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3 years ago

14

Solve [3(4-1)-7] - [6-(4-9)]

1 answer:

d1i1m1o1n [39]3 years ago

7 0

Answer:

-9

Step-by-step explanation:

You might be interested in

Find AC please, help is very much appreciated.

fenix001 [56]

Answer:

AC ≈ 67.9

Step-by-step explanation:

15² = 9 (3x + 1 + 9)

225 = 9 (3x + 10)

225 = 27x + 90

135 = 27x

x ≈ 19.29

AC = (3x + 1 + 9)

AC = 3 (19.2857...) + 1 + 9

AC = 57.857... + 1 + 9

AC = 57.857... + 10

AC ≈ 67.9

4 0

3 years ago

Find the area of the figure.

11111nata11111 [884]

Answer:

180 sq.cm

Step-by-step explanation:

A = 10 x 10 + 16 x 5 x 1/2 x 2

A = 100 + 16 x 5

A = 100 + 80

A = 180 square cm

Hope that helps!

8 0

3 years ago

Read 2 more answers

45 is what percent of 150

inn [45]

Divide 45 by 150 then multiply by 100:

45 / 150 = 0.3

0.3 x 100 = 30%

The answer is 30%

7 0

3 years ago

Read 2 more answers

Use the graph to determine the input value for

musickatia [10]

Answer: x=1.5

Step-by-step explanation:

8 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