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2 years ago

What is the inverse of the function f(x) = one-quarterx – 12? h(x) = 48x – 4 h(x) = 48x + 4 h(x) = 4x – 48 h(x) = 4x + 48

Mathematics

2 answers:

DaniilM [7]2 years ago

8 0

Answer:

h(x) = 4x+48

Step-by-step explanation:

Given the function f(x) = 1/4x -12. To get the inverse of the function, first we will let y = f(x) to have y = 1/4x - 12.

Then we will make x the subject of the formula as shown:

y = 1/4x - 12

adding 12 to both sides we have;

y + 12= 1/4x - 12 + 12

y + 12 = 1/4x

Multiplying both sides by 4 we have;

4(y+12) = x

x = 4y + 48

Substituting y for x

y = 4x+48

f^-1(x) = 4x+48

h(x) = 4x+48

Thee final expression will give the inverse of the function

Basile [38]2 years ago

4 0

Answer:

Step-by-step explanation:

Given the function f(x) = 1/4x -12. To get the inverse of the function, first we will let y = f(x) to have y = 1/4x - 12.

Then we will make x the subject of the formula as shown:

y = 1/4x - 12

adding 12 to both sides we have;

y + 12= 1/4x - 12 + 12

y + 12 = 1/4x

Multiplying both sides by 4 we have;

4(y+12) = x

x = 4y + 48

Substituting y for x

y = 4x+48

f^-1(x) = 4x+48

h(x) = 4x+48

Thee final expression will give the inverse of the function

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Answer:

Step-by-step explanation:

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Read 2 more answers

The length of the shadow of a pole on level ground increases by 90m when the angle of elevation of the sun changes from 58° to 3

viva [34]

Answer:

119.45 m

Step-by-step explanation:

Given:

When angle of elevation of the sun changes from 58° to 36° the length of shadow of a pole increases by 90 m.

To find:

Length of pole = ?

Solution:

Kindly refer to the attached image.

$\triangle ABC$ represents the 1st angle of elevation of sun i.e. 58°

$\triangle ABD$ represents the 2nd angle of elevation of sun i.e. 36°

Change in shadow is represented by CD = 90 m

Let height of pole, AB = $h$ m

Let side BC = $x$ m

Now, let us apply tangent rules in $\triangle ABC, \triangle ABD$ one by one:

$tan\theta = \dfrac{Perpendicular}{Base}\\\Rightarrow tan58^\circ=\dfrac{AB}{BC}\\\Rightarrow tan58^\circ=\dfrac{h}{x}\\\Rightarrow x = 0.624h ..... (1)$

$tan36^\circ = \dfrac{h}{x+90}$

Putting value of $x$ using equation (1):

$tan36^\circ = \dfrac{h}{0.624h+90}\\\Rightarrow 0.726\times 0.624h+0.726\times 90 = h\\\Rightarrow h-0.453h =65.34\\\Rightarrow \bold{h = 119.45\ m}$