The Math Thread - Request help with all forms of mathematics

[Demonskates]

If it's an ACRE FOOT... that means Acre * ft... which means.. we already have the answer I think...

Yep, I think so. An acre foot is not 1 acre deep. An acre foot is the volume of water,blood whatever that it would take to cover one acre a foot deep with liquid.
You and Sammy rock. Thanks all!
(By the way, with those figures, that comes to 8 or 9 miles of land, one foot deep in blood! How nasty is that?)

[Fut004]

Okay, so I received this as a Forward.
I tried to explain it to the person that it was just how math works, he's so amazed by this "Crazy" Math problem that it's bothering me.

HERE IS A MATH TRICK SO UNBELIEVABLE THAT IT WILL
STUMP YOU. PERSONALLY I WOULD LIKE TO KNOW WHO
CAME UP WITH THIS AND WHY THAT PERSON IS NOT
RUNNING THE COUNTRY.


1. GRAB A CALCULATOR. (YOU WON'T BE ABLE TO DO THIS ONE IN YOUR HEAD)
2. KEY IN THE FIRST THREE DIGITS OF YOUR PHONE NUMBER (NOT THE AREA CODE)
3. MULTIPLY BY 80
4. ADD 1
5. MULTIPLY BY 250
6. ADD THE LAST 4 DIGITS OF YOUR PHONE NUMBER
7. ADD THE LAST 4 DIGITS OF YOUR PHONE NUMBER AGAIN.
8. SUBTRACT 250
9. DIVIDE NUMBER BY 2


DO YOU RECOGNIZE THE ANSWER?
PRETTY COOL HEY!

So, I understand that Math is weird because it can be manipulated to give you the answer you are looking for (ie. If I want my GF's height to Multiple with some number to give me my height, I divide it out..), but can somebody explain how this one works? So I can then explain it to the dumbass who emailed it to me...


Thanks

[Cartesiantheater]

1. GRAB A CALCULATOR. (YOU WON'T BE ABLE TO DO THIS ONE IN YOUR HEAD)
2. KEY IN THE FIRST THREE DIGITS OF YOUR PHONE NUMBER (NOT THE AREA CODE)
3. MULTIPLY BY 80
4. ADD 1
5. MULTIPLY BY 250
6. ADD THE LAST 4 DIGITS OF YOUR PHONE NUMBER
7. ADD THE LAST 4 DIGITS OF YOUR PHONE NUMBER AGAIN.
8. SUBTRACT 250
9. DIVIDE NUMBER BY 2


DO YOU RECOGNIZE THE ANSWER?
PRETTY COOL HEY!

Let n = the first three digits of your number. let y equal the last four digits of your phone number (I assume)
3.) n * 80 = 80n
4.) 80n + 1
5.) 250 (80n + 1) = 20000n + 250
6.) 20000n + 250 + y
7.) 20000n + 250 +2y
8.) 20000n + 2y
9.) 10000n + y

now plug in n = first three digits and y = last four
You get the same answer... it's nothing but a trick...

let's look at the multiplication and divsion:
You are multiplying by 80, then by 250, and then you are dividing by 2 which is a net of multiplying by 10000.

Addition: Then, you're doing nothing but adding your last four digits ONCE, because if you add it TWICE and then divide by two, it's a net of adding it once.

SO, after all the wild goose chasing, all you are ultimately doing is just mulipying the first three digits by 10,000, and then adding the last four digits of it... that's ALL you're doing there...


...nice trick though...

EDIT- btw, the reason you have to multiply by 10000 is because when you multiply a three digit number by it you get a 7 digit number, which is how many a phone number has.

[liberdave]

Let n = the first three digits of your number. let y equal the last four digits of your phone number (I assume)
3.) n * 80 = 80n
4.) 80n + 1
5.) 250 (80n + 1) = 20000n + 250
6.) 20000n + 250 + y
7.) 20000n + 250 +2y
8.) 20000n + 2y
9.) 10000n + y

now plug in n = first three digits and y = last four
You get the same answer... it's nothing but a trick...

let's look at the multiplication and divsion:
You are multiplying by 80, then by 250, and then you are dividing by 2 which is a net of multiplying by 10000.

Addition: Then, you're doing nothing but adding your last four digits ONCE, because if you add it TWICE and then divide by two, it's a net of adding it once.

SO, after all the wild goose chasing, all you are ultimately doing is just mulipying the first three digits by 10,000, and then adding the last four digits of it... that's ALL you're doing there...


