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3 years ago

If f(x)= 3x^2 + 3x^1/2 +3x what is the value of f(-9)

Mathematics

1 answer:

Dahasolnce [82]3 years ago

4 0

Answer:

216 + 9i

Step-by-step explanation:

Given: f(x) = 3x² + 3 $x^{1/2}$ + 3x

When x = -9, f(x) equals to;

3(-9)² + 3(-9 $)^{1/2}$ + 3(-9)

243 + 3(3i) - 27

243 + 9i - 27

216 + 9i

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Which of the following would best be addressed by a scatter plot?

djyliett [7]

Answer:

Step-by-step explanation:

There has to be two variables it's being compared to, in which case this is the blood sugar level and the amount of sugar

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3 years ago

Write the cubic polynomial function f(x)in expanded form with zeros −6,−5,and −1, given that f(0)=60

RSB [31]

Answer:

$f(x)=2x^{3}+24x^{2}+82x+60$

Step-by-step explanation:

we know that

The roots of the polynomial are the values of x when the value of the polynomial f(x) is equal to zero

The roots of the polynomial function are

x=-6 -----> (x+6)=0

x=-5 -----> (x+5)=0

x=-1 -----> (x+1)=0

The equation of the cubic polynomial is

$f(x)=a(x+6)(x+5)(x+1)$

where

a is the leading coefficient

Remember that

f(0)=60

That means ------> For x=0 the value of f(x) is equal to 60

substitute the value of x and the value of y in the function and solve for a

$60=a(0+6)(0+5)(0+1)$

$60=a(6)(5)(1)$

$60=30a$

$a=2$

$f(x)=2(x+6)(x+5)(x+1)$

Applying the distributive property

Convert to expanded form

$f(x)=2(x+6)(x+5)(x+1)\\\\f(x)=2(x+6)(x^{2}+x+5x+5)\\\\f(x)=2(x+6)(x^{2}+6x+5)\\\\f(x)=2(x^{3}+6x^{2}+5x+6x^{2}+36x+30)\\\\f(x)=2x^{3}+24x^{2}+82x+60$

3 0

3 years ago

What is slope of B E with bar above A E in simplest form?

Annette [7]

This is the one
<span>C. Start Fraction 25 over 4 End Fraction
</span>

7 0

3 years ago

(1) (10 points) Find the characteristic polynomial of A (2) (5 points) Find all eigenvalues of A. You are allowed to use your ca

Yuri [45]

Answer:

Step-by-step explanation:

Since this question is lacking the matrix A, we will solve the question with the matrix

$\left[\begin{matrix}4 & -2 \\ 1 & 1 \end{matrix}\right]$

so we can illustrate how to solve the problem step by step.

a) The characteristic polynomial is defined by the equation $det(A-\lambdaI)=0$ where I is the identity matrix of appropiate size and lambda is a variable to be solved. In our case,

$\left|\left[\begin{matrix}4-\lamda & -2 \\ 1 & 1-\lambda \end{matrix}\right]\right|= 0 = (4-\lambda)(1-\lambda)+2 = \lambda^2-5\lambda+4+2 = \lambda^2-5\lambda+6$

So the characteristic polynomial is $\lambda^2-5\lambda+6=0$ .

b) The eigenvalues of the matrix are the roots of the characteristic polynomial. Note that

$\lambda^2-5\lambda+6=(\lambda-3)(\lambda-2) =0$

So $\lambda=3, \lambda=2$

c) To find the bases of each eigenspace, we replace the value of lambda and solve the homogeneus system(equalized to zero) of the resultant matrix. We will illustrate the process with one eigen value and the other one is left as an exercise.

If $\lambda=3$ we get the following matrix

$\left[\begin{matrix}1 & -2 \\ 1 & -2 \end{matrix}\right]$ .

Since both rows are equal, we have the equation

$x-2y=0$ . Thus x=2y. In this case, we get to choose y freely, so let's take y=1. Then x=2. So, the eigenvector that is a base for the eigenspace associated to the eigenvalue 3 is the vector (2,1)

For the case $\lambda=2$ , using the same process, we get the vector (1,1).

d) By definition, to diagonalize a matrix A is to find a diagonal matrix D and a matrix P such that $A=PDP^{-1}$ . We can construct matrix D and P by choosing the eigenvalues as the diagonal of matrix D. So, if we pick the eigen value 3 in the first column of D, we must put the correspondent eigenvector (2,1) in the first column of P. In this case, the matrices that we get are

$P=\left[\begin{matrix}2&1 \\ 1 & 1 \end{matrix}\right], D=\left[\begin{matrix}3&0 \\ 0 & 2 \end{matrix}\right]$

This matrices are not unique, since they depend on the order in which we arrange the eigenvalues in the matrix D. Another pair or matrices that diagonalize A is

$P=\left[\begin{matrix}1&2 \\ 1 & 1 \end{matrix}\right], D=\left[\begin{matrix}2&0 \\ 0 & 3 \end{matrix}\right]$

which is obtained by interchanging the eigenvalues on the diagonal and their respective eigenvectors

4 0

3 years ago

Y= -3/2x-2 graph graph

evablogger [386]

Answer:

To graph (like the provided picture) you must understand y=mx+b. Your m would stand for your slope which you need to determine with rise over run. Rise over run can be done by finding a place where both the x and y values are lined up. (0,-2) can be used as an example from the picture. You then want to find the next point where the x and y values line up which (-2,1) can be used. Go up 3 units and go to the LEFT 2 times which makes this a negative run. The slope would then be represented as -3/2.

Your x value stands for your x-value moving up or down while you b value stands for your y-intercept which point (0,-2) is represented as. Y represents the y value after the equation is plugged in with x and solved.

8 0

3 years ago