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3 years ago

CAN SOMEONE HELP ME PLZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZ

Mathematics

1 answer:

Anettt [7]3 years ago

7 0

The LCD is the product of (x -2)(x + 1)
To clear the fractions, multiply each term by the part of the LCD that the denominator is missing.
x(x + 1) + (x - 2)(x - 1) = -1(x - 2)(x + 1)
Distribute and foil, then combine like terms.
x^2 + x + x^2 - 3x + 2 = -x^2 + x + 2
2x^2 - 2x + 2 = -x^2 + x + 2
add x^2 to both sides
3x^2 -2x + 2 = x + 2
subtract from both sides
3x^2 - 3x + 2 = 2
subtract 2 from both sides.
3x^2 - 3x = 0
Factor out the GCF
3x(x - 1) = 0
Set each factor equal to zero and solve.
3x = 0 x - 1 = 0
x = 0 x = 1

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Find the missing information from the following relationships

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What is the value of the expression: 2 - (3x + 5y) if x = -3 and y = 4?

WARRIOR [948]

<h3>Hello there today we will solve your problem</h3>

here is our equation,

$2-(3x+5y)$ ,

Now we will plug in our numbers

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${\boxed{x=-3}}$
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nirvana33 [79]

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Step-by-step explanation:

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3 years ago

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,

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

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3 years ago

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oksian1 [2.3K]

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