...nice trick though...

It's called anunnaki math. Don't listen to CT and his logical lies.

This is the reason why this trick works....
http://www.youtube.com/watch?v=aQEOeWipIcA

[Cartesiantheater]

We were struggling with this problem for hours... and I asked people at a physics forum to help me... and one guy gave me a great answer... so great.. I must post it...

we were trying to integrate 1/ [1 + (2x)^(1/2) ]
but were getting nowhere. We couldn't do u substitution, because we kept getting an x value in the derivative... but... this guy gave us THIS brilliant tactic

1.)Let 2x = u^2 (instead of let just plain old u = something... this was the key)

Therefore, 2 dx = 2 u du

And then we solve for dx by dividing both sides by 2 ==> dx = u du

2.) we plug that into the original function and get:

u/ (1 + u) du

then we pulled some fancy algebra to separate it into two integrals...(we need to separate that into two fractions)

u/ (1 + u) = (1 + u -1)/(1 + u) (we can do this because we are adding AND subtracting 1 to the numerator [the top of the fraction])

then we can separate the terms because there is a common denominator

(1 + u)/(1 + u) - 1/ (1 + u ) which becomes 1 - 1/(1 + u)

3.) Now separate them into two separate integrals : the integral of just du minus the integral of 1/(1 +u) du.

The first on is EASY to integrate; it just becomes u. So we have u minus the integral of 1/(1 + u)

4.) We've got to get the second integral into something we can integrate easy... so... we'll use substitution again, this time v

Let v = (1 + u) thus, dv = du... very easy...

Therefore, we now have the integral of 1/v dv which we can integrate easily
It becomes the natural log of the absolute value of v (this is just a rule of integration- it's a formula, no tricks)==> ln abs (v)

5.) Finally, we fill in the substitution variables with the original variables they substitued with... and then simply subtract them

u - ln abs(v) + C (we add plus C whenever we do an indefinite integral- a rule)

we let u^2 = 2x, which means u = the sqaure root of 2x = (2x)^(1/2)

We let v = (1 + u), and since u equals something (above), then
v = [1 + (2x)^(1/2)]

Therefore, after integrating, we have:

(2x)^(1/2) - Ln abs [1 + (2x)^(1/2)]

Sorry, but that was just too brilliant not to post. After spending so much time on it... I HAD to post it

[The Silence]

X squared+2x-78.45/X squared*3/43x
-------------------------------------
X squared-5x+6.8/X squared*-(7x+2-12.87)+12

....

[Cartesiantheater]

X squared+2x-78.45/X squared*3/43x
-------------------------------------
X squared-5x+6.8/X squared*-(7x+2-12.87)+12

....

EDIT: Something just dawned on me... is that just ONE giant fraction, or is it two problems?


I'm gonna try this... but I'm not for certain exactly what this is supposed to be... a coulple of questions...

Is the +12 supposed to be under the denominator?
And in the denominator, you have x squared * 3/43x...
You ARE saying this, right? : (x^2 * 3)/ 43x right?

Anyway, we shall try it

EDIT 2: for the top part, ran into another potential problem...

Are you trying to write this: (x^2 + 2x ) - [78.45/(x^2 *3/43x)]

or are you writing this : (x^2 + 2x - 78.45)/(x^2 * 3/43x) ?

EDIT 3: We can't help you unless you use parentheses to make the problem clear; cuz we can easily solve a problem that has nothing to do with the one you need help on.

[The Silence]

All was what I was trying to do but I've never posted a math proble before sorry...

[Cartesiantheater]

All was what I was trying to do but I've never posted a math proble before sorry...

It's ok... we love math and want to help if you need it... but we just need to know exactly what problem you need solved.

[donniedarko]

Silence, can you repost the problem?

Use the following notation

x^n = x to the exponent n
+ = addition
- = subtraction
* = multiplication
/ = division
() = grouping
log(x) = the log of x in base 10
ln(x) = the natural log of x
e = the base of the natural log

Example:

(x+(x-9)^2)/(4x^3)

(log(x/5)*(x^3+2x-9))/(x^(1/2))

edit:

p.s. remember square roots are really to the exponent (1/2)

[The Silence]

(x^2+2x-78.45)/(x^2*(3/43)*x)=
------------------------------->final of above is divided by the final below
(x^2-5x+6.8)/(x^2)*-(7x+2-12.87)+12=

[Sammy56]

Ok. I am just going to write out the final answer I got Silence. I checked it on my graphing calculator, and I am confident it's right. However, if you need to explain what I did, I'll be happy to post it.

