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3 years ago

7

I was wondering how do you solve this?

1 answer:

Kruka [31]3 years ago

3 0

I am wondering the same thing about my life problems bro.

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2(-3х-6)please I need help

andrew-mc [135]

Step-by-step explanation:

$2( - 3x - 6) \\ - 6x + 12$

3 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

ABCD is a rectangle. What is the value of x?

kvv77 [185]

The answer is 33 meters
use the Pythagorean theorem
56^2+x^2=65^2
3136+x^2=4225
x^2=1089
x=33

6 0

3 years ago

Read 2 more answers

Find the sum of the geometric series for which a = 160, r = 0.5, and n = 6.

harina [27]

$\\ \sf\longmapsto S_n=\dfrac{a(1-r^n)}{1-r}$

$\\ \sf\longmapsto S_n=\dfrac{160(1-0.5^6)}{1-0.5}$

$\\ \sf\longmapsto S_n=\dfrac{160(1-0.015625)}{0.5}$

$\\ \sf\longmapsto S_n=\dfrac{160(0.984378)}{0.5}$

$\\ \sf\longmapsto S_n=80(0.984378)$

$\\ \sf\longmapsto S_n=78.75$

6 0

3 years ago

Read 2 more answers

WILL MARK BRAINLIEST PLEASE HELP I NEED SOME SMART PEOPLE WILL MARK BRAINLIEST!!!

lions [1.4K]

Answer:

The answer is $6\frac{5}{8}$

Step-by-step explanation:

-12 pounds + $5\frac{3}{8}$

-12+ $5\frac{3}{8}$

=-12+5+ $\frac{3}{8}$

=(-12+5)+ $\frac{3}{8}$

=(-7)+ $\frac{3}{8\\}$

= $-6\frac{5}{8}$

Mr.Yoo's weight dropped by $6\frac{5}{8}$

7 0

2 years ago