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4 years ago

Help me on this question? log x-log 9=1

2 answers:

6 0

$\log { x } -\log { 9 } =1\\ \\ \log { \left( \frac { x }{ 9 } \right) } =1\\ \\ \log _{ 10 }{ \left( \frac { x }{ 9 } \right) } =1\\ \\ { 10 }^{ 1 }=\frac { x }{ 9 } \\ \\ 9\cdot 10=x\\ \\ x=90$

This is because:

$\log _{ a }{ \left( \frac { x }{ p } \right) } \\ \\ =\log _{ a }{ \left( \frac { { a }^{ m } }{ { a }^{ n } } \right) } \\ \\ =\log _{ a }{ \left( { a }^{ \left( m-n \right) } \right) } \\ \\ =\left( m-n \right) \cdot \log _{ a }{ a } \\ \\ =m-n\\ \\ =\log _{ a }{ x } -\log _{ a }{ p } \\$

Alika [10]4 years ago

4 0

   log(x) - log(9) = 1

Subtracting the logs of numbers gives you the log of
the quotient of the numbers.

   Log(x) - log(9) is the log of (x/9).

So the equation says: log (x/9) = 1

Raise 10 to the power of each side: 10^(log of x/9) = 10^1

But 10^(log of x/9) is x/9, and 10^1 is 10.

So x/9 = 10

   x = 90

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4 years ago

When x-ray radiation and infrared radiation are

lys-0071 [83]

The correct answer is

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5 0

3 years ago

Read 2 more answers

A centrifuge has an angular velocity of 3,000 rpm, what is the acceleration (in unit of the earth gravity) at a point with a rad

Anna71 [15]

Answer:

$a_{r} = 1006.382g \,\frac{m}{s^{2}}$

Explanation:

Let suppose that centrifuge is rotating at constant angular speed, which means that resultant acceleration is equal to radial acceleration at given radius, whose formula is:

$a_{r} = \omega^{2}\cdot R$

Where:

$\omega$ - Angular speed, measured in radians per second.

$R$ - Radius of rotation, measured in meters.

The angular speed is first determined:

$\omega = \frac{\pi}{30}\cdot \dot n$

Where $\dot n$ is the angular speed, measured in revolutions per minute.

If $\dot n = 3000\,rpm$ , the angular speed measured in radians per second is:

$\omega = \frac{\pi}{30}\cdot (3000\,rpm)$

$\omega \approx 314.159\,\frac{rad}{s}$

Now, if $\omega = 314.159\,\frac{rad}{s}$ and $R = 0.1\,m$ , the resultant acceleration is then:

$a_{r} = \left(314.159\,\frac{rad}{s} \right)^{2}\cdot (0.1\,m)$

$a_{r} = 9869.588\,\frac{m}{s^{2}}$

If gravitational acceleration is equal to 9.807 meters per square second, then the radial acceleration is equivalent to 1006.382 times the gravitational acceleration. That is:

$a_{r} = 1006.382g \,\frac{m}{s^{2}}$

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3 years ago

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tresset_1 [31]

The answer is B. Nonrenewable

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3 years ago

An electric field of 710,000 N/C points due west at a certain spot. What is the magnitude of the force that acts on a charge of

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Answer:

The magnitude of force is $4.26\times 10^{- 6} N$