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melamori03 [73]
3 years ago
7

What is the inverse of the following function?

Mathematics
1 answer:
andrew-mc [135]3 years ago
7 0

Answer:

The inverse of this function would be h(x) = x^5 + 4

Step-by-step explanation:

In order to find the inverse, start by switching the h(x) and the x values. Once you've done that, solve for the new h(x). That will be your inverse.

h(x) = \sqrt[5]{x - 4}

x = \sqrt[5]{h(x) - 4}

x^5 = h(x) - 4

x^5 + 4 = h(x)

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3 years ago
PLEASE HURRY :,) WILL GIVE BRAINLEST
Vlad1618 [11]

Answer:

Point R is at (-40,30),  a distance of 60 units from point Q

Step-by-step explanation:

<u>Explanation </u>

<u>we will use below conditions</u>

<u>Type of transformation</u>                                 <u> change to co-ordinate point</u>

  1. vertical translation up 'd' units                (x, y) changes to (x, y +d)
  2. vertical translation down 'd' units            (x, y) changes to (x, y-d)
  3. horizontal translation left 'c' units            (x, y)  changes to (x- c, y)  

   4. horizontal translation right 'c' units         (x, y) changes to (x+ c , y)

now Given data is P( 20,-30) and in the graph Q(-40,-30)

Step1:-

<u>Now we have to find 'R' point</u>

Given data the point "R' is vertically above point Q so the point "Q' is moves vertically up with '60' units

now we will use the condition (1)

Q(-40,-30) is <u>vertical translation up '60' units </u>now changes the co-ordinate point is (-40 , -30+60)

<u>The correct translation and The point R( -40 , 30)</u>

<u>Step2:-</u>

Again in the given data  "R" is at the same distance from point Q as point P is from point Q.

<u>so given P point is (20 , -30)</u>

now we translation 'P' also

<u>vertically translation up at a distance '60' units</u> and changes the new co-ordinate (20 , -30+60) that is (20,30)  [ use above condition  is (1)]

and again<u> horizontal translation left '60</u>' now changes (20 - 60 ,30)[ use condition (3)]

There fore the new co-ordinate is R( -40 , 30)

<u>Final answer:-</u>

Point R is at (−40, 30), a distance of 60 units from point Q

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