Answer:
10
Step-by-step explanation:
You do 5C3, which is 5!/((5-3)!3!) -->
5*4*3*2*1/(2*1*3*2*1) =
10
Step-by-step explanation:
thanks you too merry Christmas
You say that there is uniform acceleration so:
vf-vi=at (final velocity minus initial velocity is equal to acceleration times time)
We know vf, vi, and t so we can solve for acceleration:
24-12=a10
12=10a
a=1.2
That is the acceleration, we will need to integrate with respect to time twice...
v=⌠a dt
v=at+vi , we know a=1.2m/s^2 and vi=12m/s
v=1.2t+12,
x=⌠1.2t+12 dt
x=1.2t^2/2+12t+xo, we can just let xo=0 for this problem...
x(t)=0.6t^2+12t
Now we know that this acceleration lasts for 10 seconds so the distance traveled in that time is:
x(10)=0.6(10^2)+12(10)
x(10)=60+120
x(10)=180 meters
well, is noteworthy that an x-intercept is when y = 0 or namely is a solution or root of the quadratic, so we know then that the x-intercepts or solutions are at (-1,0) and (3,0), that simply means that
![\bf -8=a(2)(-2)\implies -8=-4a\implies \cfrac{-8}{-4}=a\implies \boxed{2=a} \\\\[-0.35em] ~\dotfill\\\\ y=2(x+1)(x-3)\implies y=2(\stackrel{\mathbb{FOIL}}{x^2-2x-3})\implies y=2x^2-4x-6](https://tex.z-dn.net/?f=%5Cbf%20-8%3Da%282%29%28-2%29%5Cimplies%20-8%3D-4a%5Cimplies%20%5Ccfrac%7B-8%7D%7B-4%7D%3Da%5Cimplies%20%5Cboxed%7B2%3Da%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20y%3D2%28x%2B1%29%28x-3%29%5Cimplies%20y%3D2%28%5Cstackrel%7B%5Cmathbb%7BFOIL%7D%7D%7Bx%5E2-2x-3%7D%29%5Cimplies%20y%3D2x%5E2-4x-6)
This is a quadratic equation because it has degree 2.
As known as a parabola
