I=P(1+(r/n))^nt
I is the value owed
r is the rate in a decimal
n is the times compounded (annually =1, quarterly =4, monthly =12 etc.)
t is time invested
I=8000(1+(.109/1))^5x1
I=8000(1.109)^5
I=8000(1.67748)
$13419.84

6 0

3 years ago

Rewrite as a simplified fraction. 0.1 with the 1 repeating

astraxan [27]

Answer:

$\frac{1}{9}$

Step-by-step explanation:

$\frac{1}{9} =0.111111111111111111$

7 0

3 years ago

At which values of x does the function F(x) have a vertical asymptote? Check all that apply. F(x)=2/3x(x-1)(x 5).

Alekssandra [29.7K]

The function F(x) have a vertical asymptote at x =0 and x =1.

Given that

The given Function;

$\rm F(x)=\dfrac{2}{3x(x-1)(x^ 5)}$

We have to determine

At which value the function F(x) has a vertical asymptote.

According to the question

The given Function;

$\rm F(x)=\dfrac{2}{3x(x-1)(x^ 5)}$

The function is undefined when the denominator becomes 0. Find values of x for which the denominator is 0:

Then,

$\rm 3x (x-1) (x^5)= 0$

$\rm x^5 =0, \ \ x=0\\\\ x-1 = 0, \ \ x =1\\$

The value of x is 0 and 1.

Hence, the function F(x) have a vertical asymptote at x =0 and x =1.

To know more about Parabola click the link given below.

brainly.com/question/12868913

7 0

2 years ago