All categories

2 years ago

Finding the inverse function for f(x)=1/x^2

Mathematics

1 answer:

Anna [14]2 years ago

4 0

to get the inverse of any expression, we start off by doing a quick switcheroo on the variables and then solve for "y".

$\stackrel{f(x)}{y}=\cfrac{1}{x^2}\implies \stackrel{\textit{quick switcheroo}}{x=\cfrac{1}{y^2}}\implies y^2=\cfrac{1}{x}\implies y=\sqrt{\cfrac{1}{x}}\implies \stackrel{f^{-1}(x)}{y}=\cfrac{1}{\sqrt{x}}$

You might be interested in

2x - 5y = 5 and y = x + 2

Dafna1 [17]

Answer:7

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Which of these expressions can be used to calculate the monthly payment for

skad [1K]

Answer:

m=(r/n)+0.114r

Step-by-step explanation:

m=monthly payment

r=remaining amount

n=# of months remaining

3 0

3 years ago

Evaluate the expression if a = 5, b = 12 and c = 4.

Natalija [7]

3 because 2*12 24 24/8 is 3 hope this helps

4 0

3 years ago

Read 2 more answers

Find the value of p for which the given pair of linear equations have infinitely many solutions for 4X + 3 Y - 4 is equals to ze

tino4ka555 [31]

Answer:

x=8/3

y= -20/9

Step-by-step explanation:

4X + 3Y -4 =0

Linear egquations must in

$x + y = 0$

so 4X + 3Y =4

THEN

3X + 9Y = - 12

next use calculator to solve it or you can use traditional way

4 0

3 years ago

Suppose you have a light bulb that emits 50 watts of visible light. (Note: This is not the case for a standard 50-watt light bul

natali 33 [55]

Answer:

0.049168726 light-years

Step-by-step explanation:

The apparent brightness of a star is

$\bf B=\displaystyle\frac{L}{4\pi d^2}$

where

<em>L = luminosity of the star (related to the Sun) </em>

<em>d = distance in ly (light-years) </em>

The luminosity of Alpha Centauri A is 1.519 and its distance is 4.37 ly.

Hence the apparent brightness of Alpha Centauri A is

$\bf B=\displaystyle\frac{1.519}{4\pi (4.37)^2}=0.006329728$

According to the inverse square law for light intensity

$\bf \displaystyle\frac{I_1}{I_2}=\displaystyle\frac{d_2^2}{d_1^2}$

where

$\bf I_1=$ light intensity at distance $\bf d_1$

$\bf I_2=$ light intensity at distance $\bf d_2$

Let $\bf d_2$ be the distance we would have to place the 50-watt bulb, then replacing in the formula

$\bf \displaystyle\frac{0.006329728}{50}=\displaystyle\frac{d_2^2}{(4.37)^2}\Rightarrow\\\\\Rightarrow d_2^2=\displaystyle\frac{0.006329728*(4.37)^2}{50}\Rightarrow d_2^2=0.002417564\Rightarrow\\\\\Rightarrow d_2=\sqrt{0.002417564}=\boxed{0.049168726\;ly}$

Remark: It is worth noticing that Alpha Centauri A, though is the nearest star to the Sun, is not visible to the naked eye.

7 0

3 years ago