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The polynomial is 3a^3 - 8a^2 + 11a - 17.

a * b * c = -2660/27

If abc represents the number 100a + 10b +c, then a, b and c must be integers.

Set a + 5b + 15c = 100a + 10b + c

Manipulate this to get c in terms of a and b, then see what the minimum values of a and b must be for c to be the smallest possible positive integer.

Here's a helping hand:

Hi Xxmathsguy,

Welcome to our Web2.0calc forum.   I can see you have been busy here.  I hope you are enjoying yourself.   

You do not need to appologise to the guest.  Or to anyone for that matter.

Askers should always check the answers.

If they are too lazy or have so little knowledge that they can't then they are just copying and they should get it wrong.

If they are unsure then they should ask for further clarification.

The question is titled: Half Donuts. This answer should be titled Half Brains

Measures of dumbness are usually treatable with education:

However, for the stupid, education often makes the dumb dumber ...and more dangerous. 

For the teachers it makes them more suicidal.

For the observers it makes them more nauseous.  

--. .-

I would give you a hint if I had produced anything that made sense.  

Angle AEC = arctan(6/13) = 24.77514057º

I'll start by stating that I do not have a clue.

Now the discussion:

when x=0     k can be anything

But, for any other given x,  k cannot equal 2/x     

But I don't know what the answer is.     I tried graphing it but that just confused me more.  

Maybe it is    \((-\infty, \frac{2}{x}) \cup (\frac{2}{x},\infty)\)    where   \(x\ne 0\)           But I am guessing.  

oh and @ straight 8000 can onlly be counted once so if you are correct it is suppose to be 30300+301=901

melody can you help us?

0*00

technically it is 1-9 because if it is 0-9 then you have to -1 for 00*0 and the other ones @ straight

\(\frac{1}{3}x+22-22=x-10-22\)

\(\frac{1}{3}x=x-32\)

\(\frac{1}{3}x-x=x-32-x\)

\(-\frac{2}{3}x=-32\)

\(3\left(-\frac{2}{3}x\right)=3\left(-32\right)\)

\(-2x=-96\)

\(x=48\)

Again, I am so sorry guest! And @ melody I will be more careful

OMG 

I just saw your post and i realize it 

sorry guest! i'm going to fix this right away

You made a boo boo.

It is just a careless error but you should have noticed that your answer does not make sense.

Let Steven be x

\(\frac{1}{3}x+22=x-10\)

\(\frac{1}{3}x+12=x\)

\(\frac{2}{3}x=12\)

\(\frac{1}{3}x=6\)

\(x=18\)

Steven is \(\boxed{18}\) years old

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nice job guest!

\(6+b+c=15\)

\(36+b^2=c^2\)

\(c=\sqrt(36+b^2)\) that is 36 and b^2 in the square root

\(c=6+b\)

\(6+b+6+b=15\)

\(12+2b=15\)

\(2b=3\)

\(b=1.5\)

\(c=15-7.5\)

\(c=7.5\)

Someone check my work plz maybe it is wrong and guest got a different answer so idk whose is right

Why don't you put an ounce of effort into presenting your question properly?

I'm giving this my best shot at it(tel me if i am wrong) Also befor i start. The problem said"Segments EG (which is a repeating decimal)" and "EG = 40"

40 is not a repeating decimal.

EG=40 

DC=27

40*27=1080

next,BC*FH=1080 and BC=AD=18

1080/18 = 60

FH=60

Just like geno3141!

And guest, in geno3141's answer, EG is smaller than AD

I believe the question has some error it says EG=40 and AD=18

AD is a hypotenuse of EG so....

x+3,11

I agree.

the first 4 numbers = 15/4

but 15/4 is not a whole number so it is impossible

\(1152\)

I think

2^15 * 3^20 * 5^24
 

When doubled, each prime factor has an even exponent, making it a square. When tripled each prime factor has an exponent divisible by 3, making it a cube. And when multiplied by 5 each prime factor has an exponent divisible by 5 making it a power of 5.