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NemiM [27]
3 years ago
11

Consider the function represented by 9x + 3y = 12 with x as the independent variable. How can this function be written using

Mathematics
1 answer:
klio [65]3 years ago
4 0

Answer:

f(x) = - 3x + 4

Step-by-step explanation:

Note that y = f(x)

Given

9x + 3y = 12 ← Rearrange making y the subject

Subtract 9x from both sides

3y = - 9x + 12 ( divide all term by 3 )

y = - 3x + 4, thus

f(x) = - 3x + 4

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Step-by-step explanation:

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What is the slope of the line represented by the equation y = –2/3 – 5x
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3 years ago
Use the Limit Comparison Test to determine whether the series converges.
lianna [129]

Answer:

The infinite series \displaystyle \sum\limits_{k = 1}^{\infty} \frac{8/k}{\sqrt{k + 7}} indeed converges.

Step-by-step explanation:

The limit comparison test for infinite series of positive terms compares the convergence of an infinite sequence (where all terms are greater than zero) to that of a similar-looking and better-known sequence (for example, a power series.)

For example, assume that it is known whether \displaystyle \sum\limits_{k = 1}^{\infty} b_k converges or not. Compute the following limit to study whether \displaystyle \sum\limits_{k = 1}^{\infty} a_k converges:

\displaystyle \lim\limits_{k \to \infty} \frac{a_k}{b_k}\; \begin{tabular}{l}\\ $\leftarrow$ Series whose convergence is known\end{tabular}.

  • If that limit is a finite positive number, then the convergence of the these two series are supposed to be the same.
  • If that limit is equal to zero while a_k converges, then b_k is supposed to converge, as well.
  • If that limit approaches infinity while a_k does not converge, then b_k won't converge, either.

Let a_k denote each term of this infinite Rewrite the infinite sequence in this question:

\begin{aligned}a_k &= \frac{8/k}{\sqrt{k + 7}}\\ &= \frac{8}{k\cdot \sqrt{k + 7}} = \frac{8}{\sqrt{k^2\, (k + 7)}} = \frac{8}{\sqrt{k^3 + 7\, k^2}} \end{aligned}.

Compare that to the power series \displaystyle \sum\limits_{k = 1}^{\infty} b_k where \displaystyle b_k = \frac{1}{\sqrt{k^3}} = \frac{1}{k^{3/2}} = k^{-3/2}. Note that this

Verify that all terms of a_k are indeed greater than zero. Apply the limit comparison test:

\begin{aligned}& \lim\limits_{k \to \infty} \frac{a_k}{b_k}\; \begin{tabular}{l}\\ $\leftarrow$ Series whose convergence is known\end{tabular}\\ &= \lim\limits_{k \to \infty} \frac{\displaystyle \frac{8}{\sqrt{k^3 + 7\, k^2}}}{\displaystyle \frac{1}{{\sqrt{k^3}}}}\\ &= 8\left(\lim\limits_{k \to \infty} \sqrt{\frac{k^3}{k^3 + 7\, k^2}}\right) = 8\left(\lim\limits_{k \to \infty} \sqrt{\frac{1}{\displaystyle 1 + (7/k)}}\right)\end{aligned}.

Note, that both the square root function and fractions are continuous over all real numbers. Therefore, it is possible to move the limit inside these two functions. That is:

\begin{aligned}& \lim\limits_{k \to \infty} \frac{a_k}{b_k}\\ &= \cdots \\ &= 8\left(\lim\limits_{k \to \infty} \sqrt{\frac{1}{\displaystyle 1 + (7/k)}}\right)\\ &= 8\left(\sqrt{\frac{1}{\displaystyle 1 + \lim\limits_{k \to \infty} (7/k)}}\right) \\ &= 8\left(\sqrt{\frac{1}{1 + 0}}\right) \\ &= 8 \end{aligned}.

Because the limit of this ratio is a finite positive number, it can be concluded that the convergence of \displaystyle a_k &= \frac{8/k}{\sqrt{k + 7}} and \displaystyle b_k = \frac{1}{\sqrt{k^3}} are the same. Because the power series \displaystyle \sum\limits_{k = 1}^{\infty} b_k converges, (by the limit comparison test) the infinite series \displaystyle \sum\limits_{k = 1}^{\infty} a_k should also converge.

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Answer:

B

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