AnxiousLlama

Heyy! Happy birthday cal! Sorry i joined the party late! 

@Batterydude6000, I can tell you are taking an AoPS class, so I won't be sharing the answers but I'll give hints.

1 - We transform the equation into \(x = y^2 - 8y + 13\). We can then complete the square in \(y\), which turns the equation into \(x = (y-4)^2 - 3\). See if you can solve yourself from here. 

2 - Again, we can complete the square to get \(y = -2(x^2-4x)-15\). This simplifies to \(y = -2(x - 2)^2 - 7\). Try solving it from here.

3 - To complete the square, we can add \(6y \) and subtract \(2x\), then add \(1 \) to complete the square for \(x\) and add \(9\) to complete the square for \(y\). It should be pretty simple from here.

4 - Since the axis of symmetry is \(x = 4\), we can substitute that into \(y = a(x-h)^2+k\). 

5 - The simplified equation of \(y = a(x-h)^2+k\) is \(y = a(x + 3)^2 - 2\). We can also tell that the parabola passes through the point \((-1,0)\), so \(x = -1, ~ y = 0\). Try substituting that in the equation and sees where that gets you.

Hope that helped. 

AnxiousLlama

Guest, there are \(3\) answers for your problem because it's asking you how many of each flowering plant he should buy.

Remember these definitions:

Theoretical Probability: This is what is expected to happen. 

Experimental Probability: This is what actually happened in an experiment.

(A) - Since a number cube has \(6 \) even sides, the theoretical probability of rolling a side (\(3\) in this case,) is \(\frac{1}{6}\).

(B) - If the number \(3\) came up \(67\) times out of the \(450\) times rolled, then the experimental probability is \(\frac{67}{450}\).

Hi, guest! I'm answering right now! It might take some time. . . 

Yes, I think you are correct because \(\frac{1}{4}\)'s of the letters are a. This isn't as hard as combinatorics or permutations, like when you have to choose multiple objects or find out how many total ways to arrange them, etc. ... 

AnxiousLlama 

1. If you add all the values up together, you would get \(21\). Divide that by the number of days and you would get your mean (average): \(\frac{21}{5} = \) 4.2

2. Just count which number of cookies Grogg ate most in a day; the mode would be 1. 

You are correct on all of them!

Hope that helped,

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!