NotThatSmart

Alright. First, let's combine all like terms to this horrible mess. We get that

$6x^{2}-21x-31=0$

Now, we simply apply the quadratic formula and take the smaller root. Using the fomrula, we get

$x=\frac{21\pm \sqrt{(-21)^{2}-4\cdot 6(-31)}}{2\cdot 6}$

From this monstrosity, we get that

$x=\frac{\sqrt{1185}+21}{12}\\ x=\frac{-\sqrt{1185}+21}{12}$

Obviously, the first term of x is bigger, so we take that as our answer. 

We find that x is $x=\frac{\sqrt{1185}+21}{12}$

So our answer is just the biiger root. 

Thanks! :)

also, 500th answer...woop woop! :)

First, let's note something important, 

The descriminant of the quadratic is $b^2-4ac$ for any quadratic in the form of $ax^2+bx+c$

Now, combining like terms and reaaranging terms, we get

$3x^{2}+4x+10$

Thus, plugging in the numbers for the descriminant, we get

$4^2-4(10)(3)\\ 16-120\\ -104$

So -104 is our answer. 

Thanks! :)

Let's first take a look at the first equation. We want to put x in terms of y. 

So, solving the equation, we have

$x+y=10-x-y\\ x=5-y$

Now, subsituting this into the second equation, we get that

$5-y-y=\left(5-y-1\right)+y-1\\$

This simplifies to 

$5-2y=3$

Now, for this equation, we get that y is equal to 1. 

Thus, subsituting this value into x, we find that

$x=5-1=4$

Thus, the answers for the equation are

$x=4,\:y=1$

So our final answer is $(4, 1)$

Thanks! :)

First, let's note that 1/17 is a repeating decimal. 

We have that $1/17 = 0.\overline{0588235294117647}$

There are 16 repeating digits. 

Thus, we do 4000/16. We get

$4000/16=250$

Since it's divisible, it means the last digit of the repeating part is our answer, which is 7. 

So 7 is our answer. 

Thanks! :)

First, let's complete the square for both x and y to form the equation of a circle. We get

$\left(x-5\right)^2+\left(y-\left(-4\right)\right)^2=\left(4\sqrt{3}\right)^2$

This givies us a cricle with center of (5, -4) and a radius of $4\sqrt3$

Therefore, the area, which can be written as $\pi r^2$, we have

$(4\sqrt3)^2 \pi = 16(3) \pi = 48 \pi$

So 48pi is our amswer. 

Thanks! :)

Nice...Quite smart tactic here, CPhill. 

I loved the part where you did

$1/a^3 + 1/b^3 = (a^3 + b^3) / (ab)^3 \\ \frac{(a + b) (a^2 + b^2 - ab)}{(ab)^3}$

This is a very efficient way to complete the problem. 

Lol, I just brute forced it because I got bored...

At least we got that same answer!

~NTS

First, let's cobine all like terms and solve for x. Combing all like terms, we have

$x^{2}+6x-2=0$

Using the quadratic equation, we find that

$x=\sqrt{11}-3\\ x=-\sqrt{11}-3$

Now, it's time for the more tedious part of this problem. This is not efficient but it should work. 

We have the equation

$\frac{1}{\left(\sqrt{11}-3\right)^3}+\frac{1}{\left(-\sqrt{11}-3\right)^3}$

Taking the LCM on both sides to get a common denominator, we get that

$\frac{-126-38\sqrt{11}}{\left(\sqrt{11}-3\right)^3\left(-3-\sqrt{11}\right)^3}+\frac{38\sqrt{11}-126}{\left(\sqrt{11}-3\right)^3\left(-3-\sqrt{11}\right)^3}=\frac{-252}{\left(\sqrt{11}-3\right)^3\left(-3-\sqrt{11}\right)^3}$

Now, we simply have to simplfy the denominator of the two equatins. We get that

$\left(\left(\sqrt{11}-3\right)\left(-3-\sqrt{11}\right)\right)^3=(-2)^3 = -8$

Thus, we have the answer as -252/-8 = 31.5

So 31.5 is our answer. 

Thanks! :)

This is a tough questipn! It takes some thought to solve! Good one!

Now, let's note something really important first. 

The area of the koch flake can be written in the form

where a_0 was the original area and n was the iteration number. 

The limit of the area in terms of sidelength s is

Thus, plugging in 2, we have

$\frac{8\sqrt3}{5}$

Now, we square it and see what we get. We have

$\frac{64(3)}{25} = \frac{192}{25}$

So $p=192; q=25$

Thus, $p+q = 25+192 = 217$

Hmmm, I'm not sure if I'm correct. Can you explain your thought process. I think I misunderstood my own logic...

I am pretty sure I'm wrong...ummm, is there a way you can check?

Thanks! :)

~NTS

First, let's note something really important. 

If $\angle XWZ = 60°$, then we must have that $\angle WXY = 180° - 60° = 120°$ since

XY and WZ are parralel lines. 

However, this also means that $\angle ZXW$ must be less than 120 degrees. 

In the problem though, it states that $\angle ZXW = 120^\circ$

Therefore, this question is invalid/ 

This problem is not possible or solvable. 

Thanks! :)

We can write a handy equation to solve this problem. 

First, let's set a variable. Let's set x as a positive integer. 

Any number congruent to 7 modulo 29 can be written in the form

$29x+7$ since it is 7 greater than a factor of 29. 

thus, we can write the inequality

$29x+7\leq 1000$

Setting the two to equal and solving for x, we have

$29x+7=1000\\ 29x=993\\ x \approx 34.24$

Rounding down, we find x is 34. 

So our answer is 34. 

Thanks! :)