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hjlf
3 years ago
8

We’re not done just yet! Thanks for anyone who helped, means a lot to me!

Mathematics
2 answers:
olga_2 [115]3 years ago
6 0

Answer: Choice A) The number that is 5 to the left of -3 on the number line.

Explanation:

Refer to the diagram below. The value -3 is 3 units to the left of 0 on the number line. So you start at 0 and move 3 spaces to the left to arrive at -3.

Once you're at -3 on the number line, you'll move 5 more spaces to the left to arrive at -8

We can think of -3+(-5) as -3-5, both of which simplify to -8

Or we can think of it like saying -3+(-5) = -1*(3+5) = -1*8 = -8. Here I factored out a negative 1, and then added. The final result is negative since we're moving to the left in the negative territory.

bonufazy [111]3 years ago
5 0

Answer:

A

Step-by-step explanation:

if im reading it right

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4. Find the standard from of the equation of a hyperbola whose foci are (-1,2), (5,2) and its vertices are end points of the dia
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The <em>standard</em> form of the equation of the hyperbola that satisfies all conditions is (x - 2)²/4 - (y - 2)²/5 = 1 .

<h3>How to find the standard equation of a hyperbola</h3>

In this problem we must determine the equation of the hyperbola in its <em>standard</em> form from the coordinates of the foci and a <em>general</em> equation of a circle. Based on the location of the foci, we see that the axis of symmetry of the hyperbola is parallel to the x-axis. Besides, the center of the hyperbola is the midpoint of the line segment with the foci as endpoints:

(h, k) = 0.5 · (- 1, 2) + 0.5 · (5, 2)

(h, k) = (2, 2)

To determine whether it is possible that the vertices are endpoints of the diameter of the circle, we proceed to modify the <em>general</em> equation of the circle into its <em>standard</em> form.

If the vertices of the hyperbola are endpoints of the diameter of the circle, then the center of the circle must be the midpoint of the line segment. By algebra we find that:

x² + y² - 4 · x - 4 · y + 4= 0​

(x² - 4 · x + 4) + (y² - 4 · y + 4) = 4

(x - 2)² + (y - 2)² = 2²

The center of the circle is the midpoint of the line segment. Now we proceed to determine the vertices of the hyperbola:

V₁(x, y) = (0, 2), V₂(x, y) = (4, 2)

And the distance from the center to any of the vertices is 2 (<em>semi-major</em> distance, a) and the semi-minor distance is:

b = √(c² - a²)

b = √(3² - 2²)

b = √5

Therefore, the <em>standard</em> form of the equation of the hyperbola that satisfies all conditions is (x - 2)²/4 - (y - 2)²/5 = 1 .

To learn more on hyperbolae: brainly.com/question/27799190

#SPJ1

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