All categories

3 years ago

Find the amount of the annuity. Round to the nearest cent.

Mathematics

1 answer:

jasenka [17]3 years ago

8 0

FVAO=1500[((1+0.08)^(15)-1)/0.08]
FVAO=40728.17

You might be interested in

Please help Im timed!

Natalka [10]

15:10
3:2
Hope this helps :)

4 0

4 years ago

Which of the following is the x-intercept for 3y - 4x = 8?

expeople1 [14]

The answer is E because 3 x 0 - 4 x -2 = 8

3 0

4 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

Evaluate 11c3 and 11p4

In-s [12.5K]

$_{n}C_{r} = \frac{n!}{r!(n-r)!} \\ \\ _{11}C_{3}= \frac{11!}{3!*(11-3)!} =\frac{11*10*9*8*7*6*5*4*3*2*1}{3*2*1*8*7*6*5*4*3*2*1} =\frac{11*10*9}{3*2} =165 \\ \\_{11}C_{4}=\frac{11!}{4!(11-4)!} =\frac{11*10*9*8*7!}{4!*7!} =\frac{11*10*9*8}{4*3*2*1} =330$

7 0

3 years ago

Read 2 more answers

Benjamin’s and David’s ages add up to 36 years. The sum of twice their respective ages also add up to 72 years. Find their ages

nydimaria [60]

Answer:Benjamin’s age = 30 and David’s age = 6

Explanation:

Let x be the age of Benjamin
Let y be the age of David

x + y = 36

And 2x + 2y = 72

If we plug 30 for x and 6 for y we get:

30 + 6 = 36
36 = 36

2(30) + 2(6) = 72
60 + 12 = 72
72 = 72

=

5 0

3 years ago

Read 2 more answers