All categories

3 years ago

Find the inverse of g(n) = - 3 - 1

Mathematics

1 answer:

melisa1 [442]3 years ago

7 0

Answer:

$g'(n) = -\frac{n+1}{3}$

Step-by-step explanation:

Given

$g(n) = -3n - 1$

Required

Determine the inverse

$g(n) = -3n - 1$

Replace g(n) with y

$y = -3n - 1$

Swap positions of x and y

$n = -3y - 1$

Make y the subject

$3y = -n - 1$

$3y = -(n + 1)$

$y = -\frac{n+1}{3}$

Replace y with g'(n)

$g'(n) = -\frac{n+1}{3}$

You might be interested in

Eliza is a chef at a restaurant. she is making spaghetti sauce and put 14 tomatoes and 4 onions in the pot. what is the ratio of

notka56 [123]

Here,
tomato = 14
onion = 4
then,
ratio of tomato to onion = 14/4
= 7/2

7 0

3 years ago

Identify whether the following items best fit with a discrete or a continuous model. Then determine whether it is a linear (arit

astra-53 [7]

Answer:

it will be a continuous model

Step-by-step explanation:

weekend weekend assume every amount of area infected by the bed period doubles as hour passes

5 0

3 years ago

For the inverse variation equation xy = k, what is the value of x when y = 4 and k = 7? 3 28

Oliga [24]

Answer:

The value of x is $\frac{7}{4}$ .

Step-by-step explanation:

It is given that the inverse variation equation is

$xy=k$ ..... (1)

The value of y is 4 and the value of k is 7.

Substitute y=4 and k=7 in equation (1).

$x\times 4=7$

Divide both sides by 4.

$\frac{x\times 4}{4}=\frac{7}{4}$

$x=\frac{7}{4}$

Therefore the value of x is $\frac{7}{4}$ .

7 0

3 years ago

Read 2 more answers

9.6.6. The pawn shop business model. On April 9, 2011 The New York Times reported on a pawn shop that opened in an ex-Blockbuste

EleoNora [17]

The lends money would be 340

5 0

2 years ago

Read 2 more answers

Use series to verify that<br><br> <img src="https://tex.z-dn.net/?f=y%3De%5E%7Bx%7D" id="TexFormula1" title="y=e^{x}" alt="y=e^{

SVETLANKA909090 [29]

$y = e^x\\\\\displaystyle y = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y= 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \frac{d}{dx}\left( 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\frac{x^4}{4!}+\ldots\right)\\\\$

$\displaystyle y' = \frac{d}{dx}\left(1\right)+\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(\frac{x^2}{2!}\right) + \frac{d}{dx}\left(\frac{x^3}{3!}\right) + \frac{d}{dx}\left(\frac{x^4}{4!}\right)+\ldots\\\\\displaystyle y' = 0+1+\frac{2x^1}{2*1} + \frac{3x^2}{3*2!} + \frac{4x^3}{4*3!}+\ldots\\\\\displaystyle y' = 1 + x + \frac{x^2}{2!}+ \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y' = e^{x}\\\\$

This shows that $y' = y$ is true when $y = e^x$

-----------------------

Note 1: A more general solution is $y = Ce^x$ for some constant C.
Note 2: It might be tempting to say the general solution is $y = e^x+C$ , but that is not the case because $y = e^x+C \to y' = e^x+0 = e^x$ and we can see that $y' = y$ would only be true for C = 0, so that is why $y = e^x+C$ does not work.

6 0

2 years ago