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3 years ago

If f(x) = 5x, what is f–1(x)?

Mathematics

2 answers:

Gemiola [76]3 years ago

5 0

Answer:

The answer is C.

Step-by-step explanation:

Step 1: Replace f(x) with y.

f(x)=5x >>>>> y=5x

Step 2: Switch x and y.

y=5x >>>>> x=5y

Step 3: Solve for y

x=5y >>>> y= x/5

Then you have to do the inverse of dividing by 5. Which is multipilying by 1/5. This leaves you with,

y= 1/5x

Step 4: Write in standard inverse notation (Replace y with f^-1(x))

f^-1(x)=1/5x

Note: These steps work with ANY problem where you have to find the inverse function. I hope this helps :)

Ahat [919]3 years ago

4 0

To get the inverse of a function interchange
the x's with the y'x and solve for y, as follows:
y=5x-7

Interchange to get:
x=5y-7
Solve for y to get:
y=(1/5)x+(7/5)

so I think it might be c

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The price of fruits per kg are given below- Watermelon Rs 18.50 Cherry Rs 72 Grapes Rs 120.60 Apples Rs 78.40 Karan bought 412 k

Papessa [141]

Answer:

$X=-2.74*10^{-7}Rs$

Step-by-step explanation:

From the question we are told that:

Table for price per Kg

Watermelon Rs 18.50

Cherry Rs 72

Grapes Rs 120.60

Apples Rs 78.40

Karan Bought:

412 kg watermelon

1 kg 200g cherries

250 g grapes

134 kg apples

Generally the equation for Total Price of Fruits Bought is mathematically given by

\sum (Price Rs* Weight kg )

$\sum (P* W)=(412*18.5)+(1.2*72)+(0.25*120.60+(134*78.40))$

$\sum (P* W)=27374289.78$

Therefore the Amount of Money he gets back is

$X=500-\sum (P* W)$

$X=-2.74*10^{-7}Rs$

Therefore

He owes the shop keeper 2.74 million Rs

8 0

3 years ago

A ship sails 250km due North qnd then 150km on a bearing of 075°.1)How far North is the ship now? 2)How far East is the ship now

olga_2 [115]

Answer:

1) 288.8 km due North

2) 144.9 km due East

3) 323.1 km

4) 207°

Step-by-step explanation:

Bearing: The angle (in degrees) measured clockwise from north.

Trigonometric ratios

$\sf \sin(\theta)=\dfrac{O}{H}\quad\cos(\theta)=\dfrac{A}{H}\quad\tan(\theta)=\dfrac{O}{A}$

where:

$\theta$ is the angle
O is the side opposite the angle
A is the side adjacent the angle
H is the hypotenuse (the side opposite the right angle)

Cosine rule

$c^2=a^2+b^2-2ab \cos C$

where a, b and c are the sides and C is the angle opposite side c

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

Draw a diagram using the given information (see attached).

Create a right triangle (blue on attached diagram).

This right triangle can be used to calculate the additional vertical and horizontal distance the ship sailed after sailing north for 250 km.

Question 1

To find how far North the ship is now, find the measure of the short leg of the right triangle (labelled y on the attached diagram):

$\implies \sf \cos(75^{\circ})=\dfrac{y}{150}$

$\implies \sf y=150\cos(75^{\circ})$

$\implies \sf y=38.92285677$

Then add it to the first portion of the journey:

⇒ 250 + 38.92285677... = 288.8 km

Therefore, the ship is now 288.8 km due North.

Question 2

To find how far East the ship is now, find the measure of the long leg of the right triangle (labelled x on the attached diagram):

$\implies \sf \sin(75^{\circ})=\dfrac{x}{150}$

$\implies \sf x=150\sin(75^{\circ})$

$\implies \sf x=144.8888739$

Therefore, the ship is now 144.9 km due East.

Question 3

To find how far the ship is from its starting point (labelled in red as d on the attached diagram), use the cosine rule:

$\sf \implies d^2=250^2+150^2-2(250)(150) \cos (180-75)$

$\implies \sf d=\sqrt{250^2+150^2-2(250)(150) \cos (180-75)}$

$\implies \sf d=323.1275729$

Therefore, the ship is 323.1 km from its starting point.

Question 4

To find the bearing that the ship is now from its original position, find the angle labelled green on the attached diagram.

Use the answers from part 1 and 2 to find the angle that needs to be added to 180°:

$\implies \sf Bearing=180^{\circ}+\tan^{-1}\left(\dfrac{Total\:Eastern\:distance}{Total\:Northern\:distance}\right)$

$\implies \sf Bearing=180^{\circ}+\tan^{-1}\left(\dfrac{150\sin(75^{\circ})}{250+150\cos(75^{\circ})}\right)$

$\implies \sf Bearing=180^{\circ}+26.64077...^{\circ}$

$\implies \sf Bearing=207^{\circ}$

Therefore, as bearings are usually given as a three-figure bearings, the bearing of the ship from its original position is 207°

8 0

2 years ago

Read 2 more answers

Just give the answer plz 0_o

GalinKa [24]

Answer:

44.5

Step-by-step explanation:

5 0

3 years ago

Need help with math problem if do get 5 star

Luda [366]

Answer:

Step-by-step explanation:

The first co-ordinate (1) will be on the x-axis, so we go across to the first square. Then you go 3 up the y-axis because of the second co-ordinate, and that is where S is.

Hope this helps!

6 0

2 years ago

Read 2 more answers

TRIGONOMETRY - Given the complex numbers <img src="https://tex.z-dn.net/?f=z_%7B1%7D%20%3D20%20cis%20226" id="TexFormula1" title

Norma-Jean [14]

This is easily done after writing both complex numbers in exponential form:

$z_1=20\,\mathrm{cis}226^\circ=20e^{226^\circ i}$

$z_2=5\,\mathrm{cis}76^\circ=5e^{76^\circ i}$

Then

$\dfrac{z_1}{z_2}=\dfrac{20e^{226^\circ i}}{5e^{76^\circ i}}=4e^{(226-76)^\circ i}=4e^{150^\circ i}$

In trigonometric form, this is

$\dfrac{z_1}{z_2}=4\,\mathrm{cis}150^\circ=4(\cos150^\circ+i\sin150^\circ)=\boxed{-\dfrac{\sqrt3}2+\dfrac i2}$

8 0

3 years ago