All categories

2 years ago

(Algebra 2) Find the inverse of f(x) and its domain.

Mathematics

1 answer:

elixir [45]2 years ago

4 0

Let f(x)=y
f^-1(x)=x
y=1/(x+5) - 1
y+1=1/(x+5)
(x+5)=1/(y+1)
x=1/(y+1)-5
f^-1(x)=1/(x+1) -5

denominator cannot equals to zero
x+1≠0
x≠-1

answer
B

You might be interested in

What is the solution to the equation

Dafna11 [192]

$\sqrt{4t+5}=3-\sqrt{t+5}\\\\D:4t+5\geq0\ \wedge\ t+5\geq0\ \wedge\ \sqrt{t+5}\leq3\\\\t\geq-\dfrac{5}{4}\ \wedge\ t\geq-5\ \wedge\ t\leq6$

therefore
$D:x\in\left< -\dfrac{5}{4};\ 6\right>$

$\sqrt{4t+5}=3-\sqrt{t+5}\ \ \ |^2\\\\(\sqrt{4t+5})^2=(3-\sqrt{t+5})^2\ \ \ |use:(a-b)^2=a^2-2ab+b^2\\\\4t+5=3^2-2\cdot3\cdot\sqrt{t+5}+(\sqrt{t+5})^2\\\\4t+5=9-6\sqrt{t+5}+t+5\ \ \ \ |-t\\\\3t+5=14-6\sqrt{t+5}\ \ \ \ |-14\\\\3t-9=-6\sqrt{t+5}\ \ \ \ |change\ signs$
$9-3t=6\sqrt{t+5}\ \ \ \ |:3\\\\3-t=2\sqrt{t+5}\ \ \ \ |^2\\\\(3-t)^2=(2\sqrt{t+5})^2\\\\3^2-2\cdot3\cdot t+t^2=4(t+5)\\\\9-6t+t^2=4t+20\ \ \ |-4t-20\\\\t^2-10t-11=0\\\\t^2+t-11t-11=0\\\\t(t+1)-11(t+1)=0\\\\(t+1)(t-11)=0\iff t+1=0\ \vee\ t-11=0\\\\t=-1\in D\ \vee\ t=11\notin D$
Answer: t = -1.

7 0

2 years ago

Read 2 more answers

Simplify the given expression. Write your answer with positive exponents.

irina [24]

Answer:

heres your answer

Step-by-step explanation:

here you go miss.

4 0

2 years ago

A researcher is interested in the texting habits of high school students in the United States. The researcher selects a group of

vlabodo [156]

Answer:

(a)High school students in the United States.

(b)100 Students

(c)Statistic

Step-by-step explanation:

(a)Since the researcher is interested in the texting habits of high school students in the United States, the population for this study consists of all high school students in the United States.

(b)It is most often improbable to carry out research on an entire population of study. A <u>sample is taken to represent the population</u> and the results obtained from the sample are assumed to hold for the entire population.

In the given study, the researcher selects a group of 100 students to represent the High School students in the United States. This is the sample for the study.

(c)The researcher calculated the average number of text messages that each individual sends each day from the sample of 100 Students. This is an example of a Statistic. A statistic is a numerical data obtained from a sample.

Note: If the data were obtained from the entire population, it would be a <u>parameter.</u>

3 0

3 years ago

Jenny made $221 for 13 hours of work. At the same rate, how many hours would she have to work to make $85?

Sever21 [200]

Answer: Jenny would have to work 5 hours to get $85

Step-by-step explanation:

Jenny makes $17 an hour (221/13 = 17) and 85/17 is 5

5 0

2 years ago

Suppose that one person in 10,000 people has a rare genetic disease. There is an excellent test for the disease; 98.8% of the pe

nirvana33 [79]

Answer:

A)The probability that someone who tests positive has the disease is 0.9995

B)The probability that someone who tests negative does not have the disease is 0.99999

Step-by-step explanation:

Let D be the event that a person has a disease

Let $D^c$ be the event that a person don't have a disease

Let A be the event that a person is tested positive for that disease.

P(D|A) = Probability that someone has a disease given that he tests positive.

We are given that There is an excellent test for the disease; 98.8% of the people with the disease test positive

So, P(A|D)=probability that a person is tested positive given he has a disease = 0.988

We are also given that one person in 10,000 people has a rare genetic disease.

So, $P(D)=\frac{1}{10000}$

Only 0.4% of the people who don't have it test positive.

$P(A|D^c)$ = probability that a person is tested positive given he don't have a disease = 0.004

$P(D^c)=1-\frac{1}{10000}$

Formula: $P(D|A)=\frac{P(A|D)P(D)}{P(A|D)P(D^c)+P(A|D^c)P(D^c)}$

$P(D|A)=\frac{0.988 \times \frac{1}{10000}}{0.988 \times (1-\frac{1}{10000}))+0.004 \times (1-\frac{1}{10000})}$

P(D|A)= $\frac{2470}{2471}$ =0.9995

P(D|A)= $0.9995$

A)The probability that someone who tests positive has the disease is 0.9995

(B)

$P(D^c|A^c)$ =probability that someone does not have disease given that he tests negative

$P(A^c|D^c)$ =probability that a person tests negative given that he does not have disease =1-0.004

=0.996

$P(A^c|D)$ =probability that a person tests negative given that he has a disease =1-0.988=0.012

Formula: $P(D^c|A^c)=\frac{P(A^c|D^c)P(D^c)}{P(A^c|D^c)P(D^c)+P(A^c|D)P(D)}$

$P(D^c|A^c)=\frac{0.996 \times (1-\frac{1}{10000})}{0.996 \times (1-\frac{1}{10000})+0.012 \times \frac{1}{1000}}$

$P(D^c|A^c)=0.99999$

B)The probability that someone who tests negative does not have the disease is 0.99999

8 0

3 years ago