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dalvyx [7]
3 years ago
9

As part of an activity during math class, Kiera has to select a secret number containing 3 different digits from 1 to 4. How man

y different secret numbers can Kiera create?
Mathematics
1 answer:
Alenkasestr [34]3 years ago
8 0

Answer:

24.

Step-by-step explanation:

The is the number of permutations of 3 digits from 4

= 4P3

= 4! / (4- 3)!

= 4 *3 *2 *1 / 1!

= 24/1

= 24.

You might be interested in
Inverse functions always go in the same direction as the original function. true or false?
zavuch27 [327]
False , that’s why they are inverse
8 0
4 years ago
Algebra 2 will give points and brainly
aivan3 [116]

Answer:

Step-by-step explanation:

In the 3rd question, we are given the equation x ^ 3 + x ^ 2 + 2x + 24. One of the factors is x + 3. Now, we can use long division to find that the equation we have left is x^2 - 2x - 8. We can just factor this to get (x - 4) (x + 2). In the 3rd question, possible factors for the coefficient are 1, 2, -1, -2. Possible factors for the constant are 1, 7, -1, -7. Now, we can try out all of them. The possible factors are 1, 7, -1, -7, 2, 14, -2, -14.

7 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

5 0
2 years ago
Question 3. Solve each equation given<br>below.<br>5 x - 4=31​
harkovskaia [24]

Answer:

x = 7

Step-by-step explanation:

5x-4 = 31

5x = 31+4

5x = 35

x = 7

3 0
3 years ago
Consider rolling two fair dice and observing the number of spots on the resulting upward face of each one. Letting A be the even
Allushta [10]

Answer:

P(E|A)= \frac{10}{11}

Step-by-step explanation:

Given

Two rolls of die

E \to one of the outcomes is 6

A \to atleast one is 6

Required

P(E|A)

First, list out the outcome of each

E = \{(1,6),(2,6),(3,6),(4,6),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5)\}

A = \{(1,6),(2,6),(3,6),(4,6),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\}

So:

P(E|A)= \frac{n(E\ n\ A)}{n(A)}

Where:

E\ n\ A = \{(1,6),(2,6),(3,6),(4,6),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5)\}

n(E\ n\ A) = 10

n(A) = 11

So:

P(E|A)= \frac{10}{11}

7 0
3 years ago
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