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3 years ago

9

Brainliest,plz help

1 answer:

labwork [276]3 years ago

4 0

4 sqrt (32) + 6 sqrt (50)

4 sqrt (16*2) + 6 sqrt (2*25)

4 sqrt (16)*sqrt(2) + 6 sqrt (2)*sqrt(25)

4 *4*sqrt(2)+6sqrt(2)*5

16sqrt(2)+30sqrt(2)

46sqrt(2)

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A number has been round to 7700.what might the number

ankoles [38]

7699......................

3 0

3 years ago

Determine formula of the nth term 2, 6, 12 20 30,42

nalin [4]

Check the forward differences of the sequence.

If $\{a_n\} = \{2,6,12,20,30,42,\ldots\}$ , then let $\{b_n\}$ be the sequence of first-order differences of $\{a_n\}$ . That is, for n ≥ 1,

$b_n = a_{n+1} - a_n$

so that $\{b_n\} = \{4, 6, 8, 10, 12, \ldots\}$ .

Let $\{c_n\}$ be the sequence of differences of $\{b_n\}$ ,

$c_n = b_{n+1} - b_n$

and we see that this is a constant sequence, $\{c_n\} = \{2, 2, 2, 2, \ldots\}$ . In other words, $\{b_n\}$ is an arithmetic sequence with common difference between terms of 2. That is,

$2 = b_{n+1} - b_n \implies b_{n+1} = b_n + 2$

and we can solve for $b_n$ in terms of $b_1=4$ :

$b_{n+1} = b_n + 2$

$b_{n+1} = (b_{n-1}+2) + 2 = b_{n-1} + 2\times2$

$b_{n+1} = (b_{n-2}+2) + 2\times2 = b_{n-2} + 3\times2$

and so on down to

$b_{n+1} = b_1 + 2n \implies b_{n+1} = 2n + 4 \implies b_n = 2(n-1)+4 = 2(n + 1)$

We solve for $a_n$ in the same way.

$2(n+1) = a_{n+1} - a_n \implies a_{n+1} = a_n + 2(n + 1)$

Then

$a_{n+1} = (a_{n-1} + 2n) + 2(n+1) \\ ~~~~~~~= a_{n-1} + 2 ((n+1) + n)$

$a_{n+1} = (a_{n-2} + 2(n-1)) + 2((n+1)+n) \\ ~~~~~~~ = a_{n-2} + 2 ((n+1) + n + (n-1))$

$a_{n+1} = (a_{n-3} + 2(n-2)) + 2((n+1)+n+(n-1)) \\ ~~~~~~~= a_{n-3} + 2 ((n+1) + n + (n-1) + (n-2))$

and so on down to

$a_{n+1} = a_1 + 2 \displaystyle \sum_{k=2}^{n+1} k = 2 + 2 \times \frac{n(n+3)}2$

$\implies a_{n+1} = n^2 + 3n + 2 \implies \boxed{a_n = n^2 + n}$

6 0

2 years ago

If is parallel to , which statement must be true?

Igoryamba

I’m pretty sure it’s d

6 0

3 years ago

Plss do the working

djyliett [7]

Answer:

1386

Step-by-step explanation:

V=πr²*h

V=22/7*7²*9

V=1386

7 0

2 years ago

Ann Torbert purchased a truck for $11,000 on January 1, 2014. The truck will have an estimated salvage value of $1,000 at the en

alexandr402 [8]

Answer:

$Balance in accumulated depreciation=\frac{10000}{Total Estimated Value}*Units of activity$

Step-by-step explanation:

Salvage value=$1000

Purchased value=$11,000

In order to find the balance in accumulated depreciation at december 31,2015 using the units of activity we will use the following formula:

$Balance in accumulated depreciation=\frac{10000}{Total Estimated Value}*Units of activity$

In the above equation $10000 came from Purchased value - salvage Value

6 0

3 years ago

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