BuilderBoi

Subtract \({x - 5 \over x + 5} \) from both sides: \({9 \over x + 5} = {x-5 \over 2x}\)

Cross multiply: \(18x = x^2 - 25\) 

Subtract 18x from both sides: \(x^2 - 18x - 25 = 0\)

Use the quadratic formula: \(x = \color{brown}\boxed{{9 \pm \sqrt{106} }}\)

Using the distance formula yields: \(5 = \sqrt{({6 - 1})^2 + ({2 - b})^2}\). 

Simplifying this equation gives us \(25 = 25 + (2-b)^2\), meaning \((2-b)^2 = 0\). 

Can you take it from here?

 Point M is 12 units to the right of Point A, so it must be 12 units to the left of Point B. 

Can you take it from here?

Start with the fraction \({n \over 3}\). This is only terminating when n is a multiple of 3. 

Now, look at the \({n \over 1375}\) fraction. This is only terminating when n is a multiple of 11. 

So, every \(\text{lcm}(3,11) = 33\) integers, n will terminate. 

Thus, there are \(\lfloor {1000 \over 33} \rfloor = \color{brown}\boxed{30}\) integers that work. 

https://web2.0calc.com/questions/polynomial_99864

Original pizza: \(4 ^2 \pi = 16 \pi\)

New pizza: \(6^2 \pi = 36 \pi\)

So, \(36 \div 16 = 2.25 = 225\text{%}\). But, recall that we are looking for the increase, so we subtract 100, to find that it is a \(\color{brown}\boxed{125 \text{%}} \) increase. 

Try it for yourself, and see what you get 

By the Binomial Theorem, we have the system:

\(a \times n = -20\)

\({n \choose 2} \times a^2 = 180\)

Squaring the first equation gives us \(a^2n^2 = 400\), meaning \({n^2 \over {n \choose 2}} = {20 \over 9}\). Using trial and error, we find \(n = 10\), meaning \(a = -2\)

Now, we have \(c = {10 \choose 3} \times 1 \times -2x^3 = \color{brown}\boxed{-960}\)

\(\color{brown}\boxed{99}\)

No worries!!!