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I personally disagree, but I fixed it.

Thanks, Guest!

thanks, both of you!! 

The prime factorization of 300 is \(2 \times 2 \times 3 \times 5 \times 5\). 

There are \({5! \over 2! 2!} = \color{brown}\boxed{30}\) ways to order these. 

There are 24 prime numbers between 30 and 140 and multiplied together should give you:

[24  x  25] / 2 ==300 integers - each with 2 prime factors between 30 and 140.

"An equilateral triangle is a triangle whose sides are all equal. It is a specific kind of isosceles"

14 C 3 = 364

15 c 3 = 455

You need \({n \choose 3} \ge 365\)

The smallest number that satisfies this is \(\color{brown}\boxed{15}\)

Isn't an equilateral triangle also considered isoceles? 

If you agree with me that it is, then you can add 10-10-10 to the list.  

_.

you forgot 10 10 10      (eqiuilateral triangles are isosceles triangles)

See if this helps:

https://www.mathsisfun.com/geometry/triangle-inequality-theorem.html

Let's count them: 8-8-14, 9-9-12, 10-10-10 11-11-8, 12-12-6, 13-13-4, 14-14-2

In total there are \(\color{brown}\boxed{7}\)

Thanks guys.....another slightly different approach

We can subtract the second term from the right side from both sides  and we get

(1) / (x + 5)  =(x -4) /(2x)

Cross-multiply

2x = (x + 5) ( x - 4)

2x = x^2 + x -20

x^2  -x - 20   =   0

Factor

(x -5) (x + 4)   = 0

Set each factor to 0 and solve for x and we  get  that   x= 5  or x  = -4

oh sorry D:

I run a book club with n people, not including myself. Every day, for 365  days, I invite three members in the club to review a book. What is the smallest positive integer n  so that I can avoid ever having the exact same group of three members over all   365 days?
 

deleted

\(\frac{\left(x+4\right)}{\left(x+5\right)}=\frac{\left(x-4\right)}{\left(2x\right)}+\frac{\left(x+3\right)}{\left(x+5\right)}, 2x\left(x+4\right)=\left(x-4\right)\left(x+5\right)+2x\left(x+3\right), x=5, x=-4\)

\(\frac{\left(x+4\right)}{\left(x+5\right)}=\frac{\left(x-4\right)}{\left(2x\right)}+\frac{\left(x+3\right)}{\left(x+5\right)}\)

Multiply by LCM :

\(2x\left(x+4\right)=\left(x-4\right)\left(x+5\right)+2x\left(x+3\right)\)

\(2x^2+8x=3x^2+7x-20\)

Using quadratic formula to solve, 

\(x_1=\frac{-\left(-1\right)+9}{2\cdot \:1},\:x_2=\frac{-\left(-1\right)-9}{2\cdot \:1}\)

\(x = 5, -4\)

Sum of an infinite geometric  series  =     first term  /   [ 1  - common  ratio ]

The common ratio   =    5/a

So

4 / [ 1 - 5/a ]    =

4 / [ (a-5) / a ]    =

4a / [ a -5 ]

The smallest  positive a that makes the sum a perfect square is when a = 9 

4(9)  / [ 9 -5]  =  36 / 4  =   9

Whoops, thx for catching that @qjin27

https://web2.0calc.com/questions/sequence_49

Let point E be the midpoint of CD. 

Now, \(DE = EC =10\). 

We know that \(AB = OE = 12\)

Applying the Pythagorean Theorem to \(\triangle OEC\), we find the hypotenuse, or in this case, the radius to be \(\color{brown}\boxed{2 \sqrt{61}}\)

uh I think it goes to -2010, not 2007

You eventually get:

-1 + -1 + -1 + 2007(for 1005 terms)

So 

-1005 

(1 - 2) + (3 - 4) + (5 - 6) + ... + (2003 - 2004) + (2005 - 2006) + 2007

= -1+(-1)+(-1) (1005 times)

=  -1005

\(\begin{bmatrix}\left(\frac{s}{2}\right)-3t=-1\\ 3t-2s=3+s-t\end{bmatrix}\)

\(\mathrm{Substitute\:}s=-2+6t\)

\(\begin{bmatrix}3t-2\left(-2+6t\right)=3-2+6t-t\end{bmatrix}\)

\(\begin{bmatrix}-9t+4=5t+1\end{bmatrix}\)

\(t=3/14\)

\(\mathrm{Substitute\:} t\)

\(s=-2+6\cdot \frac{3}{14}\)

\(s=-5/7\)

\(=\)\((-5/7, 3/14)\)