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3 years ago

sue had 1 million dollars and ben had 1 dollars how much more did sue had after they spent 6 dollars out of there account

Mathematics

1 answer:

Naddika [18.5K]3 years ago

3 0

Ben had -5 dollars, and sue had nine hundred ninety-nine thousand nine hundred ninety-four dollars.

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3. I think of a number. I subtract 10 and add 5. I then double it. My

ella [17]

Answer:

Step-by-step explanation:

You can divide 50 by 2 which is 25

Then subtract 5

Add 10

And you get your number

It is really simple really ☺️

5 0

3 years ago

Teds needs a average of at least 70 on his three history test. he has already scored 85 and 60 on two test. what is the minimum

amm1812

Teds minimum grade he needs on his third test is 65

85 + 60 + 65 = 210

210 divide by three (as there are three tests) = 70

4 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 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 $\boxed{-1 < x < 1}$ .

6 0

2 years ago

How much times can 900 go into 8,000,000? Please help its due today!!

lakkis [162]

Answer:

8,888.888

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

5 9 1-2-3 3.4.5 5.6.7 +... to 0 = 1+3 log2.

konstantin123 [22]

Answer:

ask your teacher so you can get better grades

6 0

3 years ago