Imcool

What is the blank after the word 'Let'? That is evidence that you have copy and pasted this from somewhere. If you want us to do your AoPS homework or schoolwork, can you at least make sure the people can actually understand the question? If you repost this fixed, I will gladly answer.

Note: Im assuming that all the students have to take a class, any moderators please correct me if my assumption is wrong, I just assumed becuase otherwise I thought the problem was unsolvable.

The number of students in only math club is \(a-b\).

Therefore, the number of students in science club is \(c - (a - b) = c-a+b\)

The number of people in only science club is \(c-a\)

The number of students that are only in 1 course is \(c-b\)

Please take the time to put your answer in Latex. It helps a lot.

\(f(x) = \frac{cx}{4x - 5 - 2x + 8}\)

\( x\neq -\frac{3}{2}\)

Im going to try and brute force it, but its going to be really ugly.

\(f(f(x)) = \frac{c(\frac{cx}{4x - 5 - 2x + 8})}{4(\frac{cx}{4x - 5 - 2x + 8}) - 5 - 2(\frac{cx}{4x - 5 - 2x + 8}) + 8}\)

\(f(f(x)) = \frac{c(\frac{cx}{4x - 5 - 2x + 8})}{2(\frac{cx}{4x - 5 - 2x + 8}) +3}\)

\(x = {\frac{c^2x}{2x+3}}\cdot{\frac{2x+3}{2cx+6x+9} }\)

\(x = {\frac{c^2x}{2cx+6x+9} }\)

I think you can take it from here. Good Luck!

Im just going to brute force it. You may find other answers.

\(f(f(f(x))) = 64x - 105 - 63x + 278\)

\(f(f(px+q)) = 64x - 105 - 63x + 278\)

\(f(p^2x+pq+q) = 64x - 105 - 63x + 278\)

\(p^3x+p^2q+pq+q = 64x - 105 - 63x + 278\)

From this, we get that 

\(p^3 = 64 - 63 = 1\)

\(p = 1\)

Therefore, 

\(3q = 173\)

\(q = 57\frac{2}{3}\)

So, \(p+q=58\frac{2}{3}\)

We set the smaller base to x, so the area of the trapezoid is

\(12x+24 = 80\)

\(12x = 56\)

\(x = \frac{19}{3}\)

Next, we find the diagonals by using pythagorean theorum.

a = 2

b = 12

c = \(\sqrt{148}\) = \(2\sqrt{39}\)

so the perimeter is 2x+4+2c = 38/3 + 4 + 4 root 39, or \(\frac{50}{3}+4\sqrt{39}\)

\(2x^2 - 8x + 7 = 5x^2 - 13x + 1\)

\(3x^2 - 5x -6=0\)

\(a = \frac{5+\sqrt{97}}{6} ,b = \frac{5-\sqrt{97}}{6}\)

You can take it from here.

 Hope this helped,

 IC

Let a and b be the solutions to \(5x^2 - 11x + 4 = 4x^2 + 11x + 13\). Find \(\frac{1}{a^3} + \frac{1}{b^3}\).

For this question, I'm just going to solve for a and b, since it is not that hard.

You can use the quadratic formula, which is \(\frac{-b+or-\sqrt{b^2-4ac}}{2a}\). The +or- just stands for plus or minus. Any moderators tell me what the symbol for that is? I kind of forgot. Anyways, by plugging in the formula to the equation, first we got to simplify it.

\(5x^2 - 11x + 4 = 4x^2 + 11x + 13\)

\(x^2 - 22x -9 =0\)

\(x = \sqrt{130}+11,-\sqrt{130}+11\)

You can take it from here.

PS: I think you made a mistake, this answer looks kinda weird. Maybe it wasn't +11x it was -11x, so it would be a perfect square and easier to solve.

Well, to make it divisible by 8, the first 3 digits have to be divisible by 8. Since we know it is a palindrome, it is in the format \(abba\). We also know that \(bba\) is divisible by 8. The total number of these are 008, 112, 224, 336, 440, 448, 552, 664, 776, 880, 888, 992, which is 12 total posssible combinations.

Hoped this helped, IC

Please include a picture.

All the sides of a triangle have integer length. The perimeter of the triangle is 5 and the triangle is isosceles. How many such non-congruent triangles are there?

Well, we know that 2 sides of the triangle are the same, since it is iscoceles. that means that the 2 sides can either be 2 or 1. But, we can rule out 1, since 2 sides of a triangle musts add up to be more than the 3rd side, but if the iscoceles sides are each 1, then 1+1<3, so that is not true. That leaves only one option, which is that iscoceles sides are 2, and the other side is 1.