(-301x^3 + 381.41x^2 + 28580.27x - 77148.5145) / (3x^3 - 15x^2 + 20.4x)

[Cartesiantheater]

Great work Sammy!

[The Silence]

Sammy56 and those others whom helped...I would like to thank u for solving this problem...though yes I would like a brief explaination on the solving of the equation.

This was an essay on my upcoming MAP test...now I just need to memorize how it's done.

Good job!

[donniedarko]

I think you may have gotten it wrong Sammy. Here is my solution:



then again, my math skills are not what they used to be so if you see the error let me know.

dd

[Sammy56]

What happened is The Silence still has some ambiguity in the bottom half of his big equation and we interpreted it different. I made a post about it earlier that explained it, cause I just noticed it today. However, my computer froze and I haven't gotten the chance to redo it.

I saw the bottom half of he equation to be:
(x^2-5x+6.8)/[(x^2)*(-(7x+2-12.87)+12)]

Sorry. I was being stupid and didn't realize that it could be seen in other ways. I know I got it right the way I solved it cause of my fancy graphing calculator, but I'm not sure if that's the problem Silence meant.

So, it's up to Silence. Can you make the problem more specific again. Use some more parenthesis or brackets to show your multiplication better.

[The Silence]

(x^2+2x-78.45)/(x^2*(3/43)*x)=
------------------------------->final of above is divided by the final below
(x^2-5x+6.8)/(x^2)*-(7x+2-12.87)+12=

I'll review the bottom portion again for you Sammy56 as best I can.

[(x^2-5x+6.8)/(x^2)]*[-1(7x+2-12.87)+12]=

Hope I didn't just make it even more confusing...

[Sammy56]

Ah, the I believe donniedarko has it right. I apologize. I should have asked about the bottom part earlier.

[donniedarko]

Ah, the I believe liberdave has it right. I apologize. I should have asked about the bottom part earlier.

Liberdave!?!

*slowly walks away, shoulders slumped, grumbling about "credit where credit is due"*

[Sammy56]

Oh crap. I'm really sorry donnie! I have responded to liberdave several times in the past couple of days, and his name has just been in my head. I swear I didn't mean that.

[donniedarko]

Lol, no worries Sammy. I have a thicker skin than that.

for next time, just remember, I am the better looking one with the kind of debonair swagger dave only wishes he could have...

[liberdave]

I need some help with these problems, I need to factor them down... (I know this is easy stuff, I just forgot how to do them)

1) 5x^2 + 25x + 30 ---> (5x+2) (5x+3) (is this right?)
2) 2x^2 + 11x +9
3) 2x^2 + 11x -9

Thanks for any help

[liberdave]

Lol, no worries Sammy. I have a thicker skin than that.

for next time, just remember, I am the better looking one with the kind of debonair swagger dave only wishes he could have...

I can only prove you wrong by letting you think you're right.

[Philosopher Foelhe]

Wow, Algebra is way back in the mist. My dad would kill me if he knew how rusty I was on this stuff.

But I... think (5x) (5x) = 25x^2 Not 5x^2. I could be wrong, but I'm pretty sure that's right. So (5x+2) (5x+3) = 25x^2 + 25x + 6

Luckily, the problems you have should be pretty simple, since 5 and 2 are prime. Uh... crap, I don't have time right now. Keep in mind you have to multiply across the parenthesis. So your answer should add up (x)(x) + (x)(n) + (n)(x) + (n)(n). If that makes any sense whatsoever... sorry, I'm kind of in a hurry.

If no one has finished this by the time I get off work, I'll break it down for you.

[Cartesiantheater]

I need some help with these problems, I need to factor them down... (I know this is easy stuff, I just forgot how to do them)

Yay! A post in the math thread!!!!!! :jumps around like happy dog...::

F.O.I.L. is very useful for checking these (first times first, outside times outside, inside times inside, last times last)

Also, I basically wrote down my thought process, hoping it will help. If you just want the answers, I will go back and make them orange.

5x^2 + 25x + 30

Hmmm... first I say, we ARE talking about 5x^2 and not (5x)^2 right? I assume we are.

so, lets see...

5x^2 + 25x + 30.

