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3 years ago

What is 32.043 is in expanded form

2 answers:

4 0

In expanded form, your answer would look like this:

30 + 2 + 0.04 + 0.003

If you add up all of the numbers in their correct number value, you will get the number you originally started with, which, in this case, is 32.043. This is a great way to check your answer!

Zielflug [23.3K]3 years ago

3 0

32.043=30+2+0.04+0.003

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Brainliest to whoever can answer this question the best + 50 pts.

Gnoma [55]

3x-2y=-2--(1)
5x-3y=-4--(2)

(1)×5 and (2)×3

$\\ \tt\hookrightarrow 15x-10y=-10--(3)$

$\\ \tt\hookrightarrow 15x-9y=-12--(4)$

Subtracting

$\\ \tt\hookrightarrow -y=+2$

$\\ \tt\hookrightarrow y=-2$

Put in eq(1)

$\\ \tt\hookrightarrow 3x+4=-2$

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3 years ago

Read 2 more answers

What is an equation of the line that passes through the points (-6, 2) and (-6, 6)?

gavmur [86]

Answer:

Slope: undefined (vertical line)

X = -6

Step-by-step explanation:

Slope: (6-2)/(-6 - - 6) = 4/0

Slope = 0 (vertical line)

The x coordinates in both points= -6

x = -6

8 0

3 years ago

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

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