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3 years ago

9

2(2)+5(3)^0----------------------

2 answers:

Travka [436]3 years ago

8 0

I think the correct is 19. Not sure but don’t make me brainlist

larisa86 [58]3 years ago

6 0

Answer: 1

Step-by-step explanation: anything to the power of 0 is 1 (except 0 to the power of 0 is undefined)

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Read 2 more answers

What is the equation of the quadratic graph with a focus of (3, 1) and a directrix of y = 5?

telo118 [61]

Answer:

$f(x)=-\frac{1}{8}(x-3)^2+3$

Step-by-step explanation:

We want to find the equation of the parabola with a focus of $(3,1)$ and directrix $y=5$ .

Considering the directrix, the quadratic graph must open downwards.

The equation of this parabola is given by the formula,

$(x-h)^2=4p(y-k)$ , where $(h,k)$ is the vertex of the parabola.

The axis of this parabola meets the directrix at $(3,5)$ .

Since the vertex is the midpoint of the focus and the point of intersection of the axis of the parabola and the directrix,

$h=\frac{3+3}{2}=3$ and $k=\frac{5+1}{2}=3$ .

The equation of the parabola now becomes,

$(x-3)^2=4p(y-3)$ .

Also $|p|$ is the distance between the vertex and the directrix.

$|p|=2$

This implies that $p=-2\:or\:2$ .

Since the parabola turns downwards,

$p=-2$ .

Our equation now becomes,

$(x-3)^2=4(-2)(y-3)$ .

$(x-3)^2=-8(y-3)$ .

We make y the subject to get,

$y=-\frac{1}{8}(x-3)^2+3)$ .

This is the same as

$f(x)=-\frac{1}{8}(x-3)^2+3)$ .

4 0

3 years ago