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4 years ago

Enter the number of complex zeros for the polynomial function in the box. f(x)=x3−96x2+400

Mathematics

2 answers:

Julli [10]4 years ago

7 0

Answer:

Step-by-step explanation:

ioda4 years ago

5 0

Solution:

There is no (zero) complex zeros for the given polynomial.

Explanation:

We apply Descartes' rule of sign to identify the number of complex roots.

The given polynomial is $f(x)=x^3-96x^2+400$

Let us see the number of sign changes in f(x)

+,-,+

There are 2 sign changes in f(x). One from plus to minus and second from plus to minus. Hence, there 2 or 0 positive roots.

Now, let us see the number of sign changes in f(-x)

$f(-x)=-x^3-96x^2+400$

-,-,+

There are only one sign change. Hence, there will be 1 negative roots.

The degree of the polynomial is 3. Hence, there will be exactly 3 zeros.

Therefore, the possible numbers of zeros are

2 positive, 1 negative and 0 complex

0 positive, 1 negative and 2 complex.

Hence, possible number of complex zeros are 0 and 2.

Now, we graph the function in xy plane and see that the graph cuts the x axis at three points. It means all the zeros are real , which is the case of first possibility (2 positive, 1 negative and 0 complex).

Hence, the number of complex zeros for the given polynomial is zero.

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Tan 45°. tan² 30° tan 60° <br>slove it

makvit [3.9K]

Answer:

$\frac{\sqrt{3} }{3} or \frac{1}{\sqrt{3} }$

Step-by-step explanation:

We can break the problem into three parts,

tan45 = sin(45)/cos(45)= $\frac{\frac{\sqrt{2}}{2} }{\frac{\sqrt{2}}{2} } =1$

$tan^{2}30 =(tan30))^{2}=(sin30/cos30)^2=((1/2)/(\sqrt{3}/2))^2 =(1/\sqrt{3})^2$

$tan60=sin60/cos60=\frac{\sqrt{3}/2 }{1/2} =\sqrt{3}$

Then, $tan45*tan^230*tan60=1*(1/sqrt(3))^2*sqrt(3)=1/sqrt(3)=sqrt(3)/3$

Therefore, the solution is sqrt(3)/3

4 0

2 years ago

I am thinking of a number. My number is between 20 and 30 My number and 12 have only one common factor. What number could I be t

zhannawk [14.2K]

Answer:

21, 22 and 26

Step-by-step explanation:

To answer this question first we need to know which are the factors of 12:

$12= 2^2(3)$

So, now, we need 3 numbers that are <u>between 20 and 30 and that only have one common factor with 12</u>, in other words, they need to have just a 2 or a 3 in their factorization.

Let's take number 21:

$21= (7)(3)$ , we can see that 21 only has a 3 and a prime so therefore it has only one common factor with 12