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Marina CMI [18]
3 years ago
11

Suppose you have the following recursion formula a1 = 1, a2 = 2, and an = a(n - 1)+ a(n - 2)for integers n ≥ 3. How would you de

termine the next 3 terms?
Mathematics
1 answer:
julia-pushkina [17]3 years ago
8 0
A3=3
a4=5
a5=8

This sequence has properties similar to those of the Fibonacci sequence.
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A 25 foot ladder is leaning against a house. If the angle of Elevation of the ladder is 72 degrees:
coldgirl [10]

Answer:

1. 23.8

2. 7.7

Step-by-step explanation:

<u>for the first blank</u>

use sin because x is opposite and 25 is hypotenuse

set up an equation:

sin(72)= (x/25)

multiply both sides by 25:

25(sin(72))= (x/25)25

25(sin(72))= x

multiply 25 by sin(72):

23.8 = x

<u>for the second blank</u>

use cos because x is adjacent and 25 us hypotenuse

set up an equation:

cos(72)= (x/25)

multiply both sides by 25:

25(cos(72))= (x/25)25

25(cos(72))= x

multiply 25 by cos(72):

7.7= x

3 0
3 years ago
The answer and you got it
Ksju [112]
It would be nice If I could read it better. The answer is C if is says angle 3 and angle 5.
5 0
3 years ago
In a football or soccer game, you have 22 players, from both teams, in the field. what is the probability of having at least any
Tamiku [17]

We can solve this problem using complementary events. Two events are said to be complementary if one negates the other, i.e. E and F are complementary if

E \cap F = \emptyset,\quad E \cup F = \Omega

where \Omega is the whole sample space.

This implies that

P(E) + P(F) = P(\Omega)=1 \implies P(E) = 1-P(F)

So, let's compute the probability that all 22 footballer were born on different days.

The first footballer can be born on any day, since we have no restrictions so far. Since we're using numbers from 1 to 365 to represent days, let's say that the first footballer was born on the day d_1.

The second footballer can be born on any other day, so he has 364 possible birthdays:

d_2 \in \{1,2,3,\ldots 365\} \setminus \{d_1\}

the probability for the first two footballers to be born on two different days is thus

1 \cdot \dfrac{364}{365} = \dfrac{364}{365}

Similarly, the third footballer can be born on any day, except d_1 and d_2:

d_3 \in \{1,2,3,\ldots 365\} \setminus \{d_1,d_2\}

so, the probability for the first three footballers to be born on three different days is

1 \cdot \dfrac{364}{365} \cdot \dfrac{363}{365}

And so on. With each new footballer we include, we have less and less options out of the 365 days, since more and more days will be already occupied by another footballer, and we can't have two players born on the same day.

The probability of all 22 footballers being born on 22 different days is thus

\dfrac{364\cdot 363 \cdot \ldots \cdot (365-21)}{365^{21}}

So, the probability that at least two footballers are born on the same day is

1-\dfrac{364\cdot 363 \cdot \ldots \cdot (365-21)}{365^{21}}

since the two events are complementary.

8 0
3 years ago
Not sure if I did this correctly... please help 3. Is the one I'm kinda stuck on
denis23 [38]
You got the b part wrong. It would be 4B+4C/2=4D
6 0
3 years ago
Read 2 more answers
Triangle ABC is graphed on a coordinate grid with vertices at A (-2, -4), B (2, -4) and C (0, 2). Triangle ABC is dilated by a s
Papessa [141]

Answer:

The rule which best represents the dilation that was applied to Triangle ABC to create A’B’C’ will be:

(x, y) → (1/2x, 1/2y)

Step-by-step explanation:

Given the vertices of a triangle

  • A(-2, -4)
  • B(2, -4)
  • C(0, 2)

After the dilation, the vertices of an image triangle be

  • A' (-1, -2)
  • B '(1, -2)
  • C' (0, 1)

If we closely observe, we can notice that the image triangle vertices A'B'C' are half of the original triangle ABC.

i.e. (x, y) → (1/2x, 1/2y)

verification

A(-2, -4) → A'(-2/2, -4/2) = A'(-1, -2)

B(2, -4) → B'(2/2, -4/2) = B'(1, -2)

C(0, 2) → C'(0/2, 2/2) = C'(0, 1)

Thus, it is verified and we conclude that the rule which best represents the dilation that was applied to Triangle ABC to create A’B’C’ will be:

  • (x, y) → (1/2x, 1/2y)
6 0
2 years ago
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