2 years ago

Solve this system of linear equations. Separate

Mathematics

2 answers:

kupik [55]2 years ago

8 0

Sub in 0 for x and then solve then to the same with Y

uranmaximum [27]2 years ago

5 0

Answer:

That is tough

Step-by-step explanation:

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2 years ago

Graph y=(x+2)^2-3 <br> Show Vertex

nikklg [1K]

$\bf ~~~~~~\textit{parabola vertex form}
\\\\
\begin{array}{llll}
\boxed{y=a(x- h)^2+ k}\\\\
x=a(y- k)^2+ h
\end{array}
\qquad\qquad
vertex~~(\stackrel{}{ h},\stackrel{}{ k})\\\\
-------------------------------\\\\
y=(x+2)^2-3\implies y=\stackrel{a}{1}[x-\stackrel{h}{(-2)}]^2\stackrel{k}{-3}\quad 
\begin{cases}
h=-2\\
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\end{cases}$