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olasank [31]
3 years ago
9

I am a number between 80 and 100 my ones digit is two less than my tens digit I am a prime number what number am I

Mathematics
1 answer:
Aneli [31]3 years ago
5 0
97. the tens number here is 9 and the unit number if 7, from this 7 is less than 9 by 2
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If m_PQT=(3x+47) and m_SQR=(6x+25), find m_SQR
marishachu [46]

Answer:

6x+25

Step-by-step explanation:

here is given that

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Please help me!!<br> 20 points
vovangra [49]

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2 years ago
Is the following relation a function?
dimulka [17.4K]

Answer:

yes

Step-by-step explanation:

A function is any graph that the x values do not repeat.

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8 0
3 years ago
Let f(x) = [infinity] xn n2 n = 1. find the intervals of convergence for f. (enter your answers using interval notation. ) find
inna [77]

Best guess for the function is

\displaystyle f(x) = \sum_{n=1}^\infty \frac{x^n}{n^2}

By the ratio test, the series converges for

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

When x=1, f(x) is a convergent p-series.

When x=-1, f(x) is a convergent alternating series.

So, the interval of convergence for f(x) is the <em>closed</em> interval \boxed{-1 \le x \le 1}.

The derivative of f is the series

\displaystyle f'(x) = \sum_{n=1}^\infty \frac{nx^{n-1}}{n^2} = \frac1x \sum_{n=1}^\infty \frac{x^n}n

which also converges for |x| by the ratio test:

\displaystyle \lim_{n\to\infty} \left|\frac{x^{n+1}}{n+1} \cdot \frac n{x^n}\right| = |x| \lim_{n\to\infty} \frac{n}{n+1} = |x| < 1

When x=1, f'(x) becomes the divergent harmonic series.

When x=-1, f'(x) is a convergent alternating series.

The interval of convergence for f'(x) is then the <em>closed-open</em> interval \boxed{-1 \le x < 1}.

Differentiating f once more gives the series

\displaystyle f''(x) = \sum_{n=1}^\infty \frac{n(n-1)x^{n-2}}{n^2} = \frac1{x^2} \sum_{n=1}^\infty \frac{(n-1)x^n}{n} = \frac1{x^2} \left(\sum_{n=1}^\infty x^n - \sum_{n=1}^\infty \frac{x^n}n\right)

The first series is geometric and converges for |x|, endpoints not included.

The second series is f'(x), which we know converges for -1\le x.

Putting these intervals together, we see that f''(x) converges only on the <em>open</em> interval \boxed{-1 < x < 1}.

6 0
2 years ago
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