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! :)