![\bf \textit{vertical parabola vertex form with focus point distance} \\\\ 4p(y- k)=(x- h)^2 \qquad \begin{cases} \stackrel{vertex}{(h,k)}\qquad \stackrel{focus~point}{(h,k+p)}\qquad \stackrel{directrix}{y=k-p}\\\\ p=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix}\\\\ \stackrel{"p"~is~negative}{op ens~\cap}\qquad \stackrel{"p"~is~positive}{op ens~\cup} \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bvertical%20parabola%20vertex%20form%20with%20focus%20point%20distance%7D%20%5C%5C%5C%5C%204p%28y-%20k%29%3D%28x-%20h%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%2Ck%2Bp%29%7D%5Cqquad%20%5Cstackrel%7Bdirectrix%7D%7By%3Dk-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~%5Ccap%7D%5Cqquad%20%5Cstackrel%7B%22p%22~is~positive%7D%7Bop%20ens~%5Ccup%7D%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D)

something noteworthy is that the squared variable is the "x", thus the parabola is a vertical one, the "p" value is negative, so is opening downwards, and the h,k is pretty much the origin,
vertex is at (0,0)
the focus point is "p" or 5 units down from there, namely at (0, -5)
the directrix is "p" units on the opposite direction, up, namely at y = 5
the focal width, well, |4p| is pretty much the focal width, in this case, is simply yeap, you guessed it, 20.
Answer:
<em>13 - 6y </em>
Step-by-step explanation:
8 - 4y + (-2y) + 5 = <em>13 - 6y</em>
The values of x in the triangles and the angles in the rhombus are illustrations of tangent ratios
- The values of x in the triangles are 21.4 units, 58 degrees and 66 degrees
- The angles in the rhombus are 44 and 46 degrees, respectively
<h3>How to determine the values of x?</h3>
<u>Triangle 1</u>
The value of x is calculated using the following tangent ratio
tan(25) = 10/x
Make x the subject
x = 10/tan(25)
Evaluate
x = 21.4
<u>Triangle 2</u>
The value of x is calculated using the following tangent ratio
tan(x) = 8/5
Evaluate the quotient
tan(x) = 1.6
Take the arc tan of both sides
x = arctan(1.6)
Evaluate
x = 58
<u>Triangle 3</u>
The value of x is calculated using the following tangent ratio
tan(x) = 0.34/0.15
Evaluate the quotient
tan(x) = 2.27
Take the arc tan of both sides
x = arctan(2.27)
Evaluate
x = 66
<h3>How to calculate the angles of the rhombus?</h3>
The lengths of the diagonals are:
L1 = 2 in
L2 = 5 in
Represent the angles with x and y.
The measures of the angles are calculated using the following tangent ratios
tan(0.5x) = 2/5 and y = 90 - x
Evaluate the quotient
tan(0.5x) = 0.4
Take the arc tan of both sides
0.5x = arctan(0.4)
Evaluate
0.5x = 22
Divide by 0.5
x = 44
Recall that:
y = 90 - x
This gives
y = 90 - 44
Evaluate
y = 46
Hence, the angles in the rhombus are 44 and 46 degrees, respectively
Read more about tangent ratio at:
brainly.com/question/13347349
Answer: The value of x is 8x
Step-by-step explanation:
3x -y ⩾ 6
3x - 6 ⩾ y
now, with inequalities, what we do is, we graph the line of 3x - 6 = y, and then we shade the "true region".
if we pick a point on say hmmm (4, 0), namely x = 4 and y = 0, we can plug that in the inequality and see what we get,
3(4) - 0 ⩾ 6
12 - 0 ⩾ 6
12 ⩾ 6
is 12 really greater or equals to 6? well yes, therefore, the point (4, 0) lies on the "true region", since it's true, 12 is indeed ⩾ 6, so, where that point is, we shade.
now, the ⩾ means equals to or greater, and therefore, since the values could also equal the boundary points, the line is a solid line, because it includes the line itself, as well as the shading.
check the picture below.