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Stels [109]
3 years ago
10

How to inverse this function step by step f(x)=x-3/5

Mathematics
1 answer:
Nitella [24]3 years ago
5 0

f(x)=x3−5

Replace f(x)

with y

.

y=x3−5

Interchange the variables.

x=y3−5

Solve for y

.


Since y

is on the right side of the equation, switch the sides so it is on the left side of the equation.

y3−5=x

Add 5

to both sides of the equation.

y3=5+x

Take the cube root of both sides of the equation to eliminate the exponent on the left side.

y=3√5+x

Solve for y

and replace with f−1(x)

.


Replace the y

with f−1(x)

to show the final answer.

f−1(x)=3√5+x

Set up the composite result function.

f(g(x))

Evaluate f(g(x))

by substituting in the value of g into f

.

(3√5+x)3−5

Simplify each term.


Remove parentheses around 3√5+x

.

f(3√5+x)=3√5+x3−5

Rewrite 3√5+x3

as 5+x

.

f(3√5+x)=5+x−5

Simplify by subtracting numbers.

.

Subtract 5

from 5

.

f(3√5+x)=x+0

Add x

and 0

.

f(3√5+x)=x

Since f(g(x))=x

, f−1(x)=3√5+x is the inverse of f(x)=x3−5

.

f−1(x)=3√5+x

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MAVERICK [17]

The simplest form of the given algebraic expression √(250c³/9d⁶) is; 5c * (√10c)]/3d³

<h3>How to simplify algebra problems?</h3>

We are given the algebra problems as;

√(250c³/9d⁶)

Now 250c³ can be broken down into 250 = 25c² × 10c

Thus, we now have;

√(250c³/9d⁶) = [√(25c²) * (√10c)]/√9d⁶

When we breakdown the above expression, we have;

5c * (√10c)]/3d³

Read more about Algebra Problems at; brainly.com/question/723406

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8 0
1 year ago
A variable needs to be eliminated to solve the system of equations below. Choose the
AURORKA [14]
Two different things u can do
Substitution:
X-9y=12
+9y +9y
X=9y+12

-3(9y+12)+9y=36
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Y = -4


Or whatever this other I’ve is called

-3x-9y=36
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-3x-9y=36
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8 0
2 years ago
Y = -17.5x + 2.7x + 14.1 - 11.9x
slava [35]

Answer:

y=-26.7x+14.1

hope this helps

have a good day :)

Step-by-step explanation:

8 0
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Answer:

I know what you're asking you are asking to find the least common multiple. However all I can find is the GCF which is 1

Step-by-step explanation:

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