AnxiousLlama

Assuming that you wrote \(8- 2(\frac{5}{9})\), the answer is \(6(\frac{8}{9})\).

Hope that helped! 

AnxiousLlama

Heyy! In scenarios like this, the \(a\) coefficient isn't \(1\), so I would either factor it using decomposition or use the quadratic formula.

here's a article on decomposition (https://tentotwelvemath.com/fmp-10/factors-and-products/factoring-a-trinomial-using-decomposition/)

here's a article on the quadratic formula (https://www.purplemath.com/modules/quadform.htm)

Hoped that helped! 

AnxiousLlama

We can setup an equation:

Let \(x\) be Austin's age. 

We can make this equation: \(x -3 = \frac{1}{3} * (x+15)\) , which simplifies to \(x=12\). 

Austin is 12 years old. 

Hope this helps! 

AnxiousLlama

Oh! Sorry for the misunderstanding! 

1. \(x^2+4x+3 = (x+1)(x+3) \)

2. Cannot be simplified further

3. \(x^2 + x = x(x+1) \)

4. \(2x^2-3x+1 = \left(2x-1\right)\left(x-1\right)\)

5. \(\frac{4\left(x+6\right)}{x-2}\cdot \frac{x+10}{24\left(x+6\right)} = \frac{x+10}{6\left(x-2\right)}\)

Hope this helped!  

AnxiousLlama

\(\frac{2}{9}*\frac{4}{5}*5 = \frac{8}{9}\)

Complementary counting! 

Here is a massive hint! 

Any 2-digit number with a 3 is peachy (and all peachy numbers have at least one 3). There are nine 2-digit numbers with ones digit 3, and there are nine additional 2-digit numbers with tens digit 3 (except from 33, which we've already counted, since its ones digits is 3). Therefore, there are . . .

Got it. Thank you so much guys!