Imcool

call the distance in miles "d". Call the time in minutes "t".  speed = d/t

40 mph = d/t

55 mph = d/(t-25)

can you solve the 2 equation 2 variable? 

Then the answer will be the d, because it is asking for the distance.

You can refer to the above question as a hint, but ill leave the solution below.

consider the psychology books as one unit. so the total ways is 5!/4!, 5! because there are 5 different units, and 4! because the math books are indistinguishable. (Im assuming), so the answer is 5

You can count the 4 freshmans as one group, call it x. so the total ways to arrange 3 sophomores and x is 4! which is 24. Multiply by 4! because the freshmans can stand in different orders next to each other, 4! different ways. so 4!x4!=576

(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. 

\(110+55<180\)