All categories

3 years ago

If f(x) = 5x − 3, which expression represents its inverse function?

Mathematics

2 answers:

valentinak56 [21]3 years ago

7 0

Answer:

inverse of function is $\frac{x+3}{5}$ .

Step-by-step explanation:

Given:If f(x) = 5x − 3

To find: which expression represents its inverse function.

Solution We have given that

f(x) = 5x − 3

Step 1: take f(x) = y

y = 5x -3.

Step 2 : Interchange the x and y

x = 5y -3.

Solve for y

On adding both sides by number 3

x + 3 = 5y.

On dividing both sides by 5

y = $\frac{x+3}{5}$ .

Here y = $f(x)^{-1}$ = $\frac{x+3}{5}$ .

Therefore, inverse of function is $\frac{x+3}{5}$ .

hichkok12 [17]3 years ago

6 0

The inverse of f(x) = 5x - 3 is:
$f^{-1} (x) = \frac{3}{5} + \frac{x}{5}$

You might be interested in

1x 4 = 4 is a true statement. Which multiplication property is this an example of?

LekaFEV [45]

Answer:

Multiplicative Identity Property

Step-by-step explanation: The Product of what number times one will equal the same number.

6 0

3 years ago

Read 2 more answers

PLEASE ANSWER THE QUESTION CORRECTLY, I’LL GIVE YOU BRANLIEST! Please answer my other math questions if you solve this one, than

Alex73 [517]

Answer:

true

Step-by-step explanation:

4 0

3 years ago

Can someone help me please?

Helga [31]

You just find the lowest denominators

7 0

4 years ago

At which intervals is the function linear or nonlinear?

NeX [460]

Answer:

Linear non-linear

x<2 , x>5 2<x<5

Step-by-step explanation:

When a function is linear then the graph is a straight line

When the function is not linear then the graph is like a curve

From x = - infinity to 2 the graph is a straight line

So from - infinity to 2, the function is linear, that is x<2 function is linear

from 2 to 5 the graph is like a curve , so it is not linear

2<x<5 , the function is not linear

From x = 5 to infinity the graph is a straight line

So from 5 to infinity, the function is linear, that is x>5 function is linear

5 0

3 years ago

Rachna borrowed a certain sum at the rate of 15% per annum. if she paid at the end of the two years Rs.1290 as interest Compound

Natalija [7]

Given info:

Compound Interest = Rs.1290

Rate of Interest = 15% p.a

Time = 2 years

<h3>Formula we have to know:-</h3>

$\dag{\underline{\boxed{\sf{C.I = P \bigg(1 + \dfrac{R}{100} \bigg)^{n} - 1}}}}$

<u>Where</u>

C.I = Compound Interest

P = Principle

R = Rate of Interest

N = Time

$\textsf{ \underline{Solution-}}\\$

Here

C.I = Rs.1290

R = 15%

N = 2 years

Principle = ?

Now,Calculating the sum (Principle) borrowed by Rachna

$\quad{: \implies{\sf{C.I = \Bigg[P \bigg(1 + \dfrac{R}{100} \bigg)^{n} - 1 \Bigg]}}}$

Substituting the given values

$\quad{: \implies{\sf{1290 = \Bigg[P \bigg(1 + \dfrac{15}{100} \bigg)^{2} - 1 \Bigg]}}}$

$\quad{: \implies{\sf{1290 = \Bigg[P \bigg(\dfrac{(1 \times 100) + 15}{100} \bigg)^{2} - 1 \Bigg]}}}$

$\quad{: \implies{\sf{1290 = \Bigg[P \bigg(\dfrac{115}{100} \bigg)^{2} - 1 \Bigg]}}}$

$\quad{: \implies{\sf{1290 = \Bigg[P \bigg(\dfrac{115}{100} \times \dfrac{115}{100} \bigg) - 1 \Bigg]}}}$

$\quad{: \implies{\sf{1290 = \Bigg[P \bigg(\dfrac{13225}{10000} \bigg) - 1 \Bigg]}}}$

$\quad{: \implies{\sf{1290 = \Bigg[P \bigg( \cancel{\dfrac{13225}{10000}} \bigg) - 1 \Bigg]}}}$

$\quad{: \implies{\sf{1290 = \Bigg[P \bigg({1.3225 - 1} \bigg) \Bigg]}}}$

$\quad{: \implies{\sf{1290 = P \times {0.3225}}}}$

$\quad{: \implies{\sf{\dfrac{1290}{0.3225} = P}}}$

$\quad{: \implies{\sf{\dfrac{1290 \times 1000}{0.3225 \times 1000} = P}}}$

$\quad{: \implies{\sf{\dfrac{12900000}{3225} = P}}}$

$\quad{: \implies{\sf{\cancel{\dfrac{12900000}{3225}} = P}}}$

$\quad{: \implies{\sf{Rs.4000 = P}}}$

$\quad{\dag{\underline{\boxed{\tt{\blue{Principle} = \purple{Rs.4000 }}}}}}$

$\begin{gathered}\end{gathered}$

<u>Hence,</u>

The sum (Principle) is Rs.4000.

3 0

3 years ago