On how many points do the graphic cross each other

Mathematics

2 answers:

lukranit [14]3 years ago

5 0

I believe it would be B

Sati [7]3 years ago

3 0

These 2 equations intersect at roughly 2 points, one being the origin (0,0) and the other being (1,1) so the answer would be 2.

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What is the equation of the quadratic graph with a focus of (4,3) and a directrix of y=-6

Rufina [12.5K]

Answer:

f(x) = (1/18)x^2 - (4/9)x - 65/18

Step-by-step explanation:

Since the focus (4,3) is higher up than is the directrix y = -6, the vertex is between y = -6 and y = 3, halfway between these values, to be exact: y = -3/2. Thus, the y-coordinate of the vertex is (-6 + 3/2), or -4.5. The x-coordinate of the vertex is 4, which comes from that 4 in (4,3). Thus, the vertex is (4, -4.5) or (4, -4 1/2).

The standard equation for a vertical parabola with vertex at (h,k) is:

4p(y-k) = (x-h)^2, where p is the distance between vertex and focus or between vertex and directrix. In this case p = 4.5. Thus, the equation of this quadratic is 4(4.5)(y - [-4.5]) = (x - 4)^2, or

18(y+4.5) = (x-4)^2, or y+4.5 = (1/18)(x-4)^2.

Let's expand this and rewrite the result as a quadratic equation in standard form y = ax^2 + bx + c:

y + 4.5 = (1/18)(x^2 - 8x + 16). Multiplying all terms by 18 to remove both fractions, we get:

18y + 81 = x^2 - 8x + 16. Rearranging these terms, we get:

18y = x^2 - 8x + 16 - 81, or

18y = x^2 - 8x - 65, or

y = f(x) = (1/18)x^2 - (4/9)x - 65/18.

5 0

3 years ago

Read 2 more answers

How do you solve this whole thing please?

NemiM [27]

Answer:

Step-by-step explanation:

Let's look at the first two, and hopefully you'll be able to figure the rest out:

1. Answer below

$d = rt$

The problem asks us to solve for $r$ , so that means we need to get $r$ on one side of the equation by itself. To do so, we will need to divide both sides of the equation by $t$ :

$\frac{d}{t} = \frac{rt}{t}$

$\frac{d}{t} = r$

2. Answer below

$B = T - Lc$

The problem asks us to solve for $T$ , so that means we need to get $T$ on one side of the equation by itself. To do so, we will need to add $Lc$ to both sides of equation: