math problem?

[DA(]

ok can anyone help me solve this problem
10[3c(Cf(exp3)+(-4c))(exp7)-(-9)+H]=478
Rules: c cannot equal zero,3,8,9,a negative prime number
f cannot equal 0,8 17,c, a positive odd number
h cannot equal c,h,3 or any positive integer

if anyone can help me solve this problem, i would be greatly thankful.
if your wondering wat this is for, its for my math class, anyone that can solve this problem gets an A for the entire year in math and the math teacher said she would give 20 also

[saren]

ok can anyone help me solve this problem
10[3c(Cf(exp3)+(-4c))(exp7)-(-9)+H]=478
Rules: c cannot equal zero,3,8,9,a negative prime number
f cannot equal 0,8 17,c, a positive odd number
h cannot equal c,h,3 or any positive integer

if anyone can help me solve this problem, i would be greatly thankful.
if your wondering wat this is for, its for my math class, anyone that can solve this problem gets an A for the entire year in math and the math teacher said she would give 20 also

You're a moron. Seriously. Obviously your basic algebra class hasn't informed you of the fact that it's an unsolvable problem. Thanks.

[DA(]

why is it unsolvable smartass

[banana_sam]

it's that saren's stupid. every problem is unsolvable to him.

[arpsmack]

ok can anyone help me solve this problem
10[3c(Cf(exp3)+(-4c))(exp7)-(-9)+H]=478
Rules: c cannot equal zero,3,8,9,a negative prime number
f cannot equal 0,8 17,c, a positive odd number
h cannot equal c,h,3 or any positive integer

Ummm, h cannot equal h? You are a moron. Saren's decision stands.

[MaxX]

Ummm, h cannot equal h? You are a moron. Saren's decision stands.

I agree unless that was a typo, there is no answer the that equation...

[hba]

I don't think it's unsolvable, but it would take time to think about. If I read that right, there's a point where you have to give -4 an exponent of 7 which equals out to sixteen-thousand something. Then you have exponents as high as 22 on the c variable. I tried it on paper and I got stuck at "3c??f?? - 49152c?(?) + h = 38.8".

[MaxX]

But H ?can't be equal to H thus making the problem unsolvable. Look at this 34=3 this is obviously not true because 34!=3, therefore H cannot be equal to any real or imaginary number.

[hba]

I considered that a typo. If he's really that stupid to post a question where h can't equal itself, he should be kicked in the nuts numerous times for that.

[arpsmack]

I don't think it's unsolvable, but it would take time to think about. If I read that right, there's a point where you have to give -4 an exponent of 7 which equals out to sixteen-thousand something. Then you have exponents as high as 22 on the c variable. I tried it on paper and I got stuck at "3c??f?? - 49152c?(?) + h = 38.8".

Wow.... ummmm. If you are actually trying to solve this algebraicly, let me stop you. You can't possibly solve an equation with 3 unknowns unless you have 3 equations present. Best bet would be to simply substitute random values in. For example, if he made a typo, and actually meant that h could not equal c, f, 3 or any positive integer... then one possible solution set would be:
c = 1
f = the cube root of 4
h = 38.8

But once again let me point out the idiocy in this obviously bogus problem.

f cannot equal 0,8 17,c, a positive odd number

Why would the teacher feel the need to specify that f could not equal 17, when it later states that f cannot be a positive odd number? Textbook equations are not redundant like this.

h cannot equal c,h,3 or any positive integer

Ignoring either a typo or simply just retardedness, there is once again evidence of unneeded restrictions on the solution set. Why say that h cannot equal 3, when you later specify that h cannot be a positive integer?

Also let me point out that he used capitol C and lowercase c in the same equation, which in mathematical terms should technically be 2 separate variables (with the capitols generally representing constants). This would give us FOUR unknowns.

To put it bluntly, it looks like a 4 year old came up with this crap.

[MaxX]

I considered that a typo. If he's really that stupid to post a question where h can't equal itself, he should be kicked in the nuts numerous times for that.

Agreed.

Wow.... ummmm. If you are actually trying to solve this algebraicly, let me stop you. You can't possibly solve an equation with 3 unknowns unless you have 3 equations present. Best bet would be to simply substitute random values in. For example, if he made a typo, and actually meant that h could not equal c, f, 3 or any positive integer... then one possible solution set would be:
c = 1
f = the cube root of 4
h = 38.8

But once again let me point out the idiocy in this obviously bogus problem.

Why would the teacher feel the need to specify that f could not equal 17, when it later states that f cannot be a positive odd number? Textbook equations are not redundant like this.

Ignoring either a typo or simply just retardedness, there is once again evidence of unneeded restrictions on the solution set. Why say that h cannot equal 3, when you later specify that h cannot be a positive integer?

Also let me point out that he used capitol C and lowercase c in the same equation, which in mathematical terms should technically be 2 separate variables (with the capitols generally representing constants). This would give us FOUR unknowns.

To put it bluntly, it looks like a 4 year old came up with this crap.

After reading your post I actually took a look at the problem for a second and you are correct in saying that you cannot solve the equation algebraically. Also I believe the reason for the redundancy is to further confuse the students of the class to make the problem seem much more confusing then it truly is. As far as getting an A in the class for the entire year, the only possible way a teacher would do that would be if you found all the solutions for the problem, and from the looks of it there are going to be multiple solutions, so good luck...

[arpsmack]

You couldn't list all possible solutions, it would be infinite. What we are looking at is an equation with 3 (4?) unknowns. Basically it would boil down to something similar to y = x + z (grossly simplified of course), which when plotted in a 3-dimensional coordinate system shows infinite solutions. I suppose if you REALLY wanted to, you could go as far as to plot this... but with all the imposed limitations it would be maddening. I still think he made this up, though I have no idea why he would do that.

[Belphegor]

This could be a question that the teacher asked to see the students try and solveit when its blatently not possable. But my teacher had this one math course where 2+3 equaled 1. Not 5...

[hba]

There are equations that can make 1 = 0, but their also impossible because you have to consider certain rule modifications of math.

[Lord_Nicon]

yes... also like 1+1 not = to 2....
that is provable in one of the most disgusting paragraph proofs of all time in a college level calculus class....

[Belphegor]

This math is like a distorted math. The actual study of it is used for something I forget what.. But in normal life it is 100% useless. She said she took it for fun. And basicly normal standard digits had diffrent values. Which is possable to change... Like in hex f=16 when normally f=f.

[saren]

1 can not equal 0 or 1+1 not equal 2.

If that happens, you divided by 0 someplace, even if it was in an unsimplified form. You can't divide by zero, and using limits or newton's method still won't give you a valid way to violate the rules of mathematics. They are ratios and relations built ontop of one another, it is logically impossible for them to unravel themselves.

[hba]

Well no sh*t saren. Why did you think I said impossible?

[DeAtH_SkUlL]

Because it is

[banana_sam]

Nikolay's gay cause he bad repped me for flaming saren. Saren, how much DID you blow him last night?