Which equation has a graph that is a parabola with a vertex at (5, 3)?

Mathematics

2 answers:

Murrr4er [49]3 years ago

4 0

Answer:

$y=(x-5)^2+3$

Step-by-step explanation:

equation has a graph that is a parabola with a vertex at (5, 3)

Given the vertex is (5,3) that is (h,k)

Vertex form of equation is

$y=a(x-h)^2+k$ where (h,k) is the vertex

Given the vertex is (5,3), h=5 and k=3

Plug in the values in the vertex form of the equation

$y=a(x-h)^2+k$

$y=(x-5)^2+3$

solmaris [256]3 years ago

3 0

Answer:

y = (x - 5)^2 + 3.

Step-by-step explanation:

The vertex form of a parabola is written:

y = (x - b)^2 + c where the vertex is (b, c).

So this equation is y = (x - 5)^2 + 3.

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The first five terms of a sequence are 2, 5/3, 3/2, 7/5 and 4/3. what is the explicit formula?

katen-ka-za [31]

Answer:

an = (n+3)/(n+1)

Step-by-step explanation:

The sequence can be written as ...

4/2, 5/3, 6/4, 7/5, 8/6, ...

where the first numerator is n+3 and the first denominator is n+1. Each numerator and denominator increase by 1 from term to term, so the explicit formula can be written

an = (n+3)/(n+1)

4 0

3 years ago

Can someone please help me with this question and show the steps?

Katena32 [7]

<h2>Answer:</h2>

$D. \ 408in^2$

<h2>Step-by-step explanation:</h2>

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$s=s_{square}+4s_{triangle}$