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loris [4]
3 years ago
15

What is the length of the focal width? x^2=12y

Mathematics
2 answers:
Ghella [55]3 years ago
7 0

Answer:

The length of the Focal Width is 12

Step-by-step explanation:

Apex

Ierofanga [76]3 years ago
3 0
\bf \textit{parabola vertex form with focus point distance}\\\\
\begin{array}{llll}
\boxed{(y-{{ k}})^2=4{{ p}}(x-{{ h}})} \\\\
(x-{{ h}})^2=4{{ p}}(y-{{ k}}) \\
\end{array}
\qquad 
\begin{array}{llll}
vertex\ ({{ h}},{{ k}})\\\\
{{ p}}=\textit{distance from vertex to }\\
\qquad \textit{ focus or directrix}\\\\
4p=focal\ width
\end{array}\\\\
-------------------------------\\\\
x^2=12y\implies (x-0)^2=12(y-0)\quad 
\begin{cases}
h=0\\
k=0\\
\boxed{4p=12}
\end{cases}
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Step-by-step explanation:

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3 years ago
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4 0
4 years ago
Consider the graph of the quadratic function y = 3x2 – 3x – 6. What are the solutions of the quadratic equation 0 = 3x2 − 3x − 6
Ilya [14]
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all choices are given wrong
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4 years ago
Find x and y in <br>2x+2y=12<br>4x_4y=13
dlinn [17]
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6 0
3 years ago
Read 2 more answers
Simplify this please.
Natali5045456 [20]

Answer:

tbh i just used another app to help me help you i hope this is right the app i used is always spot on with answers :)

4 0
3 years ago
Read 2 more answers
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