:)

- anonymous

:)

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- schrodinger

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- anonymous

Is this actually a physics or math project or just a question asking for you to solve for the variable "t"?

- anonymous

you can use basic algebra to solve for t by isolating it, but it will be an equation using L. But that might be all you need to do... sort of depends on what the purpose of this question is.

- anonymous

just have to solve for t.. i think :/ sorry my computer is lagging!

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- anonymous

\[L = \frac{ 980t^{2} }{ 4\pi^{2} }\]

- anonymous

yeah, that's it :)

- phi

here is an example
http://www.khanacademy.org/math/algebra/solving-linear-equations-and-inequalities/v/solving-for-a-variable
The first step is multiply both sides by 4\(\pi\). Can you do that?

- phi

The first step is multiply both sides by 4\(\pi^2\). Can you do that?

- anonymous

hmm, not really.. :///

- phi

write \(4 \pi^2\) times on both sides of the equation

- anonymous

okay

- anonymous

like i dont understand it because it has the pi? :/ i suck at math

- phi

pi is just a number
if you multiply both sides by 4 pi^2 you get
\[ 4 \pi^2 L = 4 \pi^2 \cdot \frac{980 t^2}{4 \pi^2}\]

- phi

now use a simple rule: if you see something divided by itself , it becomes 1.
Does that ring a bell? It makes the right side "simplify"

- anonymous

no it doesnt :(

- anonymous

ahhh, this is confusing to me

- phi

Here are some examples
\[ 3\cdot \frac{4}{3} = \frac{3\cdot 4}{3}= \frac{\cancel{3}\cdot 4}{\cancel{3}}=4\]

- phi

do you see anything in your problem that "cancels"?

- anonymous

yeah, i get that but it has the pi in the one im doing? :/

- anonymous

so the pi would cancel out?:S:/

- phi

\[ 4 \pi^2 L = \cancel{4 \pi^2} \cdot \frac{980 t^2}{\cancel{4 \pi^2}} \]

- anonymous

yeaah, thats what i thought :))

- phi

it is a good rule to remember.
now multiply both sides by \(\frac{1}{980} \)

- anonymous

one over what? i can't read that:((

- phi

1/980

- phi

first write it down:
\[ 4 \pi^2 L \cdot \frac{1}{980}= 980t^2 \cdot \frac{1}{980} \]

- anonymous

k 1 sec

- anonymous

okays (:

- phi

the 980's cancel on the right side, right? (that is why we did this)
we get
\[ t^2 = \frac{4\pi^2 L}{980} \]
Do you see how we did that?

- anonymous

ohhh, yeah i did

- phi

the 4 pi^2 L is on the top, and when we multiply fractions, it is top times top and bottom times bottom

- phi

\[ \frac{4 \pi^2 L}{1} \cdot \frac{1}{980} = \frac{4 \pi^2 L\cdot 1}{1 \cdot 980} = \frac{4 \pi^2 L}{980}\]

- phi

I would simplify by dividing the top by 4, and the bottom by 4. Can you do that?

- anonymous

all of the top?:S i dont know how to simplify because the pi is there? do i use 3.14?:/

- phi

leave pi alone. if you divide the top by 4 it is doing this
\[ \frac{\cancel{4}\pi^2L}{\cancel{4}}\]

- phi

now divide the bottom by 4: 980/4 is 245
we get
\[ t^2= \frac{\pi^2 L}{245} \]

- phi

do you know how to "solve" for t?

- anonymous

multiply both sides by 245?:S

- anonymous

im not sureee :(

- phi

if you multiply both sides by 245 you would get a 245t^2 (making it more "complicated" )
Have you heard of a square root?

