Answer and Explanation:
The problem provides the following information:
Hypotenuse
=
h
y
p
=
5129
ft
Vertical drop
=
o
p
p
=
1275
ft
To find the angle of depression, the best trigonometric ratio to use is the sine ratio. After equating it with the ratio of the opposite side over the hypotenuse, get the inverse of the sine ratio as shown:
sin
θ
=
o
p
p
h
y
p
θ
=
sin
−
1
(
o
p
p
h
y
p
)
θ
=
sin
−
1
(
1275
ft
5129
ft
)
θ
=
sin
−
1
(
0.2485864691
)
θ
=
14.3938824691
≈
14.4
∘
The angle of depression is
14.4
D. 16pi ft
8 x 2 = 16 x pi = 16pi
Answer:
15 times 3 equals 60.
Step-by-step explanation:
Since Plot B is skewed to the left and Plot A is not, the statement that must be true is ...
D) Plot B is more skewed to the left than plot A.
Check the picture below.
so the focus point is there, and the directrix is above it, meaning is a vertical parabola and is opening downwards, since the parabola opens up towards the focus.
now, the vertex is half-way between those two guys, at a "p" distance from either one, if we move over the y-axis from -5 to +2, we have 7 units, half-way is 3.5 units, and that puts us at -1.5 or -1½, as you see in the picture, so the vertex is then at (-3 , -1½).
so the distance from the vertex to the focus point is then 3½ units, however since the parabola is opening downwards, "p" is negative, thus "p = 3½".
![\bf \textit{parabola vertex form with focus point distance} \\\\ \begin{array}{llll} 4p(x- h)=(y- k)^2 \\\\ \stackrel{\textit{using this one}}{4p(y- k)=(x- h)^2} \end{array} \qquad \begin{array}{llll} vertex\ ( h, k)\\\\ p=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix} \end{array} \\\\[-0.35em] \rule{34em}{0.25pt}](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bparabola%20vertex%20form%20with%20focus%20point%20distance%7D%20%5C%5C%5C%5C%20%5Cbegin%7Barray%7D%7Bllll%7D%204p%28x-%20h%29%3D%28y-%20k%29%5E2%20%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Busing%20this%20one%7D%7D%7B4p%28y-%20k%29%3D%28x-%20h%29%5E2%7D%20%5Cend%7Barray%7D%20%5Cqquad%20%5Cbegin%7Barray%7D%7Bllll%7D%20vertex%5C%20%28%20h%2C%20k%29%5C%5C%5C%5C%20p%3D%5Ctextit%7Bdistance%20from%20vertex%20to%20%7D%5C%5C%20%5Cqquad%20%5Ctextit%7B%20focus%20or%20directrix%7D%20%5Cend%7Barray%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D)
![\bf \begin{cases} h=-3\\ k=-\frac{3}{2}\\[0.7em] p=-\frac{7}{2} \end{cases}\implies 4\left( -\cfrac{7}{2} \right)\left[ y-\left(-\cfrac{3}{2} \right) \right]=\left[ x-\left( -3 \right) \right]^2 \\\\\\ -14\left( y+\cfrac{3}{2} \right)=(x+3)^2\implies y+\cfrac{3}{2} =-\cfrac{(x+3)^2}{14} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill y=-\cfrac{1}{14}(x+3)^2-\cfrac{3}{2}~\hfill](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Bcases%7D%20h%3D-3%5C%5C%20k%3D-%5Cfrac%7B3%7D%7B2%7D%5C%5C%5B0.7em%5D%20p%3D-%5Cfrac%7B7%7D%7B2%7D%20%5Cend%7Bcases%7D%5Cimplies%204%5Cleft%28%20-%5Ccfrac%7B7%7D%7B2%7D%20%5Cright%29%5Cleft%5B%20y-%5Cleft%28-%5Ccfrac%7B3%7D%7B2%7D%20%5Cright%29%20%5Cright%5D%3D%5Cleft%5B%20x-%5Cleft%28%20-3%20%5Cright%29%20%5Cright%5D%5E2%20%5C%5C%5C%5C%5C%5C%20-14%5Cleft%28%20y%2B%5Ccfrac%7B3%7D%7B2%7D%20%5Cright%29%3D%28x%2B3%29%5E2%5Cimplies%20y%2B%5Ccfrac%7B3%7D%7B2%7D%20%3D-%5Ccfrac%7B%28x%2B3%29%5E2%7D%7B14%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20~%5Chfill%20y%3D-%5Ccfrac%7B1%7D%7B14%7D%28x%2B3%29%5E2-%5Ccfrac%7B3%7D%7B2%7D~%5Chfill)