Answer:
A polynomial function that meet the conditions is x³ - 8x² + 5x + 14 = 0
Step-by-step explanation:
Hi there!
Let´s start writting a generic factored function. In this case, the function has 3 zeros so that the factored form will have 3 terms:
(x + a)(x + b)(x + c) = 0
For this expression to be 0, either (x+a) = 0 or (x+b) = 0 or (x+c) = 0
Then:
x + a = 0 ⇒ x = -a
x + b = 0 ⇒ x = -b
x + c = 0 ⇒ x = -c
Then, the values "a", "b" and "c" are equal to the zeros of the function but with opposite sign. Then, in our case:
a = 1
b = -2
c = -7
Then, the polynomial fuction will be:
(x + 1)(x - 2)(x - 7) = 0
Apply distributive property:
(x² - 2x + x -2)(x -7) = 0
(x² - x - 2)(x - 7)
x³ - x² - 2x - 7x² + 7x + 14 = 0
x³ - 8x² + 5x + 14 = 0
Then, a polynomial function that meets the conditions is:
x³ - 8x² + 5x + 14 = 0
Have a nice day!
The solution for this problem is:
To get the diagonal length of a square, multiply the length of one side by the square root of 2:If the length of one side is x.
length of diagonal = x √2
length of a diagonal = 4√2
length of a diagonal = 5.6569 feet
Simply do 297500/350000 and you would get 85%
<span>about 100.70 million.</span>
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 :
