The value 3 is an upper bound for the zeros of the function shown below.

f(x)=-3x³+20x²-36x+16

a. true
b. false

The statement that the value of 3 is an upper bound for the zeros of the function shown is;

B: False

We are given the function;

f(x) = -3x³ + 20x² – 36x + 16.

In this function, we see that there are three sign changes. Thus by Descarte's rule of signs, there exist 3 or 1 positive zeros

Let's now check f(-x) ;

f(-x) = 3x³ + 20x² + 36x + 16.

There is no sign change in this case and by Descart’s rule of signs, there is no

negative root.

The leading coefficient in the polynomial is -3, and so we must divide all terms of the main polynomial by 3. This gives;

x³ - 6.67x² + 12x - 5.33

The coefficients are 1; -6.67; -12; -5.33

Drop the leading coefficient and remove any minus signs to get: 6.67; 12; 5.33

Bound 1: the largest value is 12. We add 1 to get 13

Bound 2: adding all values gives: 6.67 + 12 + 5.33 = 24

We see that the sum of all values is greater than 1.

The smallest “bounds” value is 13.

Now, all the roots are within ±13.

Thus, 3 is not an upper bound.

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answer: the answer is one of

step-by-step explanation:   we are given to write a proper fraction where the whole is divided into an even number of equal parts less than ten, and an odd number of those parts exist.

an even number less than 10 may be one of 2, 4, 6 or 8. so, the whole may be one of these four evens.

now, an odd number of these parts must exist, so the numerator must be an odd number, one of 1, 3, 5 or 7. it cannot be 9 because then the fraction will become improper.

thus, our fraction may be one of the following

8y^50

step-by-step explanation: