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3 years ago

The sum of 4 consecutive integers is 354. What is the second number in this sequence?

Mathematics

1 answer:

Ulleksa [173]3 years ago

8 0

354 / 4 = 88.5

so2 numbers below that and 2 numbers above that:

87 + 88 + 89 +90 =

2nd number is 88

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Expand.<br> Your answer should be a polynomial in standard form.<br> (x-3)(x-4)=

nekit [7.7K]

Answer:

Step-by-step explanation:

So, it's x*x+x*-3+-4*x+3*-3

That simplifies to x^2 - 7x + 12

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3 years ago

How many digits follow the decimal point in the product of 23.893 and 47.21 ?

DENIUS [597]

The product is 1,127.98853 ⇨ 5 digits follow after the decimal point. Hope that helps

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3 years ago

Read 2 more answers

3×-y=5,2×+y=5 pleaseplease

Tanya [424]

Answer:

x = 2

y = 1

Step-by-step explanation:

3x - y = 5..............(1)× 1

2x + y = 5...….......(2)× 1

3x - y = 5..............(3)

2x + y = 5.............(4)

Cancel y

3x + 2x = 5 + 5

5x = 10

Divide both sides by 5

x = 2

Substitute for the value of y

3x - y = 5

3(2) - y = 5

6 - y = 5

6 - 5 = y

1 = y

y = 1

6 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

Raj recorded the scores of six of his classmates in the table.

WINSTONCH [101]

98-70 is 28 meaning that the range is 28

4 0

3 years ago