first off, let's notice the graph touches the x-axis at -1 and 3, namely, those are the zeros/solutions/roots of the polynomial and therefore, the factors come from those points.
now, at -1, the graph doesn't cross the x-axis, instead it <u>simply bounces off</u> of it, that means the zero of x = -1, has an even multiplicity, could be 4 or 2 or 6, but let's go with 2.
at x = 3, the graph does cross the x-axis, meaning it has an odd multiplicity, could be 3 or 1, or 7 or 9, but let's use 1.
![\bf \begin{cases} x=-1\implies &x+1=0\\ x=3\implies &x-3=0 \end{cases}~\hspace{5em}\stackrel{\textit{even multiplicity}}{(x+1)^2}\qquad \stackrel{\textit{odd multiplicity}}{(x-3)^1}=\stackrel{y}{0} \\\\\\ (x^2+2x+1)(x-3)=y\implies x^3+2x^2+x-3x^2-6x-3=y \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill x^3-x^2-5x-3=y~\hfill](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Bcases%7D%20x%3D-1%5Cimplies%20%26x%2B1%3D0%5C%5C%20x%3D3%5Cimplies%20%26x-3%3D0%20%5Cend%7Bcases%7D~%5Chspace%7B5em%7D%5Cstackrel%7B%5Ctextit%7Beven%20multiplicity%7D%7D%7B%28x%2B1%29%5E2%7D%5Cqquad%20%5Cstackrel%7B%5Ctextit%7Bodd%20multiplicity%7D%7D%7B%28x-3%29%5E1%7D%3D%5Cstackrel%7By%7D%7B0%7D%20%5C%5C%5C%5C%5C%5C%20%28x%5E2%2B2x%2B1%29%28x-3%29%3Dy%5Cimplies%20x%5E3%2B2x%5E2%2Bx-3x%5E2-6x-3%3Dy%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20~%5Chfill%20x%5E3-x%5E2-5x-3%3Dy~%5Chfill)
Answer: See attached
Step-by-step explanation:
[] Since the inequality uses greater than (>) and <em>not</em> greater than or equal to (≥), the line will be dashed.
[] The inequality is greater than (>) so we will shade above the line as values "above" are greater.
Answer:
1950.54
Step-by-step explanation:
1950.54
Answer: About 45.
Step-by-step explanation: There's always a pattern in these kinds of questions, you can see for 9th to 13th August, the number has always been between 20 to 40. And since on 13th August the graph is on a rise again, it would be around 45. 90 is just not realistic considering there's only an increase of around 5-10 tickets every day.
Answer:
see explanation
Step-by-step explanation:
(a)
OC = OB ( both radii of the circle )
Thus Δ BOC is isosceles with congruent base angles.
∠ BOC = ∠ BCO = 50°
(b)
∠ ACB = 90° ( angle in a semicircle ), then
∠ ACO = 90° - 50° = 40°
OA = OC ( both radii of the circle )
Thus Δ ACO is isosceles with congruent base angles.
∠ BAC = ∠ ACO = 40°