-first thing I notice is that all three terms are multiples of five, so, factoring out the five we have:

5 (x^2 + 5x + 6)

-next, leaving the five outside for a second...

x^2 + 5x + 6 ==> (x + something) (x + something) ==> (x + 3)(x + 2) works there, so then just put the five back on...

==> 5 (x + 3)(x + 2)

That right there is the quadratic factored completely out. If you want to put the five back in, you have to destribute it to every term. We can also use the quadratic formula to test (which I will do right now==> gives you roots of x = -3 and x = -2, so therefore the factors x +3 and x +2 must be in this one, so therefore what I wrote above is correct). The correct answer for this factoring is [5(x + 3)(x + 2).

This is slightly different from your answer, so look at this below though.

---> (5x+2) (5x+3) (is this right?)

The reason this is not correct is because you must redistribute the five to all terms. If you want to write it as two binomials multiplied together it should be this:

5 (x + 3) (x + 2) = (5x +15) (5x + 10)

EDIT- whoops!

The above is inaccurate! But the rest is right! What it REALLY should be when you redistribute it is just... (x + 3) (5x + 10) OR it could be (x + 2) ( 5x + 15)... it should be obvious where I messed up in the above...


Ok, next one:

2x^2 + 11x +9

You cannot just factor out a constant on this one (like we did the five on the first one) because 11 is a prime number. So... here's how I'd do it:

I know I have a 2 as a constant for my x^2 term, and I see that both signs are positive again, so the very first thing I will write is:

(2x + [something that I don't know yet] ) (x + [something else I don't know yet])

Next I'll lood at the next two terms (with coefficient 11 and 9). So, if I am going to get an 11x as my middle term, I've got to find a way to relate it to 9. What is the relationship between 11 and 9? well, 11 is 2 plus 9.

How can I get a 2 plus 9 out of this? Well, remember that the last term (the 9) will be the one that our factors are MULTIPLIED, right? And 11 is the term that the two factors we get from nine must ADD UP TO, right? I know this might be hazy, so look at this easy one:

If we have x^2 + 2x + 1, how to we factor it? well, we take two factors of 1 (1 and 1) and ADD them together to get the middle term (2) and MULTIPLY them together to get the last term (1) ==> so the factoring of this easy one would be (x + 1)(x + 1), right?

So, looking at the 11 and 9, we know if we had 2x + 9x we'd be set. The only problem is the coefficient of 2 on the x^2 term. But let's take a shot. What are the factors of 9? They are 9, 3, and 1. Those are the ONLY ones. So pick two. What if we pick 9 and 1? Pluging them in we might get...

(2x + 1) (x + 9)

And then testing it by multiplying it back out we get (F.O.I.L.==> first times first, outside times outside, inside times inside, last times last):

2x^2 + 18x + x + 9 = 2x^2 + 19x + 9, which is WRONG. But don't panic. What if we SWAP the 1 and the 9? Then we'd have:

(2x + 9)(x + 1)

And then testing THAT by multiplying we get:

2x^2 + 2x + 9x + 9

which equals: 2x^2 + 11x + 9

BINGO!



Number two defeated, we move on to number three...

2x^2 + 11x -9

EDIT- I can't figure this one out without the quadratic formula, but I'll keep what I wrote so you can follow my train of thought (who knows? it may help)

Now, perhaps we are in luck, as these are the same NUMBERS but different signs. When you see a positive AND a negative, that means you will have one positive and one negative atleast when you factor. So the first thing I would write is:

(2x + something) (x - something)


Let's try the factors of 9 we used before, 9 and 1:

(2x + 9) (x - 1)

= 2x^2 -2x + 9x - 9 = 2x^2 + 7x -9 which is WRONG.

what happens if we SWITCH the numbers?

(2x + 1) (x - 9)

= 2x^2 - 18x + x - 9 ==> 2x^2 -17x - 9 which is WRONG.

Hmm.. is there anything else we can try without trying new numbers?

YES! We can switch the SIGNS!

signs switched but with the same original placement of 9 and 1:

(2x - 9) (x + 1)

= 2x^2 + 2x -9x -1 ==> 2x^2 -7x -9 which AGAIN is WRONG...

Now we're in a hole, aren't we?

... or ARE we? There exist quadradicts that cannot be solved. The trusty quadratic formula will tell us (and will solve it if the values are difficult):

==> here's what I've found with the quadratic formula.

The quadratic has the roots of (-11/4) + ((193)^(1/2))/4 AND (-11/4) - ((193)^(1/2))/4.

