Answer:
the domain of the function f(x) is 
the range of the function f(x) is 
Step-by-step explanation:
Consider the parent function 
The domain og this function is
the range of this function is 
The function
is translated function
7 units to the right and 9 units up, so
the domain of the function f(x) is 
the range of the function f(x) is 
A square root is a number that produces a specific number when multiplied by itself. Basically it's the same number multiplied by itself.
hope this helps:)
To solve this problem, we must use the order of operations outlined by PEMDAS, which tells us that we should simplify or compute parentheses first, then exponents, multiplication, division, addition, and finally subtraction.
Using this method, we have to perform the multiplication inside the parentheses first.
(20 * 40) * 14
800 * 14
Finally, we must perform the final operation to simplify this expression, which is multiplication.
800 * 14 = 11200
Therefore, your final answer is 11200.
Hope this helps!
<span>16/20 is 80%.</span>
You find out by multiplying the denominator: 20 * 5 = 100
Then, you do the same to the numerator which is: 16 * 5 = x
So, x = 80%
Hope this helps! :D
Answer:
We have been given a unit circle which is cut at k different points to produce k different arcs. Now we can see firstly that the sum of lengths of all k arks is equal to the circumference:

Now consider the largest arc to have length \small l . And we represent all the other arcs to be some constant times this length.
we get :

where C(i) is a constant coefficient obviously between 0 and 1.

All that I want to say by using this step is that after we choose the largest length (or any length for that matter) the other fractions appear according to the above summation constraint. [This step may even be avoided depending on how much precaution you wanna take when deriving a relation.]
So since there is no bias, and \small l may come out to be any value from [0 , 2π] with equal probability, the expected value is then defined as just the average value of all the samples.
We already know the sum so it is easy to compute the average :
