![\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.
6x+7y=4x+4y6x+7y=4x+4y6, x, plus, 7, y, equals, 4, x, plus, 4, y Complete the missing value in the solution to the equation. (((
Dmitrij [34]
Answer:
<h3>The missing value in the given solution is x=6</h3><h3>Therefore the solution is (6,-4)</h3>
Step-by-step explanation:
Given equation is 
<h3>To find the missing value in the solution to the equation :</h3>
- Let missing value in the solution be x
- Then the solution of the given equation is (x,-4)
Substitute the value of y=-4 in the given equation (1) we get
( adding the like terms )
Therefore x=6
<h3>Therefore the missing value in the given solution is x=6</h3><h3>Therefore the solution is (6,-4)</h3>
Answer:
3xz5+2x+4y−2z
Step-by-step explanation:
2x−5y+3z5x+9y−2z
=2x+−5y+3xz5+9y+−2z
Combine Like Terms:
=2x+−5y+3xz5+9y+−2z
=(3xz5)+(2x)+(−5y+9y)+(−2z)
=3xz5+2x+4y+−2z
Answer:
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Here, is the answer of comparison
