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2 years ago

Professor Andrews found that as the number of days absent increases, students' grades decrease. Professor Andrews has found:

Mathematics

1 answer:

n200080 [17]2 years ago

3 0

Answer:

The function relating student absent days to grades has a negative slope (grades decrease as absences increase).

Step-by-step explanation:

There isn't much information in the question, but here are a couple of observations. Prof. Andrews found:

The function relating student absent days to grades has a negative slope (grades decrease as absences increase).
Evidence suggesting his lectures do convey useful information
Students don't enjoy his class

Just kidding on the last 2, but all statements could be valid based on what little information is provided.

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(11/10)^2= pls help!!!!!!!!

AURORKA [14]

The answer is going to be 121/100 in exact form
in decimal form 1.21
in mixed number form 1 21/100

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3 years ago

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PLEASE HURRY

Aleks04 [339]

Answer:

Step-by-step explanation:

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4 years ago

Jervane is planning on buying a coffee and several donuts for a road trip. She wants to spend no more than $20 in total. Coffee

max2010maxim [7]

Answer:

Step-by-step explanation:

Given:

Total money Jervane can spend = $20.

Cost of a coffee = $3.

Cost of a donut = $2.

To find: Maximum number of donuts Jervane can buy.

Solution:

Let the number of donuts she can buy be $x$ .

She wants to spend no more than $20 in total.

So,

$3+2x\leq 20$

$\Rightarrow 2x\leqslant 20-3$

$\Rightarrow 2x\leqslant 17$

$\Rightarrow x\leqslant \frac{17}{2}$

$\Rightarrow x\leqslant 8 \frac{1}{2}$

As the number of donuts can only be a natural number.

Hence, the maximum number of donuts that Jervane can buy are 8.

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3 years ago

CAN SOMEONE HELP????????????????????????????????????????????????????????????????????????????

LuckyWell [14K]

Slope is -1/2

You can get that by isolating the x by moving it to the other side then divide the y value by the other side.

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3 years ago

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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 closed 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 closed-open 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 open interval $\boxed{-1 < x < 1}$ .

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2 years ago