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anastassius [24]
3 years ago
14

Choose all the fractions that are equivalent to1/7. (Click all that apply.)

Mathematics
1 answer:
Alchen [17]3 years ago
8 0

Answer:

2/14, 4/28, 5/35, 7/70 or just 2/14

Step-by-step explanation:

Multiply both the numerator and denominator of 1/7 by 2, to get 2/14, or 2:14

And multiply the numerator and denominator of 1/7 by 3, to get 3/21, or 3:21. So 2:14 and 3:21 are two ratios that are equal to 1:7.

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A grading machine can grade 48 tests in 1 minute. How
S_A_V [24]
\bf \begin{array}{ccll}
tests&minutes\\
\text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\
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3 0
3 years ago
Find the radius of convergence, r, of the series. ∞ xn 2n − 1 n = 1 r = 1 find the interval, i, of convergence of the series. (e
Bingel [31]
Assuming the series is

\displaystyle\sum_{n\ge1}\frac{x^n}{2n-1}

The series will converge if

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We have

\displaystyle\lim_{n\to\infty}\left|\frac{\frac{x^{n+1}}{2(n+1)-1}}{\frac{x^n}{2n-1}}\right|=|x|\lim_{n\to\infty}\frac{\frac1{2n+1}}{\frac1{2n-1}}=|x|-\lim_{n\to\infty}\frac{2n-1}{2n+1}=|x|

So the series will certainly converge if -1, but we also need to check the endpoints of the interval.

If x=1, then the series is a scaled harmonic series, which we know diverges.

On the other hand, if x=-1, by the alternating series test we can show that the series converges, since

\left|\dfrac{(-1)^n}{2n-1}\right|=\dfrac1{2n-1}\to0

and is strictly decreasing.

So, the interval of convergence for the series is -1\le x.
6 0
3 years ago
5 < 4x = 2> 3x PLEASE HELP
Galina-37 [17]
Looks fishy to me, the way this problem combines <, = and > symbols.

But anyway.  Subtract 3x from both sides.  You'll get

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Then 5 < x = true, so you must find x values that are larger than 5.

That would be (5, infinity)

Please go back and ensure that you have copied down the original problem exactly as it appears.

8 0
3 years ago
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