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3 years ago

State how many imaginary and real zeros the function has.

1 answer:

7 0

Hello,
Degre 5==> even number of imaginary roots and odd number of reals.

Real roots must be a divisor of 7.
Using Excel, we find

 x P(x)
-7 0 
 -6 1369
 -5 1352
 -4 867
 -3 400
 -2 125
 -1 24
0 7
1 32
2 225
3 1000
4 3179
5 8112
6 17797
7 35000

So: one real root: 7 and 4 imaginary roots.

Answer B

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Select the equation of the line to the equation y=-4x-6 and that passes through the point (1,2)

omeli [17]

I don't see a list of options, but I know that the answer is y = -4x + 6.

Point - Slope Form: y - y₁ = m(x - x₁)

y - 2 = -4(x - 1)

y - 2 = -4x + 4
+ 2 + 2
--------------------
y = -4x + 6

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3 years ago

If the expression 5+2x was evaluated for x=4 it’s value would be

iris [78.8K]

Answer: 13

Step-by-step explanation:

5+2x

When x=4

5+2(4)

5+8

= 13

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3 years ago

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Gus wrote the following riddle: I am a number between 30 and 60. My ones digit is three less than my tens digit. I am a prime nu

Neporo4naja [7]

Answer: 41

Step-by-step explanation:

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3 years ago

How do I do any of this?!

evablogger [386]

U should get the app "photomath" it will help u out alot!!!

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3 years ago

Find the characteristic polynomial of A. Use x for the variable in your polynomial.You do not need to factor your polynomial. A

Vesna [10]

Answer: The required characteristic polynomial of the given matrix A is $x^3+6x+11x+6=0.$

Step-by-step explanation: We are given to find the characteristic polynomial of the following 3 × 3 matrix A with unknown variable x :

$A=\left[\begin{array}{ccc}0&0&1\\4&-3&4\\-2&0&-3\end{array}\right].$

We know that

for any square matrix M, the characteristic polynomial is given by

$|M-xI|=0,$ where I is an identity matrix of same order as M.

Therefore, the characteristic polynomial of matrix A is

$|A-xI|=0\\\\\\\Rightarrow \left|\left[\begin{array}{ccc}0&0&1\\4&-3&4\\-2&0&-3\end{array}\right]-x\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right]\right|=0\\\\\\\Rightarrow \left|\left[\begin{array}{ccc}-x&0&1\\4&-3-x&4\\-2&0&-3-x\end{array}\right] \right|=0\\\\\\\Rightarrow -x(3+x)^2+1(0-6-2x)=0\\\\\Rightarrow (x+3)(-3x-x^2-2)=0\\\\\Rightarrow (x+3)(x^2+3x+2)=0\\\\\Rightarrow x^3+6x+11x+6=0.$

Thus, the required characteristic polynomial of the given matrix A is $x^3+6x+11x+6=0.$

6 0

3 years ago