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puteri [66]
3 years ago
11

how do I find out if someone answer a question that I asked I never see it posted and I never get to notifications​

Mathematics
2 answers:
Anastasy [175]3 years ago
8 0

Answer:

Step-by-step explanation:

If you find out and read this let me know I also have no clue

Whitepunk [10]3 years ago
5 0

It should ask for your email to let you know if someone answered your question

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Which of the following are parallel to
Bond [772]
Y = 3x
y = 3x - 1
y = 3x - 4

i’m pretty sure in order for lines to be parallel, they have to have the same slope but different y- intercepts
7 0
3 years ago
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-4(1 - 8n) - 4(8n + 4)<br> Simplify with Distributive property and combining like terms
tangare [24]
Jffjjddjkdfjjffjdkdkjdfkdkdkdkdkdjdkdjdkdkdjdjjdkddkjd
6 0
3 years ago
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A = 4 0 0 1 3 0 −2 3 −1 Find the characteristic polynomial for the matrix A. (Write your answer in terms of λ.) Find the real ei
Illusion [34]

Answer:

Step-by-step explanation:

We are given the matrix

A = \left[\begin{matrix}4&0&0 \\ 1&3&0 \\-2&3&-1 \end{matrix}\right]

a) To find the characteristic polynomial we calculate \text{det}(A-\lambda I)=0 where I is the identity matrix of appropiate size. in this case the characteristic polynomial is

\left|\begin{matrix}4-\lambda&0&0 \\ 1&3-\lambda&0 \\-2&3&-1-\lambda \end{matrix}\right|=0

Since this matrix is upper triangular, its determinant is the multiplication of the diagonal entries, that is

(4-\lambda)(3-\lambda)(-1-\lambda)=(\lambda-4)(\lambda-3)(\lambda+1)=0

which is the characteristic polynomial of A.

b) To find the eigenvalues of A, we find the roots of the characteristic polynomials. In this case they are \lambda=4,3,-1

c) To find the base associated to the eigenvalue lambda, we replace the value of lambda in the expression A-\lambda I and solve the system (A-\lambda I)x =0 by finding a base for its solution space. We will show this process for one value of lambda and give the solution for the other cases.

Consider \lambda = 4. We get the matrix

\left[\begin{matrix}0&0&0 \\ 1&-1&0 \\-2&3&-5 \end{matrix}\right]

The second line gives us the equation x-y =0. Which implies that x=y. The third line gives us the equation -2x+3y-5z=0. Since x=y, it becomes y-5z =0. This implies that y = 5z. So, combining this equations, the solution of the homogeneus system is given by

(x,y,z) = (5z,5z,z) = z(5,5,1)

So, the base for this eigenspace is the vector (5,5,1).

If \lambda = 3 then the base is (0,4,3) and if \lambda = -1 then the base is (0,0,1)

3 0
3 years ago
If m ≤ f(x) ≤ M for a ≤ x ≤ b, where m is the absolute minimum and M is the absolute maximum of f on the interval [a, b], then
agasfer [191]

Answer:

smaller value is 1.2

larger value is 6

Step-by-step explanation:

f(x)=\frac{3}{1+x^{2} }

when x=a=0, one has

f(0)=\frac{3}{1+0^{2} } \\f(0)=3

now, when x=b=2, one has

f(2)=\frac{3}{1+2^{2} } \\f(2)=\frac{3}{5}

Therefore, the absolute minumun is

m=\frac{3}{5}

and the absolute maximun is

M=3

The approximation to the integral is

\frac{3}{5}(2-0)\leq \int\limits^2_0 {f(x)} \, dx  \leq 3(2-0)

hence

\frac{3}{5}(2)\leq \int\limits^2_0 {f(x)} \, dx  \leq 3(2)\\\frac{6}{5} \leq \int\limits^2_0 {f(x)} \, dx  \leq 6\\1.2 \leq \int\limits^2_0 {f(x)} \, dx  \leq 6

7 0
4 years ago
The box plot below shows the total amount of time, in minutes, the students of a class surf the Internet every day:
pashok25 [27]

The information that provided by the graph are median and the range. The information that cannot be gotten from the graph is the mode.

The interquartile range is 25.

The outlier would distort the value of the median and mean.

<h3>How can a box plot be interpreted?</h3>

The information that can be derived from the box plot are the minimum value, maximum value, range, median, first quartile and third quartile.

The interquartile range is the difference between the first quartile and third quartile.

Interquartile range = 60 - 35  = 25

To learn more about box plots, please check: brainly.com/question/27215146

#SPJ1

5 0
3 years ago
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