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Thank you, CPhill

\(V=\dfrac 4 3 \pi r^3\\ \dfrac{dV}{dr} = 4 \pi r^2\\ \text{ok a bit of magic here}\\ dV = 4\pi r^2 dr\\ \dfrac{dV}{dt} = 4\pi r^2 \dfrac{dr}{dt} \\ 2 = 4 \pi (3)^2 \dfrac{dr}{dt}\\ \dfrac{dr}{dt} = \dfrac{1}{18\pi}~in/hr\)

What is the smallest positive integer n for which

9n-2 and 7n+3

share a common factor greater than 1?

\(\text{Euclidean algorithm:}~ \boxed{gcd(a-nb,b)=gcd(a,b)} \)

\(\begin{array}{|rcll|} \hline && gcd(9n-2,7n+3) \\ &=& gcd(9n-2-1\cdot(7n+3) ,7n+3)=gcd(2n-5,7n+3) \\ &=& gcd(2n-5,7n+3-3\cdot(2n-5))=gcd(2n-5,n+18) \\ &=& gcd(2n-5-2\cdot(n+18),n+18)=gcd(-41,n+18) \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline gcd(-41,\underbrace{n+18}_{=41~(\Rightarrow n>0!)}) \qquad -41~ \text{is prime} \\\\ n+18 &=& 41 \\ n &=& 41-18 \\ n &=& 23 \\ \hline \end{array}\)

The common factor greater than 1 is \(\mathbf{41}\).

and the smallest positive integer n is \(\mathbf{23}\)

Hint: \(gcd(-41,41) = 41\)

The major fallacy is still 2 = 1. 

If \(a=b=0\), technically 0 + 0 = 0 is correct, and (0+0)(0-0) does equal 0(0-0). 

What if \(a=b=0\)?

We all do that sometimes... :P

Ok then... :P

LOL!!!!

Maybe we have a "Conjunctive Contraction"   ....[ or is it....a "Contractive Conjunction ?? " ]

We also do not know if the 5m wire was cut into 2 equal length wires or 2 random wires. 

Actually, since the wire is cut into two pieces (that are equal, I assume), the perimeter of the square is 5/2 meters or 250 cm. Each side of the square would then be 250/4 cm, which equals 62.5 cm. \(62.5^2\)  is 3906. The total area of the square would then be 3906.25 \(cm^2\) or 0.390625 \(m^2\). 

The minimum area would be 0...just don't place the wire.

I think you meant to state the question differently...

Hope this sort of helps,

- PartialMathematician

The fallacy is the transition from Step 4 to Step 5

4    (a + b) ( a - b)   = b ( a - b)

But if   a   = b.....then we have

5   (a + b) ( 0)  = b ( 0) 

So....6   becomes

0    =    0         which is the logical conclusion, not    " a + a   =   a "

Ok then...

Please do not ask or post offensive or disrespectful questions on this website. 

You can put that into the "LaTeX" box to make the GCD part look nicer. 

In the proof, 0 is restricted because the denominator of a fraction cannot be 0 (cannot divide by 0). If we set x to 1, then the part \(3f(\dfrac{x-1}{x})\) cancels out and turns into 0 because \(3f(0) = 0\). Also, if you look around in the proof, you will find \(x\)  and \(1-x\) many times as the denominator, which cannot equal 0. Simply stated, 0 and 1 are restricted, AKA not in the domain of the function. 

Hope this helps,

- PartialMathematician

Thanks, EP.....sometimes....my brain actually works!!!

Mmmmmm....not sure about all perfect squares , GA

9    =   2 + 3 + 4

25 = 3 + 4 + 5 + 6 + 7

36 = 11 + 12 + 13

81 = 26 + 27 + 28

etc.....

Ok, Guest, I'll see if I can help you. 

Ok, thanks, CoolStuffYouTube. 

You probably saved a grade...

"F" could also be an altitude, median, or perpendicular bisector...we really need an image to help you. 

I am guessing "F" is the incenter of the triangle. (i think I've seen and solved this problem before...). But just to clarify, could you please send an image?