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KiRa [710]
3 years ago
8

Which set of ordered pairs represents a function?

Mathematics
2 answers:
Gennadij [26K]3 years ago
8 0
C - 1
a - 2

are ur answers
if i helped, please mark brainliest! i am trying to gain :)
bagirrra123 [75]3 years ago
6 0
1. C 
becasue there are no repeating x values in a function.

2. A 
because the values of f(X) are always 5 more than the values of x.

hope this helps!
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How do you write 7.369E6 in standard form?
melamori03 [73]

\\ \bull\sf\dashrightarrow 7.369E6

\\ \bull\sf\dashrightarrow 7.369\times 10^6

\\ \bull\sf\dashrightarrow 7369\times 10^{-3}\times 10^6

\\ \bull\sf\dashrightarrow 7369\times 10^3

\\ \bull\sf\dashrightarrow 7369000

3 0
3 years ago
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
Ad libitum [116K]

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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2 years ago
Pls help me with math due tonight
qaws [65]

Answer:

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Hint: Click on the point (0,0) on the graph. Drag all the way up to (7,9) and stop. This should automatically create your graph that represents the relationship.

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Dilbert invests $10,000 at 2% simple interest for 1 year. How
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If f (x) = StartFraction x minus 3 Over x EndFraction and g(x) = 5x – 4, what is the domain of (f circle g) (x)?
alexandr1967 [171]

Answer:

<h2>The domain is all real values of x except \frac{4}{5}.</h2>

Step-by-step explanation:

The function, f(x) = \frac{x - 3}{x}.

g(x) = 5x - 4.

Hence, (f circle g) (x) = \frac{5x - 4 - 3}{5x - 4}.

We need to find the domain of (f circle g) (x) .

The domain means the set of values of x, for which we will get a real value of (f circle g) (x).

The function will not give a real value when, 5x - 4 = 0 that is x = \frac{4}{5}.

Hence, the domain of the function will be all real values of x rather than \frac{4}{5}.

5 0
3 years ago
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