NotThatSmart

First, let's simplify the equation a bit. We get

\(x^2 - 16x + y^2 + 22y = 86\)

Completing the square for BOTH x and y, we get the equation

\(x^2 - 16x + 64 + y^2 + 22y + 121 = 86 + 64 + 121 \\ (x - 8)^2 + (y + 11)^2 = 271\)

This is a  circle centered at (8, - 11) with r^2  = 217

Now, we can find the area. We simply have

\(Area = pi * r^2 = 271 pi ≈ 851.4\)

Thanks! :)

doeansodc has the right idea, but just mae a careless error. 

He simplified it correctly to get \(x^2+6x>16-x+14+x^2 \)

Now, canceling out the x^2, we get

\(6x>16-x+14 \)

Isolating x, we have

\(7x > 30\\ x > 30/7\)

Thus, in interval notation, we have

\((30/7, \infty)\)

Thanks! :)

First, let's simplify the right hand side of the equation a little bit. 

We get that \(-9t+12-4t+23+4t^2-1=-13t+4t^2+34\)

Now, we simply move all terms to one side. Note that most terms cancel out. 

we get the equation

\(13t-34\le \:0\)

Solving for t, we get

\(13t\le \:34\\t\le \:\frac{34}{13}\)

In interval notation, this is

\((-inf, \frac{34}{13}]\)

Thanks! :)

We can accomplish this problem first by Multiplying by (z + 1)^2 on both sides. Then,we solve for x. 

\((3 - z)(z + 1) \geq 2(z + 4)(z + 1)^2\text{ and }z + 1 \neq 0\\ (z + 1)(2(z + 4)(z + 1) -(3-z)) \leq 0\text{ and }z + 1 \neq 0\\ (z + 1)(2z^2 +11z+5)\leq 0 \text{ and }z + 1 \neq 0\\ (z+1)(2z+1)(z+5)\leq 0 \text{ and }z + 1 \neq 0\\ z \leq -5 \text{ or }-1 < z \leq -\dfrac12\)

Writing this in interval notation, we get

\(z \in (-\infty, -5] \cup \left(-1, -\dfrac12\right]\)\

This might be wrong...but it's probably correct LOL :)

Thanks! :)

Ok! This one is pretty easy! 

To make this problem way easier, we have a graph!

Anyways, we have triangles in this graph!

First off, let's note that the yellow part is the part that would satisfy the statements made in the problem. 

Now, we can note we have a lot of triangles. There are a few ays we could fo this, but I like to note that we have multiple 30-60-90 triangles. Each unshaded region is made up of 2, for more a more specific answer. 

Now, we can easily tell that there are 18 30-60-90 triangles in this graph. (I counted them all)

If we count the triangles only in the shaded region, we can count 12 of them!

Meaning, we just have 

Thanks! :) :) :)

We can easily use variables to accomplish this question. 

Suppose the percent of the remainder of games needed to win is x%. 

We have the equation

\(60\% \cdot 30\% + (1 - 60\%) x\% = 50\%\\ 0.18 + 0.004x = 0.5\\ x = 80\)

Thus, our answer is 80%. 

Thanks! :)

*800th answer