solve variables in terms of variables

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February 5th 2007, 06:01 PM
dazed&confuzed

solve variables in terms of variables

Help would be appreciated.

1.)

Solve this formula for the variable $\mu$ :

z = $\frac{x-\mu}{\mu}$

2.)

Solve this formula for the variable R:

$\frac{E}{e}$ = $\frac{R+r}{R}$

3.)

The intensity of light as measured from a light bulb varies inversely as the square of the distance "d" from the bulb. Suppose that I is 90W/m^2 when the distance is 5 m. What would the light intensity be 10m from the bulb?

4.) Write a radical expression with an index of 3 and a radicand of (x - 5)^9. Simplify it as much as possible.

Help with an explanation of the steps would be very much appreciated. I'm completely lost and didn't get a chance to meet with the instructor after class because of work.
February 5th 2007, 06:54 PM
AlvinCY

Question 1: Solve this formula for the variable $\mu$ :

$z = \frac{x-\mu}{\mu}$
$z = \frac{x}{\mu} - \frac{\mu}{\mu}$
$z = \frac{x}{\mu} - 1$
$z + 1 = \frac{x}{\mu}$
$\mu = \frac{x}{z + 1}$

Question 2: Solve this formula for the variable R:

$\frac{E}{e} = \frac{R+r}{R}$
$\frac{E}{e} = \frac{R}{R} + \frac{r}{R}$
$\frac{r}{R} = \frac{E}{e} - 1$
$\frac{r}{R} = \frac{E - e}{e}$
$r \times e = R \times (E - e)$
$R = \frac {r\,e}{E - e}$

Question 3: The intensity of light as measured from a light bulb varies inversely as the square of the distance "d" from the bulb. Suppose that $I = 90 W/m^2$ when the distance is 5 m. What would the light intensity be 10m from the bulb?

We know that I varies inversely to the square of the distance, d. In another words:

$I = k \times \frac{1}{d^2}$ for some constant k.

We can work k out by substituting $I = 90, d = 5$

$90 = k \times \frac{1}{25}$
$k = 2250$ , therefore,
$I = 2250 \times \frac{1}{d^2}$

For d = 10, $I = 2250 \times \frac{1}{100} = 22.5 W/m^2$

Question 4: Write a radical expression with an index of 3 and a radicand of (x - 5)^9. Simplify it as much as possible.

This question simply refers to: $\sqrt[3]{(x - 5)^9}$

$\sqrt[3]{(x - 5)^9}$
$=[(x - 5)^9]^\frac{1}{3}$
$=(x - 5)^3$ This would be the simplest form without expanding that.
February 5th 2007, 10:30 PM
dazed&confuzed

Thanks for the response.

I have a question regarding number 3. How did you come about I = 2250? Thats the step that I got lost on.
February 5th 2007, 10:42 PM
AlvinCY

Quote:

Originally Posted by AlvinCY

Question 3: The intensity of light as measured from a light bulb varies inversely as the square of the distance "d" from the bulb. Suppose that $I = 90 W/m^2$ when the distance is 5 m. What would the light intensity be 10m from the bulb?

We know that I varies inversely to the square of the distance, d. In another words:

$I = k \times \frac{1}{d^2}$ for some constant k.

We can work k out by substituting $I = 90, d = 5$

$90 = k \times \frac{1}{25}$
$k = 2250$ , therefore,
$I = 2250 \times \frac{1}{d^2}$

For d = 10, $I = 2250 \times \frac{1}{100} = 22.5 W/m^2$

I didn't say I = 2250...

Ok, let me take you through it,

$I = k \times \frac{1}{d^2}$ for some constant k.

This is what we're given in the question, Intensity varies indirectly to the square of the distance, so we know $I$ and $\frac{1}{d^2}$ are related... and so I've introduced a factor of $k$ in there

Make sense so far?

Then we're given that $I = 90$ when $d = 5$ , so I substituted $I$ and $d$ into the equation above, yielding:

$90 = k \times \frac{1}{25}$ , so $k = 90 \times 25 = 2250$ ...

So now that we know $k$ we can replace
$I = k \times \frac{1}{d^2}$ with
$I = 2250 \times \frac{1}{d^2}$ by substituting $k$ into the equation.

And now it's a matter of finding $I$ when $d=10$ using this new equation!

Does that make sense? If not, post here again... and explain to me what you don't understand.
February 5th 2007, 10:46 PM
dazed&confuzed

Ooooo okay. I just wasn't paying attention to details. I realized what you were doing after I typed that response, I should have looked at it again. Thanks for the response though.