xCorrosive

Combine like terms to get..

\(3x^2+6x-9\)

Then isolate the $x^2$ and $x$ terms and we get \(3(x^2+2x) - 9\). 

Then complete the square inside the parentheses to get \(3(x^2+2x+1) - 9 - 1 \cdot 3\) .

Then factor the parentheses and simplify to get \(3(x+1)^2 - 12\).

You should be able to finish from here.  

As the first term is 1, the smallest common difference between the terms is 1 because the terms are all integers.

This series would go 1, 2, 3... 89, 90, 91 ... and so on. 

\(91 \cdot (1 - \frac{4}{7}) = 91 \cdot \frac{3}{7} = 39\)

Thus, Tom had 39 cards left. 

Assuming that bags need to be completely full, we get $\dfrac{\frac{11}{2}}{\frac{1}{3}} = \frac{11}{2} \cdot 3 = 16.5$, meaning that $\boxed{16}$ bags will be filled and $\boxed{\frac{1}{6}}$ pounds will be left for the packer to eat. 

$BYX$ ~ $ABC$ by $AAA$ similarity. Because $ABC$ has $45-45-90$ angles it follows that $BYX$ has $45-45-90$ angles (and is isosceles).

The diagram below explains $45-45-90$ triangles. 

Thus, we can see that $BY = \frac{12}{\sqrt{2}} = 6\sqrt{2}$. 

Similarly, we can see that $BA = 18\sqrt{2}$ by triangle $ABC$. 

Subtracting $BC - BY = 12\sqrt{2} = AY$. 

Thus the ratio is $\frac{BY}{AY} = \frac{6\sqrt{2}}{12\sqrt{2}} = \boxed{\frac{1}{2}}$, which is our answer. 

Let $f(x)=3x+2$ and $g(x)=ax+b$, for some constants $a$ and $b$. If $ab=20$ and $f(g(x))=g(f(x))$ for $x=0,1,2...9$ find the sum of all possible values of $a$.

For nested functions, we take the interior function and substitute it into $x$ of the outer function. 

$f(g(x)) = 3ax + 3b + 2$

$g(f(x)) = 3ax + 2a + b$

$3ax + 3b + 2 = 3ax + 2a + b \rightarrow 2a = 2b+2$

Thus, we need to solve the system of equations 

$ab = 20$

$2a = 2b + 2$

We can either solve for a variable and substitute into the other equation, or we can guess and check to see that our solutions are 

$a = 5, b = 4$

$a = -4, b= -5$

Thus the sum of all possible values of $a$ is $5 - 4 = \boxed{1}$. 

We calculate, applying a similar method for $f(4)$ and $f(5)$. 

$f(3) = 3f(1) - 2f(2) = -3 - 16 = -19$

$f(4) = 24 + 38 = 62$

$f(5) = -57 + 124 = \boxed{67}$. 

$(a)$

Since each burger costs 3 times that of a cupcake, naturally it follows that $15 \cdot 3 = \boxed{45}$. 

$(b)$

Let the cost of a burger be $x$. Writing an equation gives $15x + \frac{75}{3}x = 40x$, for the number of burgers Ahmad could buy with $\frac{2}{5}$ of his total money. $\frac{3}{5} \cdot \frac{4}{5} = \frac{12}{25}$, meaning that $\frac{4}{5}$ of his remaining money is $\frac{12}{25}$ of his total money. 

We can see that $\frac{2}{5} \cdot 1.2 = \frac{12}{25}$, meaning that he brought $40 \cdot 1.2 = 48$ burgers with his remaining money. 

Adding the burgers from before together gives $15 + 48  = \boxed{63}$ burgers in total. 

Note that...

$1-0 = 1$

$4-1 = 3$

$9-4 = 5$

$16-9 = 7$

$25-16 = 9$

$...$ 

Thus, every positive odd number can be written as the difference between two perfect squares. We consider squares in the range of $1 - 99$, since $100^2 = 10,000$. An even number squared will never be odd, and vice versa an odd number squares will always be even. 

Thus, taking $\frac{99+1}{2} = \boxed{50}$, which is our answer. 

Thank you!

In the diagram below, circles represent students and bars represent the "walls" of different rooms. 

We can see that there are 4 configurations of "bars". Zero bars, one bar, two bars, and three bars, which represent different ways to separate the students respectively. 

There are 3 slots that the bars can be in, so counting the different numbers of combinations for each of the scenarios gives $1 + 3 + 3 + 1 = \boxed{8}$

Alternativly, 1 + 3 choose 1 + 3 choose 2 + 3 choose 3 = 8 also works (if anyone knows how to type Combination in LaTeX, please tell me. 

Factor out $2^2$ from each of the numbers to get...

$\boxed{2\sqrt{2-\sqrt{3}}}$

Never mind this equation seems exceedingly obvious unless there's some trick I'm missing. 

$\begin{align*} \sum^{2020}_{n=1} [\frac{1}{n^2} - \frac{1}{(n+1)^2}] = \frac{1}{1} - \frac{1}{4} + \frac{1}{4} - \frac{1}{9} + \frac{1}{9} ... -\frac{1}{2020^2} + \frac{1}{2020^2} - \frac{1}{2021^2} \end{align*} $

Thus, terms cancel and..

$$\begin{align*} 1 - \frac{1}{2021^2} \end{align*} $$

Try typing without $\LaTeX$ (the dollar marks).