NotThatSmart

I'm sure it was a coincidence, but the problem posted before this was the exact same!

First, we want to get rid of the deniminators. We can do this by multiplying both sides by the LCM of the denominators. 

The LCM is (x-5)(x-2). 

We have

\(x\left(x-2\right)=2\left(x-5\right)+2x\left(x-5\right)\left(x-2\right)\)

We now get a quadratic. 

We get 

\(x^2-2x=2x^3-14x^2+22x-10\\ 2x^3-15x^2+24x-10=0\)

This where it gets tricky. There are no integer roots for this equation. 

We could use the Newton-Raphson equation, and we eventually get

\(x\approx \:0.67798\dots ,\:x\approx \:1.34697\dots ,\:x\approx \:5.47503\dots \)

I hope this was clear enough!

Thanks! :)

First, let's combine all like terms. We get

\(-15x^2-kx+45\)

In order for the solutions not to be real, the descriminant must be less than 0 (negative number).

We have 

\((-15)^2 - 4(-k)(45) < 0 \\ 225+180k<0\\ k<-1.25\)

Therefore, the largest integer must be -2. 

Thanks! :)

First, let's combine all like terms to get a real quadratic. 

We get \(x^{2}+8x+9\).

The descriminant of the quadratic is \(b^2-4ac\) for those in \(ax^2+bx+c\)

We have \(8^2-4(1)(9) = 64-36=28\)

So 28 is our answer!

Thanks! :)

We can easily set up an equation to do this problem. 

We can do

\(\Delta EFG \cdot \frac{2h}{3}=18\\ \Delta EFG =\frac{3\cdot 18}{2h}\\ V_{EFGH}=\Delta EFG\cdot \frac{1}{3}h\\ [EFGH]=\frac{3\cdot 18}{2h}\cdot \frac{1}{3}h\\ [EFGH]=9\)

So 9 is our answer!

Thanks! :)

Let's first notice something really important. 

The list \(P_1 P_2 + P_2 P_3 + P_3 P_4 + \dots + P_9 P_{10} + P_{10} P_1\) are just the sides of a regular decagon!

The sidelength of one side the decagon can be represented as \(\frac{\text{radius}}{2 ( -1 + \sqrt{5})}\)

Plugging in all the values we know, we have

\(10 (1/2) ( -1 + \sqrt{5}) = 5 ( -1 + \sqrt{ 5}) ≈ 6.18\)

So about 6.18 is our answer!

Thanks! :)

Let's first move all terms to one side of the equation and combine all like terms.

We get

\(x^2 - 21x + y^2 + 13y = 25\)

Now, let's complete the square for x and y on the left side the equation. 

\(x^2 - 21x + 441/4 + y^2 + 13y + 169/4 = 25 + 441/4 + 169/4\)

\((x - 21/2)^2 + (y + 13/2)^2 = 355 / 2\)

This is the equation for a circle. 

According to the circle rules, this circle has a center at \((21/2 , -13/2)\)and radius \(\sqrt{355 / 2}\)

The largest possible value is basically the radius added onto the x coordinate of the center. 

We have \(x = (21/2) + \sqrt{355 / 2}\)

Thanks! :)

Let's note something important really quickly. 

If we have \(XWZ = 60°\), then we must have \( WXY = 180° - 60° = 120°\) since XY and WZ are parralels. 

\(\angle ZXW\) has to be less than 120 degrees, but in the problem, it clearly states it is 120 degrees. 

This is an invalid statement, so therefore, this problem is not solvable. 

Thanks! :)

We want to put \(\log_{8a} b\) in terms of a and b. 

It's actually not that hard. 

We just have

\(\frac{ log(b)}{ log (8a)} = \frac{log( b)}{ log (8) + log (a) }\)

That's the best I could do.

Thanks! :)