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Morgarella [4.7K]
3 years ago
13

What time is 5 1/2 after 9:22 pm

Mathematics
1 answer:
kow [346]3 years ago
4 0
2:52 am I am pretty sure
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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 d
Firlakuza [10]
\left\{\begin{array}{ccc}a_1=1\\a_2=2\\a_n=a_{n-1}+a_{n-2}&for\ n\geq3\end{array}\right\\\\a_3=a_{3-1}+a_{3-2}=a_2+a_1\to a_3=2+1=3\\\\a_4=a_{4-1}+a_{4-2}=a_3+a_2\to a_4=3+2=5\\\\a_5=a_{5-1}+a_{5-2}=a_4+a_3\to a_5=5+3=8\\\vdots
6 0
3 years ago
Find The Area Of The Shape Shown Below
Vesnalui [34]

Answer:

<u>32</u>

Step-by-step explanation:

I proceeded to divide the shape up into 2 triangle and 1 square. The added sum of the 2 divided triangles is 16 and the square is 16. Add those to to get <u>32</u>.

The following shape is a trapezoid

area formula:

A=

A=

A= (4)(12+4)

A= (4)(16)

A= (4)(16)

A= (64)

A=32

Also, is that Khan Academy?

8 0
3 years ago
Read 2 more answers
What is -2 and 1/2 as a decimal<br>​
olganol [36]

Answer:

-2.5

Step-by-step explanation:

Just change 1/2 to .5 and there you have it. Negative two isn't a fraction or percentage so it doesn't need to be changed.

5 0
3 years ago
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The size of the hermit crab’s shell depends on
elixir [45]

Answer:

Length of crab in x years = x + 2

Step-by-step explanation:

Since the crab grows 1 inch per year, and it would grow:

1 inch in the 1st year

2 inches in the 2nd year

3 inches in the 3rd year

4 inches in the 4th year

.

.

.

And so on.

However, in order to find the total length, you need to remember that the crab is originally 2-inches long. So the total length after x years would be the number of inches it grew over the years (x) + the length before the growth (2).

Thus, Length of crab in x years = x + 2

5 0
3 years ago
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If you are given a3=2 a5=16, find a100.
ra1l [238]

I suppose a_n denotes the n-th term of some sequence, and we're given the 3rd and 5th terms a_3=2 and a_5=16. On this information alone, it's impossible to determine the 100th term a_{100} because there are infinitely many sequences where 2 and 16 are the 3rd and 5th terms.

To get around that, I'll offer two plausible solutions based on different assumptions. So bear in mind that this is not a complete answer, and indeed may not even be applicable.

• Assumption 1: the sequence is arithmetic (a.k.a. linear)

In this case, consecutive terms <u>d</u>iffer by a constant d, or

a_n = a_{n-1} + d

By this relation,

a_{n-1} = a_{n-2} + d

and by substitution,

a_n = (a_{n-2} + d) + d = a_{n-2} + 2d

We can continue in this fashion to get

a_n = a_{n-3} + 3d

a_n = a_{n-4} + 4d

and so on, down to writing the n-th term in terms of the first as

a_n = a_1 + (n-1)d

Now, with the given known values, we have

a_3 = a_1 + 2d = 2

a_5 = a_1 + 4d = 16

Eliminate a_1 to solve for d :

(a_1 + 4d) - (a_1 + 2d) = 16 - 2 \implies 2d = 14 \implies d = 7

Find the first term a_1 :

a_1 + 2\times7 = 2 \implies a_1 = 2 - 14 = -12

Then the 100th term in the sequence is

a_{100} = a_1 + 99d = -12 + 99\times7 = \boxed{681}

• Assumption 2: the sequence is geometric

In this case, the <u>r</u>atio of consecutive terms is a constant r such that

a_n = r a_{n-1}

We can solve for a_n in terms of a_1 like we did in the arithmetic case.

a_{n-1} = ra_{n-2} \implies a_n = r\left(ra_{n-2}\right) = r^2 a_{n-2}

and so on down to

a_n = r^{n-1} a_1

Now,

a_3 = r^2 a_1 = 2

a_5 = r^4 a_1 = 16

Eliminate a_1 and solve for r by dividing

\dfrac{a_5}{a_3} = \dfrac{r^4a_1}{r^2a_1} = \dfrac{16}2 \implies r^2 = 8 \implies r = 2\sqrt2

Solve for a_1 :

r^2 a_1 = 8a_1 = 2 \implies a_1 = \dfrac14

Then the 100th term is

a_{100} = \dfrac{(2\sqrt2)^{99}}4 = \boxed{\dfrac{\sqrt{8^{99}}}4}

The arithmetic case seems more likely since the final answer is a simple integer, but that's just my opinion...

3 0
2 years ago
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