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3 years ago

Find the inverse f⁡(x)=2⁢x/x+3 x≠-3

Mathematics

1 answer:

zysi [14]3 years ago

7 0

Answer:

f^(-1)(x) = 3x/(2-x)

Step-by-step explanation:

f(x) = 2x/(x+3)

f(x)*(x+3)=2x

x(2-f(x))=3f(x)

x=3f(x)/(2-f(x))

f^(-1)(x) = 3x/(2-x)

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Please explain how do you get the answer. Also tell me am I correct or wrong.

Zigmanuir [339]

Your doing everything correctly

3 0

4 years ago

Read 2 more answers

A solution initially contains 200 bacteria. 1. Assuming the number y increases at a rate proportional to the number present, wri

GuDViN [60]

Answer:

1. $\frac{dy}{dt}=ky$

2.543.6

Step-by-step explanation:

We are given that

y(0)=200

Let y be the number of bacteria at any time

$\frac{dy}{dt}$ =Number of bacteria per unit time

$\frac{dy}{dt}\proportional y$

$\frac{dy}{dt}=ky$

Where k=Proportionality constant

2. $\frac{dy}{y}=kdt$ ,y'(0)=100

Integrating on both sides then, we get

$lny=kt+C$

We have y(0)=200

Substitute the values then , we get

$ln 200=k(0)+C$

$C=ln 200$

Substitute the value of C then we get

$ln y=kt+ln 200$

$ln y-ln200=kt$

$ln\frac{y}{200}=kt$

$\frac{y}{200}=e^{kt}$

$y=200e^{kt}$

Differentiate w.r.t

$y'=200ke^{kt}$

Substitute the given condition then, we get

$100=200ke^{0}=200 \;because \;e^0=1$

$k=\frac{100}{200}=\frac{1}{2}$

$y=200e^{\frac{t}{2}}$

Substitute t=2

Then, we get $y=200e^{\frac{2}{2}}=200e$

$y=200(2.718)=543.6=543.6$

e=2.718

Hence, the number of bacteria after 2 hours=543.6

4 0

4 years ago

How do you convert a fraction to decimal

muminat

To convert a fraction to a decimal, divide the numerator by the denominator (divide the top by the bottom)

7 0

4 years ago

Read 2 more answers

Can someone please help me on this one? I will give BRAINLIEST to whoever has the correct answer.

Reptile [31]

I think the first one is 4x=3

6 0

3 years ago

1.What are the zeros of the polynomial function?

Lorico [155]

Let's to the first example:

f(x) = x^2 + 9x + 20

Ussing the formula of basckara

a = 1
b = 9
c = 20

Delta = b^2 - 4ac

Delta = 9^2 - 4.(1).(20)

Delta = 81 - 80

Delta = 1

x = [ -b +/- √(Delta) ]/2a

Replacing the data:

x = [ -9 +/- √1 ]/2

x' = (-9 -1)/2 <=> - 5

Or

x" = (-9+1)/2 <=> - 4
_______________

Already the second example:

f(x) = x^2 -4x -60

Ussing the formula of basckara again

a = 1
b = -4
c = -60

Delta = b^2 -4ac

Delta = (-4)^2 -4.(1).(-60)

Delta = 16 + 240

Delta = 256

Then, following:

x = [ -b +/- √(Delta)]/2a

Replacing the information

x = [ -(-4) +/- √256 ]/2

x = [ 4 +/- 16]/2

x' = (4-16)/2 <=> -6

Or

x" = (4+16)/2 <=> 10
______________

Now we are going to the 3 example

x^2 + 24 = 14x

Isolating 14x , but changing the sinal positive to negative

x^2 - 14x + 24 = 0

Now we can to apply the formula of basckara

a = 1
b = -14
c = 24

Delta = b^2 -4ac

Delta = (-14)^2 -4.(1).(24)

Delta = 196 - 96

Delta = 100

Then we stayed with:

x = [ -b +/- √Delta ]/2a

x = [ -(-14) +/- √100 ]/2

We wiil have two possibilities

x' = ( 14 -10)/2 <=> 2

Or

x" = (14 +10)/2 <=> 12
________________

To the last example will be the same thing.

f(x) = x^2 - x -72

a = 1
b = -1
c = -72

Delta = b^2 -4ac

Delta = (-1)^2 -4(1).(-72)

Delta = 1 + 288

Delta = 289

Then we are going to stay:

x = [ -b +/- √Delta]/2a

x = [ -(-1) +/- √289]/2

x = ( 1 +/- 17)/2

We will have two roots

That's :

x = (1 - 17)/2 <=> -8

Or

x = (1+17)/2 <=> 9

Well, this would be your answers.

7 0

4 years ago