proyaop

The entries in row n of Pascal's triangle are 1, n,..., n, 1. Additionally, the sum of row n's elements is 2^n. 

Additionally, row n has n + 1 terms, so we know from the problem that \({2^n\over{n+1}}=2\). 

Cross multiplying, you get \(2^n=2n+2\). The question here is trivial by bash => you will find that n = 3 works. (If you are a quick guess + checker, you might've found that right of the bat quickly, as row 3 is 1, 3, 3, 1 which has an average entry of 2.)

Determine all of the following for \(f(x)\cdot{g(x)}\), where \(f(x) = -x^2+8x-5\), and \(g(x) = x^3 - 11x^2 + 2x\).

\(f(x)\cdot{g(x)}=(-x^2+8x-5)(x^3-11x^2+2x)=-x^5+19x^4-95x^3+71x^2-10x\)

Obviously, you don't need to actually expand the whole thing... you can just expand certain parts of the trinomial multiplication.

But from the expression above, I leave you to find the 5 terms!

f(x) = 5x^2 + 11x after combining like terms. 

f(x) = f(k - x) = 5(k - x)^2 + 11(k - x) = 5k^2 - 10kx + 5x^2 + 11k - 11x = 5x^2 + 11x

5k^2 + 11k = 22x + 10kx for all real x. 

\(5k^2+11k=x(10k+22)\)

Our goal is to get an equation that is true without variables, generally 0 = 0, to show that any value of x will not affect the output.

We can see that by setting 10k + 22 = 0, k = -11/5, we get k(5k + 11) = x(0) => k(0) = x(0), which is a true equation.

So, the answer is k = -11/5

The coordinates of the intersection points happen when f(x) = g(x):

\(x^3+x^2-3x+5=x^4-7x^3+5x^2-18x+17\)

\(x^4-8x^3+4x^2-15x+12=0\)

This is one monster of an equation... you can use desmos.com/calculator to graph it, but you should get x values = 0.747 and 7.707, plugging those back into the original equations, your points of intersection are: (0.747, 3.734), (7.707, 499.116)

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This diagram is not to scale and misleading. Because ABCD is a square, PR is collinear with NL. NL also bisects AM and OC, so AP : PM = 1 : 1, and thus the center of the circle, say point X, XP : PL = 1 : 1, likewise XR : RN = 1 : 1. Thus, PR = NL/2 = 12/2 = 6.

For 1) and 2), the equation would be (most likely) in the form y = mx + b, unless the problem otherwise specifies so.

3) Slope of AB = rise/run = 1/-3

Slope of CD = rise/run = 1/-3 too... if C is (-2, -2), then D has to go up 1 and left 3 => (-5, -1)

(Monic) Quadratic with roots 5 and -5 can be factored as (x - 5)(x + 5) = 0. Expanding this, you get x^2 - 25 from difference of squares.

Thus, the first blank = 0, and the second blank = -25.

\(f(x) = px + q\), then for each cyclic succession (next f(x)), substitute px + q for the next x.

\(f(f(x))=p(px+q)+q=p^2x+pq+q\)

\(f(f(f(x))) = p(p(px + q) + q) + q =p(p^2x+pq+q)+q=p^3x+p^2q+pq+q\)

\(p^3x+p^2q+pq+q=64x-105-30x+70+32x-12=66x-47\)

I have done most of the work for you here, so just set \(p^3x=66x\), solve for p, and then plug that into the equation \(p^2q+pq+q=-47\) to get q, and then calculate p + q for your final answer!