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2 years ago

12

Z/3 <= 3 solve for z

1 answer:

Molodets [167]2 years ago

4 0

Answer:

assuming <= is less than or equal to

then z/3 is less than or equal to 3

z had to be less than 9

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Step-by-step explanation:

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3 years ago

Find an equivalent fraction with the lowest common denominator. <br> 9/10+2/3

mafiozo [28]

Answer: 1-17/30, or 47/30

Step-by-step explanation:

Find the common denominator. In this case, it’s 30. Multiply the numerators to fit the new denominator. (10•3=30, so 9•3=27, and 2•10=20.)

Your new fractions are 27/30 and 20/30. Now add the numerators.

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Your new fraction is 47/30.

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3 years ago

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

3 years ago