Alan

Write this as:

\(y=(x-9)(x-3)(x+8)\)

Expand the brackets to get:

\(y=x^3-4x^2-69x+216\)

I get the following solutions (one of us has probably typed in something incorrectly!):

.

Expand the brackets in f(x)=a(x-h)²+k    to get   f(x) = a*x² - 2a*h*x + a*h² + k  

Compare this with f(x) = a*x² + b*x + c  term by term

x²    Both are the same

x     We must have b = -2*a*h    so   h = -b/(2a)

constant    We must have c = a*h² + k   so  k = c - a*h²  or  k = c - b²/(4a)

Call the number N

N/4  -  5 = 15   so  N/4  =  20   so  N = 4*20  or  N = 80

Neither sin nor cos can have values beyond the range -1 to + 1.

Presumably the applied force matches the component of gravitational force along the slope (the ice suggests zero friction).

100 = Mass*g*sin(25°)    (g = 32.2 ft/sec/sec)

Rearrange to find the mass.

This is known as the harmonic series and the partial sum is given by:

\(\Sigma_{k=1}^n\frac{1}{k}=\gamma+\psi_0(n+1)\)

where \(\gamma\)  is the Euler-Mascheroni constant (0.5772....)

and \(\psi_0(x)\)  is the digamma function

(see http://mathworld.wolfram.com/HarmonicSeries.html)

This is far from elementary mathematics!!

tan(pi/4) = 1

sec(pi/4) = sqrt(2)

tan(0) = 0

sec(0) = 1

You can finish it from here (use the calculator on the home page).

"Here how did you get 3^2  ? it must be ( 5^2 - 3^2 ) according to ( Sf^2 - Si^2 ) ! ​ "

Sf and Si are the distances from the unstretched position.  So Sf = 2-5 (i.e. 3)  and Si = 5-5 (i.e. 0)

I get the following: