Answer:
help
Step-by-step explanation:
![\bf ~~~~~~\textit{parabola vertex form} \\\\ \begin{array}{llll} \stackrel{\textit{we'll use this one}}{y=a(x- h)^2+ k}\\\\ x=a(y- k)^2+ h \end{array} \qquad\qquad vertex~~(\stackrel{2}{ h},\stackrel{-1}{ k}) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \begin{cases} h=2\\ k=-1 \end{cases}\implies y=a(x-2)^2-1 \\\\\\ \textit{we also know that } \begin{cases} y=0\\ x=5 \end{cases}\implies 0=a(5-2)^2-1\implies 1=9a \\\\\\ \cfrac{1}{9}=a\qquad therefore\qquad \boxed{y=\cfrac{1}{9}(x-2)^2-1}](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~%5Ctextit%7Bparabola%20vertex%20form%7D%20%5C%5C%5C%5C%20%5Cbegin%7Barray%7D%7Bllll%7D%20%5Cstackrel%7B%5Ctextit%7Bwe%27ll%20use%20this%20one%7D%7D%7By%3Da%28x-%20h%29%5E2%2B%20k%7D%5C%5C%5C%5C%20x%3Da%28y-%20k%29%5E2%2B%20h%20%5Cend%7Barray%7D%20%5Cqquad%5Cqquad%20vertex~~%28%5Cstackrel%7B2%7D%7B%20h%7D%2C%5Cstackrel%7B-1%7D%7B%20k%7D%29%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20%5Cbegin%7Bcases%7D%20h%3D2%5C%5C%20k%3D-1%20%5Cend%7Bcases%7D%5Cimplies%20y%3Da%28x-2%29%5E2-1%20%5C%5C%5C%5C%5C%5C%20%5Ctextit%7Bwe%20also%20know%20that%20%7D%20%5Cbegin%7Bcases%7D%20y%3D0%5C%5C%20x%3D5%20%5Cend%7Bcases%7D%5Cimplies%200%3Da%285-2%29%5E2-1%5Cimplies%201%3D9a%20%5C%5C%5C%5C%5C%5C%20%5Ccfrac%7B1%7D%7B9%7D%3Da%5Cqquad%20therefore%5Cqquad%20%5Cboxed%7By%3D%5Ccfrac%7B1%7D%7B9%7D%28x-2%29%5E2-1%7D)
now, let's expand the squared term to get the standard form of the quadratic.

The correct answer is option A. Erica is correct in saying that the two lines are not necessarily the same and we should also look at the y-intercepts before determining how many solutions there were. <span>Two lines with equal slopes could be the same line, but only if they have the same y-intercept.</span>
<span>4. Simplify the expression.
sine of x to the second power minus one divided by cosine of negative x</span>
<span>(1−sin2(x))/(sin(x)−csc(x))<span>
</span>sin2x+cos2x=1</span>
<span>1−sin2x=cos2x<span>
</span>cos2(x)/(sin(x)−csc(x))</span>
<span>csc(x)=1/sin(x)</span>
<span>cos2(x)/(sin(x)− 1/sin(x))= cos2(x)/((sin2(x)− 1)/sin(x))</span>
<span>sin2(x)− 1=-cos2(x)</span>
<span>cos2(x)/(( -cos2(x))/sin(x))
=-sin(x)</span>
<span>
the answer is the letter a)
-sin x
</span><span>
5. Find all solutions in the interval [0, 2π). (6 points)sin2x + sin x = 0</span> using a graphical tool
the solutions
x1=0
x2=pi
<span>x3=3pi/2
the answer is the letter </span><span>
D) x = 0, π, three pi divided by two</span>
Answer:
Step-by-step explanation:
A = LW + ½πr²
A = 15(12) + ½(3.14)(12²/4)
A = 180 + 56.52
A = 236.52 m²