Answer:
8² = 4² + 5² - 2(4)(5) cos(P)
which is the third option
Explanation:
The general rule of cosine is:
a² = b² + c² - 2bc*cos(A)
In our triangle:
a is substituted by p = 8 cm
b is substituted by n = 4 cm
c is substituted by m = 5 cm
A is substituted by P which we want to find
Replace the variables in the general equation with the givens as follows:
p² = n² + m² - 2mn*cos(P)
8² = 4² + 5² - 2(4)(5) cos(P)
Hope this helps :)
320 guests / 5 servings per can = 64 cans required
![\bf \textit{perpendicular, negative-reciprocal slope for slope}\quad \cfrac{a}{b}\\\\ slope=\cfrac{a}{{{ b}}}\qquad negative\implies -\cfrac{a}{{{ b}}}\qquad reciprocal\implies - \cfrac{{{ b}}}{a}\\\\ -------------------------------\\\\](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bperpendicular%2C%20negative-reciprocal%20slope%20for%20slope%7D%5Cquad%20%5Ccfrac%7Ba%7D%7Bb%7D%5C%5C%5C%5C%0Aslope%3D%5Ccfrac%7Ba%7D%7B%7B%7B%20b%7D%7D%7D%5Cqquad%20negative%5Cimplies%20%20-%5Ccfrac%7Ba%7D%7B%7B%7B%20b%7D%7D%7D%5Cqquad%20reciprocal%5Cimplies%20-%20%5Ccfrac%7B%7B%7B%20b%7D%7D%7D%7Ba%7D%5C%5C%5C%5C%0A-------------------------------%5C%5C%5C%5C)
![\bf \boxed{5i+12j}\implies \begin{array}{rllll} \ \textless \ 5&,&12\ \textgreater \ \\ x&&y \end{array}\quad slope=\cfrac{y}{x}\implies \cfrac{12}{5} \\\\\\ slope=\cfrac{12}{{{ 5}}}\qquad negative\implies -\cfrac{12}{{{ 5}}}\qquad reciprocal\implies - \cfrac{{{ 5}}}{12} \\\\\\ \ \textless \ 12, -5\ \textgreater \ \ or\ \ \textless \ -12,5\ \textgreater \ \implies \boxed{12i-5j\ or\ -12i+5j}](https://tex.z-dn.net/?f=%5Cbf%20%5Cboxed%7B5i%2B12j%7D%5Cimplies%20%0A%5Cbegin%7Barray%7D%7Brllll%7D%0A%5C%20%5Ctextless%20%5C%205%26%2C%2612%5C%20%5Ctextgreater%20%5C%20%5C%5C%0Ax%26%26y%0A%5Cend%7Barray%7D%5Cquad%20slope%3D%5Ccfrac%7By%7D%7Bx%7D%5Cimplies%20%5Ccfrac%7B12%7D%7B5%7D%0A%5C%5C%5C%5C%5C%5C%0Aslope%3D%5Ccfrac%7B12%7D%7B%7B%7B%205%7D%7D%7D%5Cqquad%20negative%5Cimplies%20%20-%5Ccfrac%7B12%7D%7B%7B%7B%205%7D%7D%7D%5Cqquad%20reciprocal%5Cimplies%20-%20%5Ccfrac%7B%7B%7B%205%7D%7D%7D%7B12%7D%0A%5C%5C%5C%5C%5C%5C%0A%5C%20%5Ctextless%20%5C%2012%2C%20-5%5C%20%5Ctextgreater%20%5C%20%5C%20or%5C%20%5C%20%5Ctextless%20%5C%20-12%2C5%5C%20%5Ctextgreater%20%5C%20%5Cimplies%20%5Cboxed%7B12i-5j%5C%20or%5C%20-12i%2B5j%7D)
if we were to place <5, 12> in standard position, so it'd be originating from 0,0, then the rise is 12 and the run is 5.
so any other vector that has a negative reciprocal slope to it, will then be perpendicular or "orthogonal" to it.
so... for example a parallel to <-12, 5> is say hmmm < -144, 60>, if you simplify that fraction, you'd end up with <-12, 5>, since all we did was multiply both coordinates by 12.
or using a unit vector for those above, then
A(-5,5)
B(4,5)
C(2,0)
D(-5,-2)
AB,BC,CD,DA
AB = [4-(-5)),5-5]=[9,0]
Lenght
![AB= \sqrt{9^2+0^2}= \sqrt{81} =9](https://tex.z-dn.net/?f=AB%3D%20%5Csqrt%7B9%5E2%2B0%5E2%7D%3D%20%5Csqrt%7B81%7D%20%20%3D9)
BC = [2-4,0-5]=[-2,-5]
Lenght
![BC=\sqrt{(-2)^2+(-5)^2}=\sqrt{4+25}=\sqrt{29}](https://tex.z-dn.net/?f=BC%3D%5Csqrt%7B%28-2%29%5E2%2B%28-5%29%5E2%7D%3D%5Csqrt%7B4%2B25%7D%3D%5Csqrt%7B29%7D)
CD = [-5-2,-2-0]=[-7,-2]
Lenght
![CD=\sqrt{(-7)^2+(-2)^2}=\sqrt{49+4}=\sqrt{53}](https://tex.z-dn.net/?f=CD%3D%5Csqrt%7B%28-7%29%5E2%2B%28-2%29%5E2%7D%3D%5Csqrt%7B49%2B4%7D%3D%5Csqrt%7B53%7D)
DA =[-5-(-5),-2-5]=[0,-7]
Lenght
![DA=\sqrt{0^2+(-7)^2}=\sqrt{49}=7](https://tex.z-dn.net/?f=DA%3D%5Csqrt%7B0%5E2%2B%28-7%29%5E2%7D%3D%5Csqrt%7B49%7D%3D7)
sorted from longest to shortest:
AB, CD,DA,BC