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Tcecarenko [31]
2 years ago
6

Find the rule.

Mathematics
1 answer:
mezya [45]2 years ago
7 0

The <em>first</em> series uses a <em>linear</em> function with - 1 as <em>first</em> element and 1 as <em>common</em> difference, then the rule corresponding to the series is y = |- 1 + x|.

The <em>second</em> series uses a <em>linear</em> function with - 3 as <em>second</em> element as 2 as <em>common</em> difference, then the rule corresponding to the series is y = |- 3 + 2 · x|.

<h3>What is the pattern and the function behind a given series?</h3>

In this problem we have two cases of <em>arithmetic</em> series, which are sets of elements generated by a condition in the form of <em>linear</em> function and inside <em>absolute</em> power. <em>Linear</em> <em>functions</em> used in these series are of the form:

y = a + r · x      (1)

Where:

  • a - Value of the first element of the series.
  • r - Common difference between two consecutive numbers of the series.
  • x - Index of the element of the series.

The <em>first</em> series uses a <em>linear</em> function with - 1 as <em>first</em> element and 1 as <em>common</em> difference, then the rule corresponding to the series is y = |- 1 + x|.

The <em>second</em> series uses a <em>linear</em> function with - 3 as <em>second</em> element as 2 as <em>common</em> difference, then the rule corresponding to the series is y = |- 3 + 2 · x|.

To learn more on series: brainly.com/question/15415793

#SPJ1

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which expressions are equivalent to the first one? I don't understand how to determine that so please explain. Thanks!​
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9514 1404 393

Answer:

  (a) -(x+7)/y

  (b) (x+7)/-y

Step-by-step explanation:

There are several ways you can show expressions are equivalent. Perhaps the easiest and best is to put them in the same form. For an expression such as this, I prefer the form of answer (a), where the minus sign is factored out and the numerator and denominator have positive coefficients.

The given expression with -1 factored out is ...

  \dfrac{-x-7}{y}=\dfrac{1(x+7)}{y}=\boxed{-\dfrac{x+7}{y}} \quad\text{matches A}

Likewise, the expression of (b) with the minus sign factored out is ...

  \dfrac{x+7}{-y}=\boxed{-\dfrac{x+7}{y}}

On the other hand, simplifying expression (c) gives something different.

  \dfrac{-x-7}{-y}=\dfrac{-(x+7)}{-(y)}=\dfrac{x+7}{y} \qquad\text{opposite the given expression}

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Another way you can write the expression is term-by-term with the terms in alpha-numeric sequence (so they're more easily compared).

  Given: (-x-7)/y = (-x/y) +(-7/y)

  (a) -(x+7)/y = (-x/y) +(-7/y)

  (b) (x+7)/(-y) = (-x/y) +(-7/y)

  (c) (-x-7)/(-y) = (x/y) +(7/y) . . . . not the same.

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Of course, you need to know the use of the distributive property and the rules of signs.

  a(b+c) = ab +ac

  -a/b = a/(-b) = -(a/b)

  -a/(-b) = a/b

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<u>Summary</u>: The given expression matches (a) and (b).

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<em>Additional comments</em>

Sometimes, when I'm really stuck trying to see if two expressions are equal, I subtract one from the other. If the difference is zero, then I know they are the same. Looking at (b), we could compute ...

  \left(\dfrac{-x-7}{y}\right)-\left(\dfrac{x+7}{-y}\right)=\dfrac{-y(-x-7)-y(x+7)}{-y^2}\\\\=\dfrac{xy+7y-xy-7y}{-y^2}=\dfrac{0}{-y^2}=0

Yet another way to check is to substitute numbers for the variables. It is a good idea to use (at least) one more set of numbers than there are variables, just to make sure you didn't accidentally find a solution where the expressions happen to be equal. We can use (x, y) = (1, 2), (2, 3), and (3, 5) for example.

The given expression evaluates to (-1-7)/2 = -4, (-2-7)/3 = -3, and (-3-7)/5 = -2.

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(b) evaluates to (1+7)/-2 = -4, (2+7)/-3 = -3, (3+7)/-5 = -2, same as given

(c) evaluates to (-1-7)/-2 = 4, different from given

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