Ask question

Ask question

All categories

rosijanka [135]

3 years ago

8

La diferencia entre un numero y ocho

2 answers:

cupoosta [38]3 years ago

4 0

Answer:

hzyuzj ahuaaia hs7w8aijw. hu7sisj. shususijs bhsusj

Andreas93 [3]3 years ago

3 0

Answer:

lo siento, no lo sé

Step-by-step explanation:

You might be interested in

Tess needs to buy sand for her children’s playground. The sand costs $8 per ton, and there is a delivery cost of $50 with each o

Butoxors [25]

Answer:

All positive rational numbers greater than 50

Step-by-step explanation:

Since the delivery cost is fixed per trip, the cost incurred can be represented by the equation

y(x)=8x+50 so in case Tess wants only 1 tone, the cost will be y(x)=8(1)+50=58

Therefore, it's proper to conclude that all numbers in the possible range are positive rational numbers greater than 50

6 0

2 years ago

Which of these is true? A) 357 > 571 B) 432 > 571 C) 571 > 432 D) 571 < 357

elena55 [62]

Answer is C) 571>432

4 0

3 years ago

In ΔABC, BC = 13, CA = 20 and AB = 19. Which statement about the angles of ΔABC must be true?

I am Lyosha [343]

Answer:

m∠B > m∠C > m∠A

Step-by-step explanation:

The angle opposite the largest side in a triangle is the largest, and the angle opposite the shortest side is the smallest.

4 0

1 year ago

HELP HELP HELP HELP HELP HELP

mixas84 [53]

Answer:

mean: 14

median: 13.5

range: 8

Step-by-step explanation:

Hope this helps! Sorry if I am late.

4 0

3 years ago

Read 2 more answers

Find a power series for the function, centered at c, and determine the interval of convergence. f(x) = 9 3x + 2 , c = 6

san4es73 [151]

Answer:

$\frac{9}{3x + 2} = 1 - \frac{1}{3}(x - \frac{7}{3}) + \frac{1}{9}(x - \frac{7}{3})^2 - \frac{1}{27}(x - \frac{7}{3})^3 ........$

The interval of convergence is: $(-\frac{2}{3},\frac{16}{3})$

Step-by-step explanation:

Given

$f(x)= \frac{9}{3x+ 2}$

$c = 6$

The geometric series centered at c is of the form:

$\frac{a}{1 - (r - c)} = \sum\limits^{\infty}_{n=0}a(r - c)^n, |r - c| < 1.$

Where:

$a \to$ first term

$r - c \to$ common ratio

We have to write

$f(x)= \frac{9}{3x+ 2}$

In the following form:

$\frac{a}{1 - r}$

So, we have:

$f(x)= \frac{9}{3x+ 2}$

Rewrite as:

$f(x) = \frac{9}{3x - 18 + 18 +2}$

$f(x) = \frac{9}{3x - 18 + 20}$

Factorize

$f(x) = \frac{1}{\frac{1}{9}(3x + 2)}$

Open bracket

$f(x) = \frac{1}{\frac{1}{3}x + \frac{2}{9}}$

Rewrite as:

$f(x) = \frac{1}{1- 1 + \frac{1}{3}x + \frac{2}{9}}$

Collect like terms

$f(x) = \frac{1}{1 + \frac{1}{3}x + \frac{2}{9}- 1}$

Take LCM

$f(x) = \frac{1}{1 + \frac{1}{3}x + \frac{2-9}{9}}$

$f(x) = \frac{1}{1 + \frac{1}{3}x - \frac{7}{9}}$

So, we have:

$f(x) = \frac{1}{1 -(- \frac{1}{3}x + \frac{7}{9})}$

By comparison with: $\frac{a}{1 - r}$

$a = 1$

$r = -\frac{1}{3}x + \frac{7}{9}$

$r = -\frac{1}{3}(x - \frac{7}{3})$

At c = 6, we have:

$r = -\frac{1}{3}(x - \frac{7}{3}+6-6)$

Take LCM

$r = -\frac{1}{3}(x + \frac{-7+18}{3}+6-6)$

r = -\frac{1}{3}(x + \frac{11}{3}+6-6)

So, the power series becomes:

$\frac{9}{3x + 2} = \sum\limits^{\infty}_{n=0}ar^n$

Substitute 1 for a

$\frac{9}{3x + 2} = \sum\limits^{\infty}_{n=0}1*r^n$

$\frac{9}{3x + 2} = \sum\limits^{\infty}_{n=0}r^n$

Substitute the expression for r

$\frac{9}{3x + 2} = \sum\limits^{\infty}_{n=0}(-\frac{1}{3}(x - \frac{7}{3}))^n$

Expand

$\frac{9}{3x + 2} = \sum\limits^{\infty}_{n=0}[(-\frac{1}{3})^n* (x - \frac{7}{3})^n]$

Further expand:

$\frac{9}{3x + 2} = 1 - \frac{1}{3}(x - \frac{7}{3}) + \frac{1}{9}(x - \frac{7}{3})^2 - \frac{1}{27}(x - \frac{7}{3})^3 ................$

The power series converges when:

$\frac{1}{3}|x - \frac{7}{3}| < 1$

Multiply both sides by 3

$|x - \frac{7}{3}|$

Expand the absolute inequality

$-3 < x - \frac{7}{3}$

Solve for x

$\frac{7}{3} -3 < x$

Take LCM

$\frac{7-9}{3} < x$

$-\frac{2}{3} < x$

The interval of convergence is: $(-\frac{2}{3},\frac{16}{3})$

6 0

2 years ago