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4 years ago

8

Determine wether the following relationships are functions (6,1)(4,2)(6,-3)(2,5)

1 answer:

IRISSAK [1]4 years ago

5 0

Answer:

non- function

Step-by-step explanation:

any X being REPEATED (the first number of each pair) is non-function.

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Eighty-one million four hundred twelve thousand six hundred fifty

Jobisdone [24]

Answer:

81,412,650

Step-by-step explanation:

im not sure if this is what ur asking for??

8 0

4 years ago

Determine the number of solutions for each equation

Kay [80]

8. There is no solution. The absolute value of a number is always positive.
9. There are two solutions. Since the absolute value of a number is always positive, a solution can either be 4 or -4.
10. There is no solution because of the same reason for number 8.
11. There is one solution. There really is not -0.
12. There are two solutions. Since the absolute value is always positive, the solution can be either 15 or -15.

3 0

3 years ago

Find a power series representation for the function. (Give your power series representation centered at x = 0.) f(x) = x2/(x4 +

ella [17]

Answer:

Given the function: $f(x) =\frac{x^2}{x^4+16}$

A geometric series is of the form of :

$\sum_{n=0}^{\infty} ar^n$

Now, rewrite the given function in the form of $\frac{a}{1-r}$ so that we can express the representation as a geometric series.

$\frac{x^2}{x^4+16}$

Now, divide numerator and denominator by $x^4$ we get;

$\frac{\frac{1}{x^2}}{1+\frac{16}{x^4}}$ = $\frac{\frac{1}{x^2}}{1+(\frac{4}{x^2})^2}$

Therefore, we now depend on the geometric series which is;

$\frac{1}{1+x} =\sum_{n=0}^{\infty} (-1)^n x^n$

let $x \rightarrow x^2$ then,

$\frac{1}{1+x^2} =\sum_{n=0}^{\infty} (-1)^n x^{2n}$

to get the power series let $x \rightarrow \frac{4}{x^2}$

so,

$\frac{1}{1+(\frac{4}{x^2})^2} =\sum_{n=0}^{\infty} (-1)^n (\frac{4}{x^2})^{2n}$

Multiply both side by $\frac{1}{x^2}$ we get;

$\frac{\frac{1}{x^2}}{1+(\frac{4}{x^2})^2} =\frac{1}{x^2} \cdot \sum_{n=0}^{\infty} (-1)^n (\frac{4}{x^2})^{2n}$

or

$\frac{\frac{1}{x^2}}{1+(\frac{4}{x^2})^2} =x^{-2} \cdot \sum_{n=0}^{\infty} (-1)^n (16)^n (x^{-2})^{2n}$

or

$\frac{\frac{1}{x^2}}{1+(\frac{4}{x^2})^2} =\sum_{n=0}^{\infty} (-1)^n (16)^n x^{-4n} \cdot x^{-2}$

Using $x^n \cdot x^m = x^{n+m}$

we have,

$\frac{\frac{1}{x^2}}{1+(\frac{4}{x^2})^2} =\sum_{n=0}^{\infty} (-1)^n (16)^n x^{-4n-2}$

therefore, the power series representation centered at x =0 for the given function is: $\sum_{n=0}^{\infty} (-1)^n (16)^n x^{-4n-2}$

6 0

3 years ago

I give brainliest

xxMikexx [17]

Functions are accurate through having one x intercepts which is how you can find which are functions or not.
Answer- B,D,E

4 0

3 years ago

What is the difference in lengths of the plants that measured 1/4 and 1/8?

Natasha2012 [34]

Answer:

1/8

Step-by-step explanation:

1/4 = 2/8 2/8-1/8=1/8

7 0

4 years ago