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olganol [36]
3 years ago
11

Evaluate (4√81)³.............​

Mathematics
1 answer:
DochEvi [55]3 years ago
8 0
Your answer would be 27

By the way please mark brainliest
You might be interested in
A. Use composition to prove whether or not the functions are inverses of each other.
kogti [31]

A. In a composition of two functions the first function is evaluated, and then the second function is evaluated on the result of the first function. In other word, you are going to evaluate the second function in the first function.

Remember that you can evaluate function at any number just replacing the variable in the function with the number. For example, let's evaluate our function f(x) at x=1:

f(x)=\frac{1}{x-3}

f(1)=\frac{1}{1-3}

f(1)=\frac{1}{-2}

Similarly, to find the composition of f(x) andg(x), we are going to evaluate f(x) at g(x). In other words, we are going to replace x in f(x) with \frac{3x+1}{x}:

f(x)=\frac{1}{x-3}

f(g(x) = f(\frac{3x+1}{x} ) = \frac{1}{\frac{3x+1}{x} -3}

Remember that two functions are inverse if after simplifying their composition, we end up with just x. Let's simplify and see what happens.

f(g(x)=\frac{1}{\frac{3x+1}{x} -3}

f(g(x)=\frac{1}{\frac{3x+1-3x}{x} }

f(g(x)=\frac{1}{\frac{1}{x} }

f(g(x)=x

Now let's do the same for g(f(x)):

g(\frac{1}{x-3} )=\frac{3(\frac{1}{x-3})+1}{x}

g(\frac{1}{x-3} )=\frac{\frac{3}{x-3}+1}{x}

g(\frac{1}{x-3} )=\frac{\frac{3+x-3}{x-3}}{x}

g(\frac{1}{x-3} )=\frac{\frac{x}{x-3}}{x}

g(\frac{1}{x-3} )=\frac{x}{x(x-3)}

g(f(x))=\frac{x}{x(x-3)}

We can conclude that g(x) is the inverse function of f(x), but f(x) is not the inverse function of g(x).

B. The domain of a function is the set of all the possible values the independent variable can have. In other words, the domain are all the possible x-values of function.

Now, interval notation is a way to represent and interval using an ordered pair of numbers called the end points; we use brackets [ ] to indicate that the end points are included in the interval and parenthesis ( ) to indicate that they are excluded.

Notice that when x=0, g(x)=\frac{3(0)+1}{0} =\frac{0}{0}, so when x=0, g(x) is not defined; therefore we have to exclude zero from the domain of f(g(x)).

We can conclude that the domain of the composite function f(g(x)) in interval notation is (-∞,0)U(0,∞)

Now let's do the same for g(f(x)).

Notice that the composition is not defined when its denominator equals zero, so we are going to set its denominator equal to zero to find the values we should exclude from its domain:

x(x-3)=0

x=0 and x-3=0

x=0 and x=3

Know we know that we need to exclude x=0 and x=3 from the domain of g(f(x)).

We can conclude that the domain of the composition function g(f(x)) is (-∞,0)U(0,3)U(3,∞)

4 0
3 years ago
Read 2 more answers
Hello can you please help me posted picture of question
andrew-mc [135]
The correct option is (B): (3x+2)(3x+1)

Explanation:
Add all the boxes:
9x^2 (boxes)
9x (boxes)
The constants = 2

So it becomes:
9x^2 + 9x + 2
9x^2 + 6x + 3x + 2
3x(3x+2) + 1(3x+2)

(3x+2)(3x+1)


5 0
3 years ago
23 hours after 8:00 pm
Sophie [7]
A day has 24 hours, so if you add 23 hours, you are going 1 hour less than a full day.  When you are ignoring what day it is, it appears to just be moving back one hour.  Therefore, the answer is 7:00 pm.  Comment if you have any questions!
5 0
3 years ago
Read 2 more answers
Anyone know this thanks
PolarNik [594]
I think the answer is
No
No
Yes
But I'm not sure about the last one
3 0
3 years ago
Why is there no set path that I've scientific inquiry must follow
sveta [45]
Because you follow wherever the evidence leads you.

4 0
3 years ago
Read 2 more answers
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