(2x - 3y)(4x - y) simplify for me

PLEASE ANSWER + BRAINLIEST!!!

grandymaker [24]

Width is 11k^2

Length is 5k^3 + 2k^2

Look at: https://www.wyzant.com/resources/answers/392371/what_is_the_length_and_width_of_the_rectangle

5 0

3 years ago

Julio and his sister bought 8 books and m magazines for $1 each, and then they split the cost. The amount of money that Julio sp

Ivahew [28]

Answer: No. both are independent quantity. Therefore, the number of books purchased does not affect the value of m.

Step-by-step explanation:

Given: Julio and his sister bought 8 books and m magazines for $1 each.

The amount of money that Julio spent is represented by the expression 1/2(m+8)

Thus, the amount she and her sister will pay is dependent on the number of books and the number of magazines, but the number of magazines and the number books are independent quantity.

Therefore, the number of books purchased does not affect the value of m.

3 0

3 years ago

Am I correct for 2 and 3?

Novay_Z [31]

yes. you are correct

8 0

3 years ago

Read 2 more answers

<img src="https://tex.z-dn.net/?f=%284x%20-%206y%29" id="TexFormula1" title="(4x - 6y)" alt="(4x - 6y)" align="absmiddle" class=

Natalija [7]

Answer:

I’m pretty sure it’s -2xy if it’s not I’m really sorry

Step-by-step explanation:

4-6=-2+xy=-2xy

6 0

2 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