I=2,400×0.07
I=168
,,,,,,,,,,,,,,,,,,,,,,,,,,,,,
The graph of the function will have zero of -2 and 8 respectively
<h3>Quadratic function</h3>
Quadratic functions are function that has a leading degree of 2. Given the quadratic equation
y = x^2 - 6x - 16
Factorize
y = x^2 - 6x - 16
y = x^2 - 8x + 2x - 16
y = x(x - 8) + 2(x-8)
y = (x+2)(x-8)
The graph of the function is attached below
Learn more on quadratic graph here: brainly.com/question/9643976
#SPJ1
Check the picture below, so the parabola looks more or less like so, with a "p" distance of 3 and the vertex at the origin, keeping in mind the vertex is half-way between the focus point and the directrix.
![\textit{horizontal parabola vertex form with focus point distance} \\\\ 4p(x- h)=(y- k)^2 \qquad \begin{cases} \stackrel{vertex}{(h,k)}\qquad \stackrel{focus~point}{(h+p,k)}\qquad \stackrel{directrix}{x=h-p}\\\\ p=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix}\\\\ \stackrel{"p"~is~negative}{op ens~\supset}\qquad \stackrel{"p"~is~positive}{op ens~\subset} \end{cases} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Ctextit%7Bhorizontal%20parabola%20vertex%20form%20with%20focus%20point%20distance%7D%20%5C%5C%5C%5C%204p%28x-%20h%29%3D%28y-%20k%29%5E2%20%5Cqquad%20%5Cbegin%7Bcases%7D%20%5Cstackrel%7Bvertex%7D%7B%28h%2Ck%29%7D%5Cqquad%20%5Cstackrel%7Bfocus~point%7D%7B%28h%2Bp%2Ck%29%7D%5Cqquad%20%5Cstackrel%7Bdirectrix%7D%7Bx%3Dh-p%7D%5C%5C%5C%5C%20p%3D%5Ctextit%7Bdistance%20from%20vertex%20to%20%7D%5C%5C%20%5Cqquad%20%5Ctextit%7B%20focus%20or%20directrix%7D%5C%5C%5C%5C%20%5Cstackrel%7B%22p%22~is~negative%7D%7Bop%20ens~%5Csupset%7D%5Cqquad%20%5Cstackrel%7B%22p%22~is~positive%7D%7Bop%20ens~%5Csubset%7D%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)

Answer:
sleepy hollow
Step-by-step explanation:
1.5M like this
Answer:
it is 1/3
Step-by-step explanation:
also you can type (1/9)x3 into g o o g l e
pls mark brainliest (it is the crown)