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nataly862011 [7]
3 years ago
14

6) Which relation describes a function? A) {(0, 0), (0, 2), (2, 0), (2, 2)} B) {(−2, −3), (−3, −2), (2, 3), (3, 2)} C) {(2, −1),

(2, 1), (3, −1), (3, 1)} D) {(2, 2), (2, 3), (3, 2), (3, 3)}
Mathematics
1 answer:
navik [9.2K]3 years ago
6 0

Answer:

B) {(−2, −3), (−3, −2), (2, 3), (3, 2)}

Step-by-step explanation:

B) {(−2, −3), (−3, −2), (2, 3), (3, 2)}

each input of x (-2), (-3), (2), (3) does not have duplicated output value

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Irina-Kira [14]

Answer:  see attachment

<u>Step-by-step explanation:</u>

Put the numbers in order from smallest to biggest:

1, 2, 2, 3, 3, 3,         3, 4, 4, 4, 5, 6

        ↓              ∨                ↓

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Q1 median = (2+3)/2 = 2.5  <---- left side of box plot

Median = (3+3)/2 = 3          <---- Vertical Line of box plot

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3 0
3 years ago
Find the radius of convergence, r, of the series. ? n2xn 7 · 14 · 21 · ? · (7n) n = 1
defon
I'm guessing the series is supposed to be

\displaystyle\sum_{n=1}^\infty\frac{n^2x^n}{7\cdot14\cdot21\cdot\cdots\cdot(7n)}

By the ratio test, the series converges if the following limit is less than 1.

\displaystyle\lim_{n\to\infty}\left|\frac{\frac{(n+1)^2x^{n+1}}{7\cdot14\cdot21\cdot\cdots\cdot(7n)\cdot(7(n+1))}}{\frac{n^2x^n}{7\cdot14\cdot21\cdot\cdots\cdot(7n)}}\right|

The first n terms in the numerator's denominator cancel with the denominator's denominator:

\displaystyle\lim_{n\to\infty}\left|\frac{\frac{(n+1)^2x^{n+1}}{7(n+1)}}{n^2x^n}\right|

|x^n| also cancels out and the remaining factor of |x| can be pulled out of the limit (as it doesn't depend on n).

\displaystyle|x|\lim_{n\to\infty}\left|\frac{\frac{(n+1)^2}{7(n+1)}}{n^2}\right|=|x|\lim_{n\to\infty}\frac{|n+1|}{7n^2}=0

which means the series converges everywhere (independently of x), and so the radius of convergence is infinite.
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