- anonymous

yeaah i have

- phi

square roots are how you "undo" squares:
\[ \sqrt{x^2}= x\]

- phi

I would take the square root of both sides

- anonymous

can you draw it out for me so i can see what you mean? it makes more sense when you do that:D

- phi

I've noticed it's easier to do work when someone else is doing it, too.
try, it won't hurt

- anonymous

it really does help when someone else shows you, but ill draw it on here to see if im right :) im prob not thou ...haha

- phi

Just draw a big square root over each side

- anonymous

|dw:1354152167387:dw|

- phi

yes, by the t^2 is inside the root sign. and so is the 245.
now use the idea that sqrt(x^2) is x so sqrt(t^2) is ?

- anonymous

245? haha.. im not sure :/ or 15? :S

- phi

no, not a number. square root of x*x is x square root of t*t is t
(I guess you have to memorize that)
we get
\[ t= \sqrt\frac{ \pi^2 L}{245} \]

- phi

now, you can "simplify" (just to show off how much math you know)
use the rule that sqrt(a * b) is the same as sqrt(a) * sqrt(b)
so the top is sqrt(pi^2) * sqrt(L)
you can simplify sqrt(pi^2) , right?

- anonymous

yes?.....

- phi

use the same rule as for sqrt(x^2) (or sqrt(t^2) for that matter)

- anonymous

i dont understand when you say sqrt)x^2) really..

- phi

we start with
\[ t= \sqrt\frac{ \pi^2 L}{245} \]
we can "break up" the square root by separating it:
\[ t = \frac{ \sqrt{\pi^2}\sqrt{L}}{\sqrt{245}} \]
the rule "sqrt(x^2) = x means when you see "something" to the 2nd power inside a square root , you can replace it with just "something"
There is a good reason to learn this, but first what is
\[ \sqrt{\pi^2} \] ?

- phi

or, another way to think of it, sqrt undoes the square

- anonymous

ooh, that kinda makes more sense. what do you mean what is it?

- anonymous

so it would be pi^2?

- anonymous

or sqaure root pie? :S

- anonymous

or neither? hahaa

- phi

sqrt(pi^2) is not pi^2 (that would mean sqrt is doing nothing)
sqrt(pi^2) is not sqrt(pi) (that would mean pi^2 is the same as pi)
sqrt(pi^2) "makes the square go away"

- phi

if
\[ \sqrt{x^2}= x\]
and
\[ \sqrt{t^2}=t\]
what is
\[ \sqrt{\pi^2}= ?\]

- anonymous

pi..lol

- phi

yes, so now you have
\[ t = \frac{ \sqrt{\pi^2}\sqrt{L}}{\sqrt{245}} \]
\[ t = \frac{ \pi\sqrt{L}}{\sqrt{245}} \]
finally we can rewrite sqrt(245) as sqrt(5*7*7)
or
\[ \sqrt{7^2}\cdot \sqrt{5} \]
do I dare ask what is sqrt(7*7)?

- anonymous

hahah!

- anonymous

49

- phi

7*7 = 49
but sqrt(7*7) is 7 (or sqrt(49) is 7)
7*7 is 7^2 and the square root undoes the square.

- anonymous

yeaah

- phi

we now have
\[ t = \frac{ \pi\sqrt{L}}{\sqrt{245}} \]
\[ t = \frac{ \pi\sqrt{L}}{7\sqrt{5}}

- phi

\[t = \frac{ \pi\sqrt{L}}{7\sqrt{5}} \]

- phi

The last step. People (why?) do not like sqrt in the denominator, so we multiply top and bottom by sqrt(5)
\[ t = \frac{ \pi\sqrt{L}}{7\sqrt{5}}\cdot \frac{\sqrt{5}}{\sqrt{5}}= \frac{ \pi\sqrt{5L}}{35}\]

- phi

that is your answer

- anonymous

wow, that is a loooot of work!! but thank you so much.. when i re-read over it tomorrow, i'll prob be awesome at it ;) haha....ill prob get it! :D are you a teacher?

- phi

The reason you learn this is to
1. See if you *can* learn it.
2. If you can learn it, it changes how you think (in a good way)

- phi

I am an engineer for a living..

- anonymous

thats awesome :D

Looking for something else?

Not the answer you are looking for? Search for more explanations.