Therefore, the factored out answer is:

(x -11/4 + ((193)^(1/2))/4) (x - 11/4 - ((193)^(1/2))/4), I believe.


Here's my work for the quadratic formula:

The quadratic formula is

-b plus or minus the square root of ( b^2 - 4ac) ALL divided by (2a)
where a is the coeficient on the x^2 term, b is the coefficient on the x term, and c is the constant.

-(11) + sqrt ((11^2) - 4 (2) (-9))
-------------------------------- = x
2 (2)

AND

-(11) - sqrt ((11^2) - 4 (2) (-9))
-------------------------------- = x
2 (2)

*remember plus or minus

which becomes (first the PLUS one):

-11 + sqrt (193)
--------------- = x
4

which separated out becomes

-11/4 + [sqrt (193)]/4 = x

Therefore, one of your values is (brining all of that stuff on the left to the same side as the x which leaves us with ONE term being:

(x - (x -11/4 + ((193)^(1/2))/4)

Now the negative:



-(11) - sqrt ((11^2) - 4 (2) (-9))
-------------------------------- = x
2 (2)

-11/4 - [sqrt (193)]/4

which gives us the second binomial of...


(x - 11/4 - ((193)^(1/2))/4)



I can also show you how to do this using Completing the square:

Here's a walk through of how it's done:

link

Move the loose number over to the other side.

Divide through by whatever is multiplied on the squared term.


Take half of the coefficient (don't forget the sign!) of the x-term, and square it.
Add this square to both sides of the equation.

Convert the left-hand side to squared form, and simplify the right-hand side. (This is where you use that sign that you kept track of earlier.)

Square-root both sides, remembering the "±" on the right-hand side. Simplify as necessary.


Solve for "x =".


Remember that the "±" means that you have two values for x.

So here's doing it using completing the square:


step one, get the loose number (the constant) on the other side:

2x^2 + 11x -9 = 0 becomes 2x^2 + 11x = 9

(the "= 0" is because when we are solving for roots [a tool that helps us factor] we are really trying to find out WHEN the quadratic is EQUAL to zero; because of this, we LET it equal zero, and this is how we find the roots, which we then use to factor [by basically switching them back to the x side]

step two, Divide through by whatever is multiplied on the squared term:

2x^2 + 11x = 9 becomes x^2 + (11/2)x = 9/2 (notice that 2x^2 just becomes x^2)

step three, Take half of the coefficient of the x-term, and square it; Add this square to both sides of the equation (the x-term is 11/2. Half of that is 11/4. 11/4 squared = 121/16:

x^2 + (11/2)x = 9/2 becomes x^2 + (11/2)x + 121/16 = 9/2 + 121/16

step four, Convert the left-hand side to squared form, and simplify the right-hand side.

You will always get a perfect square on the LEFT doing this, and the constant added to x in the parenthesis will ALWAYS be the square root of the number you squared and then added [see below]):

x^2 + (11/2)x + 121/16 = 9/2 + 121/16 becomes

(x + 11/4)^2 = 9/2 + 121/16

*note that we did NO algebra here; we only REWROTE the left side into a squared term. this always happens when you Complete the Square. That's why it's called "Completing the Square."

step five, Square-root both sides, remembering the "±" on the right-hand side. Simplify as necessary.

We are now trying to solve for x. The first thing we must do is get rid of the exponent. To do that, we take the square root of both sides (but before we do, first I want to combine the terms on the right- just add them up, remembering to find a common denomenator: 9/2 + 121/16 becomes 193/16):

(x + 11/4)^2 = 193/16 then becomes (taking the square root of both sides)

x + 11/4 = plus or minus the sqrt of (193/16)

*we always have plus or minus on terms that don't have variables in them

next, solve for x, by subtracting 11/4 from both sides:

x + 11/4 = plus or minus the sqrt of (193/16) becomes:

x = -11/4 plus or minus the sqrt of (193/16)

then, we notice that in the square root of 193/16, 16 is a perfect square (4 * 4 = 16). Therefore, we can pull that one out and then have:

x = -11/4 plus or minus [(193)^(1/2)]/4

which is the same roots obtained from the quadratic formula.


Therefore, the quadratic written in factored form is (again):


(x -11/4 + ((193)^(1/2))/4) (x - 11/4 - ((193)^(1/2))/4).



Hope that didn't confuse you. The third problem cannot be solved without one of the two methods: the quadratic formula or completing the square.