Imcool

(x^2 - 2x - 5) f(x) = 2x^4 + 19x^3 + kx^2 - 15x - 1, what does this mean?

1. -4/3, I just graphed it in desmos and fiddled with some stuff.

2. So the distance of the Center, (x,y), to B and A are the same.

We know that y is 5 from tangent rules.

(x,5)

The distance from (x,5) to (9,0) is x, since to B is also x

\(\sqrt{(9-x)^2+25}=x\)

\((9-x)^2+25=x^2\)

\(x^2-18x+106=x^2\)

\(x = \frac{53}{9}\)

\((53/9,5)\)

We set the 2 colors as x and y. You can have xxx,yyy,xxy,yyx. Notice that all other combinations of xxy and yyx can be repeated by just rotating the triangle.

The final answer will be \(\fbox{4}\)

x = 2,4,6,8,10

If x = 10,

u,v,w,y,z = 0

1 CHOICE

If x = 8,

     If there is one 2:

     5 CHOICE

     If there is two 1:

     10 CHOICE

15 CHOICE

You keep doing this for all possible combinations, for example x=6, you can do 1111,112,22,13,4

\(f(x) + b \cdot g(x)\)

\(f(x) = 2x^5 - 6x^4 - 4x^3 + 12x^2 + 7x - 5\)

\(g(x) = x^{15} - 3x^{14} - 2x^{13} - 6x^{12} + 14x - 10\)

I believe it is degree one. Since if we set \(x^{10}=2\), than all the x's cancel out except for x.

\(f(x) = \frac{cx}{(4x-5)(3x+2)}\)

\(f(f(x)) = f(\frac{cx}{(4x-5)(3x+2)})\)

\( f(\frac{cx}{(4x-5)(3x+2)}) = \frac{c(\frac{cx}{(4x-5)(3x+2)})}{(4(\frac{cx}{(4x-5)(3x+2)})-5)(3(\frac{cx}{(4x-5)(3x+2)})+2)}\)

\( x = \frac{c(\frac{cx}{(4x-5)(3x+2)})}{(4(\frac{cx}{(4x-5)(3x+2)})-5)(3(\frac{cx}{(4x-5)(3x+2)})+2)}\)

\(a = \frac{cx}{(4x-5)(3x+2)}\)

\( x = \frac{ca}{(4a-5)(3a+2)}\)

Blahblahblah, you are eventually gonna end up with c = (4x-5)(3x+2), I think.. then just plug it in!

I believe that there is no answer. This is because opposite sides of quadrilaterals always add up to 180 degrees. 

\(110+55<180\)

\(\sqrt{\frac{3x + 4y}{6x + 5y + 4z}} + \sqrt{\frac{y + 2z}{5x + 6y + 4z}} + \sqrt{\frac{2z + 3x}{x + 2y + 3z}}\)

So, largest value of that.

\(u=3x+4y,v=y+2z,w=3x+2z\)

we have

\(\sqrt{\frac{u}{u+v+w}}+\sqrt{\frac{v}{u+v+w}}+\sqrt{\frac{w}{u+v+w}}\)

and also

\(\sqrt{\frac{u}{u+v+w}}+\sqrt{\frac{v}{u+v+w}}+\sqrt{\frac{w}{u+v+w}}\le\sqrt{3}\)

You can google why. I can't write it in latex since idk how.

Answer: \({\sqrt{3}}\)

Just as a side note, I'm not criticizing, but most of the time Vxrtate, you answer the question by rationalizing the denominator. In other words, you should put \(\frac{8\sqrt{3}}{3}\)unless it specifically states otherwise.

The answer is 9

In trapezoid PQRS, \(\overline{PQ} \parallel \overline{RS}\).  Let X be the intersection of diagonals \(\overline{PR},\overline{QS}\).  The area of triangle PQX is 2 and the area of triangle RSX is 2.  Find the area of trapezoid PQRS.

There are many solutions. The area could be 9, it also could be 12.5, ETC. Not enought info, I guess.

1/3.

Check out this similar question.

In triangle PQR, let X be the intersection of the angle bisector of \(\angle P\) with side \(\overline{QR}\), and let Y be the foot of the perpendicular from X to side \(\overline{PR}\).  If PQ = 8, QR = 5, and PR = 1, then compute the length of \(\overline{XY}\).

Ookie Dookie

No answer. Simply logic

Let me repeat the question, and you try to figure out whats wrong.

PQ = 8, QR = 5, PR = 1.

2 sides of a triangle has to add up to more than the third side, but 

\(5+1<8\)

so this triangle is not even possible in the first place.

Find all points (x,y) that are 5 units away from the point (2,7) and that lie on the line y = 5x - 28

If we plug it into desmos and draw a circle of \(\left(x-2\right)^{2}+\left(y-7\right)^{2}=25\), then we can see all the points of intersection, and there are 2.

\((7,7)\)

\((6.615,5.077)\) That is not the value, it is just an approximation.

\(y= 1/5\cdot f(5x + 5) - 5\)

\(y= 1/5\cdot f(5x + 5) - 5, x = -2/5\)

\(y= 1/5\cdot 4- 5\)

\(y = 6-5\)

\(y = 1\)

The point will be \((-2/5,1)\)

Check out this solution from Heureka!

Amazing solution by the way.

CPhill if you are here do you mind checking my work? I think its wrong but I think I used the right steps..