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inna [77]
2 years ago
14

Number 1 is the question

Mathematics
1 answer:
ozzi2 years ago
3 0
You’ll have to adjust the plotting a bit to fit your graph that you have, but this is the main idea.

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The Fibonacci sequence is defined by $F_1 = F_2 = 1$ and $F_{n + 2} = F_{n + 1} + F_n$. Find the remainder when $F_{1999}$ is di
zavuch27 [327]

Answer:

The remainder is 1

Step-by-step explanation:

Given the Fibonacci sequence

F_1 = F_2 = 1, and

F_(n + 2) = F_(n + 1) + F_n

We want to find the remainder when F_(1999) is divided by 5.

Let us write the first 20 numbers of the sequence in (mod 5). They are

F_1 = 1,

F_2 = 1,

F_3 = 2,

F_4 = 3,

F_5 = 5 = 0 (mod 5),

F_6 = 3,

F_7 = 3,

F_8 = 1

F_(9) = 4

F_(10) = 0

F_(11) = 4

F_(12) = 4

F_(13) = 3

F_(14) = 2

F_(15) = 0

F_(16) = 2

F_(17) = 2

F_(18) = 4

F_(19) = 1

F_(20) = 0

We have: 1, 1, 2, 3, 0, 3, 3, 1, 4, 0, 4, 4, 3, 2, 0, 2, 2, 4, 1, 0

Now, 1999 = 19(mod 20)

The 19th number in the sequence is 1.

So, the remainder is 1.

6 0
2 years ago
3/5 ÷ 1 1/4 =<br> (20 points)
Anna11 [10]

3/5 ÷ 1 1/4 = 0.48

0.48 will be your anwer


7 0
2 years ago
X²-10x+25=0 quadratic equation by factoring​
lisov135 [29]

Answer:

x = 5

Step-by-step explanation:

{x}^{2}  - 10x + 25 = 0 \\  {x}^{2}  - 5x - 5x + 25 = 0 \\ x(x  -  5) - 5(x - 5) = 0 \\ (x - 5)(x - 5) = 0 \\ ( x- 5) = 0 \: or \: (x - 5) = 0 \\ x = 5 \: or \: x = 5 \\ x = 5

4 0
3 years ago
A small combination lock on a suitcase has 4 ​wheels, each labeled with the 10 digits 0 to 9. how many 4 digit combinations are
tamaranim1 [39]
This is the easiest way to solve this problem:

Imagine this represents how many combinations you can have for each of the 4 wheels (each blank spot for one wheel): __ __ __ __

For the first situation it says how many combos can we make if no digits are repeated.
We have 10 digits to use for the first wheel so put a 10 in the first slot 
10 __ __ __
Since no digit can be repeated we only have 9 options for the second slot
10 9_ __ __
Same for the third slot, so only 8 options
<u>10</u> <u> 9 </u> <u> 8 </u> __
4th can't be repeated so only 7 options left
<u>10</u> <u> 9 </u> <u> 8 </u> <u> 7 
</u><u>
</u>Multiply the four numbers together: 10*9*8*7 = 5040 combinations


For the next two do the same process as the one above.

If digits can be repeated? You have ten options for every wheel so it would look like this: <u>10</u> <u>10</u> <u>10</u> <u>10
</u>
10*10*10*10 = 10,000 combinations

If successive digits bust be different?
We have 10 for the first wheel, but second wheel only has 9 options because 2nd number can't be same as first. The third and fourth wheels also has 9 options for the same reason.

<u>10</u> <u> 9</u><u> </u> <u> 9 </u> <u> 9 </u>

10*9*9*9 = 7290 combinations






8 0
2 years ago
X-3&lt;9<br><br> What’s the answer to this everbody should know
Murrr4er [49]

Answer:

x<12

Step-by-step explanation:

Let's solve your inequality step-by-step.

x−3<9

Step 1: Add 3 to both sides.

x−3+3<9+3

x<12

Answer:

x<12

5 0
2 years ago
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