years ago

Read 2 more answers

What are the roots to<br> the equation<br> 2x² + 6x-1=0?

lys-0071 [83]

Answer:

The roots of equations are as m = $\frac{-3}{2} + \frac{\sqrt{11} }{2}$ And n = $\frac{-3}{2} - \frac{\sqrt{11} }{2}$

Step-by-step explanation:

The given quadratic equation is 2 x² + 6 x - 1 = 0

This equation is in form of a x² + b x + c = 0

Let the roots of the equation are ( m , n )

Now , sum of roots = $\frac{ - b}{a}$

And products of roots = $\frac{c}{a}$

So, m + n = $\frac{ - 6}{2}$ = - 3

And m × n = $\frac{ - 1}{2}$

Or, (m - n)² = (m + n)² - 4mn

Or, (m - n)² = (-3)² - 4 ( $\frac{ - 1}{2}$ )

Or, (m - n)² = 9 + 2 = 11

I.e m - n = $\sqrt{11}$

Again m + n = - 3 And m - n = $\sqrt{11}$

Solving this two equation

(m + n) + ( m - n) = - 3 + $\sqrt{11}$

I.e 2 m = - 3 + $\sqrt{11}$

Or, m = $\frac{-3}{2} + \frac{\sqrt{11} }{2}$

Similarly n = $\frac{-3}{2} - \frac{\sqrt{11} }{2}$

Hence the roots of equations are as m = $\frac{-3}{2} + \frac{\sqrt{11} }{2}$ And n = $\frac{-3}{2} - \frac{\sqrt{11} }{2}$ Answer

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4 years ago

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Sveta_85 [38]

Answer:

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3 years ago