It’s all there in the picture will give brainlest to correct answer

Three consecutive odd integers are such that the sum of the first and second is 31 less than 3 times the third. Find the integer

Otrada [13]

1st:x

2nd x+2

3rd: x+4

(x)+(x+2) = 3(x+4)-31

distribute

2x+2 = 3x+12-31

2x+2 =3x-19

subtract 2x from each side

2 = x-19

add 19 to each side

21 =x

Answer: 21,23,25

3 0

3 years ago

Write the trinomial as a square of a binomial or as an expression opposite to a square

timurjin [86]

(3a-7)^2

1. 9a^2-21a-21a+49
2. 3a(3a-7)-7(3a-7)
3. (3a-7)(3a-7)
4. Combine like terms

3 0

2 years ago

The Miller family likes to paddle along the North River.

miss Akunina [59]

Answer:

4.7 miles per day.

If number two is meaning to add 4.7(.5)= 2.3(3) plus the original 14 miles over a three day travel from number one. Then the total distance for the six day trip would be 14+ 2.3(3)= 20.9

If it is meaning a total of 6 days traveling at half the speed

4.7÷2= 2.3 2.3 × 6= 13.8 miles

3 0

2 years ago

Write in expanded form 5,002,190

makvit [3.9K]

5,000,000+2,000+100+90

5 0

3 years ago

Read 2 more answers

What is a quick and easy way to remember explicit and recursive formulas?

Oliga [24]

I always found derivation to be helpful in remembering. Since this question is tagged as at the middle school level, I assume you've only learned about arithmetic and geometric sequences.

First, remember what these names mean. An arithmetic sequence is a sequence in which consecutive terms are increased by a fixed amount; in other words, it is an additive sequence. If $a_n$ is the $n$ th term in the sequence, then the next term $a_{n+1}$ is a fixed constant (the common difference $d$ ) added to the previous term. As a recursive formula, that's

$a_{n+1}=a_n+d$

This is the part that's probably easier for you to remember. The explicit formula is easily derived from this definition. Since $a_{n+1}=a_n+d$ , this means that $a_n=a_{n-1}+d$ , so you plug this into the recursive formula and end up with

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

You can continue in this pattern, since every term in the sequence follows this rule:

$a_{n+1}=a_{n-1}+2d$
$a_{n+1}=(a_{n-2}+d)+2d$
$a_{n+1}=a_{n-2}+3d$
$a_{n+1}=(a_{n-3}+d)+3d$
$a_{n+1}=a_{n-3}+4d$

and so on. You start to notice a pattern: the subscript of the earlier term in the sequence (on the right side) and the coefficient of the common difference always add up to $n+1$ . You have, for example, $(n-2)+3=n+1$ in the third equation above.

Continuing this pattern, you can write the formula in terms of a known number in the sequence, typically the first one $a_1$ . In order for the pattern mentioned above to hold, you would end up with

$a_{n+1}=a_1+nd$

or, shifting the index by one so that the formula gives the $n$ th term explicitly,

$a_n=a_1+(n-1)d$

Now, geometric sequences behave similarly, but instead of changing additively, the terms of the sequence are scaled or changed multiplicatively. In other words, there is some fixed common ratio $r$ between terms that scales the next term in the sequence relative to the previous one. As a recursive formula,

$a_{n+1}=ra_n$

Well, since $a_n$ is just the term after $a_{n-1}$ scaled by $r$ , you can write

$a_{n+1}=r(ra_{n-1})=r^2a_{n-1}$

Doing this again and again, you'll see a similar pattern emerge:

$a_{n+1}=r^2a_{n-1}$
$a_{n+1}=r^2(ra_{n-2})$
$a_{n+1}=r^3a_{n-2}$
$a_{n+1}=r^3(ra_{n-3})$
$a_{n+1}=r^4a_{n-3}$

and so on. Notice that the subscript and the exponent of the common ratio both add up to $n+1$ . For instance, in the third equation, $3+(n-2)=n+1$ . Extrapolating from this, you can write the explicit rule in terms of the first number in the sequence:

$a_{n+1}=r^na_1$

or, to give the formula for $a_n$ explicitly,

$a_n=r^{n-1}a_1$

6 0

3 years ago