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4 years ago

Find a rational zero of the polynomial function and use it to find all the zeros of the function. f(x) = x4 + 3x3 - 5x2 - 9x - 2

Mathematics

1 answer:

olchik [2.2K]4 years ago

3 0

Polynomials in the fourth degree are called quartic equations. In solving the roots of polynomials, there are techniques available. For quadratic equations, you use the quadratic formula. For cubic equations, you use the scientific calculator. But for quartic equations and higher, it is very complex. The method is very lengthy and can get very messy because you introduce a lot variables. So, I suggest you do the easiest method to estimate the roots.

Graph the equation by plotting arbitrary points. The graph looks like that in the figure. The points at which the curve passes the x-axis are the solution which are encircled in red.In approximation, the rational roots or zero's are -3.73, -1, -0.28 and 2.

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A small combination lock on a suitcase has 3 wheels, each labeled with the 10 digits from 0 to 9. If an opening combination is

Artyom0805 [142]

Answer:

0.14% probability of a person guessing the right combination

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes.

The order in which the numbers are selected is important. For example, 1,3,2 is a different combination than 3,1,2. So we use the permutations formula to solve this question.

Permutations formula:

The number of possible permutations of x elements from a set of n elements is given by the following formula:

$P_{(n,x)} = \frac{n!}{(n-x)!}$

Desired outcomes:

One right combination, so $P = 1$

Total outcomes:

10 numbers from a set of 3. So

$P_{(10,3)} = \frac{10!}{(10-3)!} = 720$

What is the probability of a person guessing the right combination?

$p = \frac{D}{T} = \frac{1}{720} = 0.0014$

0.14% probability of a person guessing the right combination

5 0

4 years ago

Arrange these numbers from least to greatest:

Sindrei [870]

-3/7 is obviously the least because it is a negative number.Then 2/5 because if you think about it if 2/5 is 2 of 5 equal pieces. 2/3 is 2 of 3 pieces.So if you think about a pizza if you cut the pizza into 3 pieces and ate 2 imagine how much you would have left and then think about cutting the pizza into 5 pieces and only eating 2 . Think about that. So the answer in the end is -3/7,2/5, and then 2/3. Hope I helped!

8 0

3 years ago

Find the measure of

Ghella [55]

sorry been trying to find something to say but i don't have a answer for it wish i could help

4 0

4 years ago

Read 2 more answers

What is equivalent to the equation to 1/4 (8 + 6z + 12) ?

Fed [463]

1/4 (8 + 6z + 12)

------------------------------------------------------------
Combine like terms :
------------------------------------------------------------
1/4 (6z + 20)

------------------------------------------------------------
Apply distributive property :
------------------------------------------------------------
1/4(6z) + 1/4(20)
3/2z + 5

------------------------------------------------------------
Answer: 3/2z + 5
------------------------------------------------------------

7 0

3 years ago

Section 5.2 Problem 21:

Fittoniya [83]

Answer:

$y(x)=e^{-2x}[3cos(\sqrt{6}x)+\frac{2\sqrt{6}}{3}sin(\sqrt{6}x)]$ (See attached graph)

Step-by-step explanation:

To solve a second-order homogeneous differential equation, we need to substitute each term with the auxiliary equation $am^2+bm+c=0$ where the values of $m$ are the roots:

$y''+4y'+10y=0\\\\m^2+4m+10=0\\\\m^2+4m+10-6=0-6\\\\m^2+4m+4=-6\\\\(m+2)^2=-6\\\\m+2=\pm\sqrt{6}i\\\\m=-2\pm\sqrt{6}i$

Since the values of $m$ are complex conjugate roots, then the general solution is $y(x)=e^{\alpha x}[C_1cos(\beta x)+C_2sin(\beta x)]$ where $m=\alpha\pm\beta i$ .

Thus, the general solution for our given differential equation is $y(x)=e^{-2x}[C_1cos(\sqrt{6}x)+C_2sin(\sqrt{6}x)]$ .

To account for both initial conditions, take the derivative of $y(x)$ , thus, $y'(x)=-2e^{-2x}[C_1cos(\sqrt{6}x+C_2sin(\sqrt{6}x)]+e^{-2x}[-C_1\sqrt{6}sin(\sqrt{6}x)+C_2\sqrt{6}cos(\sqrt{6}x)]$

Now, we can create our system of equations given our initial conditions:

$y(x)=e^{-2x}[C_1cos(\sqrt{6}x)+C_2sin(\sqrt{6}x)]\\\\y(0)=e^{-2(0)}[C_1cos(\sqrt{6}(0))+C_2sin(\sqrt{6}(0))]=3\\\\C_1=3$

$y'(x)=-2e^{-2x}[C_1cos(\sqrt{6}x+C_2sin(\sqrt{6}x)]+e^{-2x}[-C_1\sqrt{6}sin(\sqrt{6}x)+C_2\sqrt{6}cos(\sqrt{6}x)]\\\\y'(0)=-2e^{-2(0)}[C_1cos(\sqrt{6}(0))+C_2sin(\sqrt{6}(0))]+e^{-2(0)}[-C_1\sqrt{6}sin(\sqrt{6}(0))+C_2\sqrt{6}cos(\sqrt{6}(0))]=-2\\\\-2C_1+\sqrt{6}C_2=-2$

We then solve the system of equations, which becomes easy since we already know that $C_1=3$ :

$-2C_1+\sqrt{6}C_2=-2\\\\-2(3)+\sqrt{6}C_2=-2\\\\-6+\sqrt{6}C_2=-2\\\\\sqrt{6}C_2=4\\\\C_2=\frac{4}{\sqrt{6}}\\ \\C_2=\frac{4\sqrt{6}}{6}\\ \\C_2=\frac{2\sqrt{6}}{3}$

Thus, our final solution is:

$y(x)=e^{-2x}[C_1cos(\sqrt{6}x)+C_2sin(\sqrt{6}x)]\\\\y(x)=e^{-2x}[3cos(\sqrt{6}x)+\frac{2\sqrt{6}}{3}sin(\sqrt{6}x)]$

7 0

2 years ago