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4 years ago

Round the number 63,825 On the hundreds place

Mathematics

2 answers:

anzhelika [568]4 years ago

7 0

Answer:

63,800

Step-by-step explanation:

Mrac [35]4 years ago

6 0

Answer:

63,800

hope this helps :3

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Which of the following trinomials can be factored using the square of the difference?

ArbitrLikvidat [17]

Answer:

d) 16x² -56x +49

Step-by-step explanation:

The square of a binomial difference will have the form ...

(ax -b)² = a²x² -2abx +b²

The sign of the linear term will be negative, which eliminates choices A and C.

The linear term will be double the product of the roots of the other two terms.

In choice B, the linear term has a factor of 5 that appears nowhere else, so that cannot be a perfect square.

Choice D factors as ...

16x² -56x +49 = (4x -7)² . . . . the square of a difference

8 0

3 years ago

Consider the matrix A. A = 1 0 1 1 0 0 0 0 0 Find the characteristic polynomial for the matrix A. (Write your answer in terms of

dusya [7]

Answer with Step-by-step explanation:

We are given that a matrix

$A=\left[\begin{array}{ccc}1&0&1\\1&0&0\\0&0&0\end{array}\right]$

a.We have to find characteristic polynomial in terms of A

We know that characteristic equation of given matrix $\mid{A-\lambda I}\mid=0$

Where I is identity matrix of the order of given matrix

I= $\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right]$

Substitute the values then, we get

$\begin{vmatrix}1-\lambda&0&1\\1&-\lambda&0\\0&0&-\lambda\end{vmatrix}=0$

$(1-\lambda)(\lamda^2)-0+0=0$

$\lambda^2-\lambda^3=0$

$\lambda^3-\lambda^2=0$

Hence, characteristic polynomial = $\lambda^3-\lambda^2=0$

b.We have to find the eigen value for given matrix

$\lambda^2(1-\lambda)=0$

Then , we get $\lambda=0,0,1-\lambda=0$

$\lambda=1$

Hence, real eigen values of for the matrix are 0,0 and 1.

c.Eigen space corresponding to eigen value 1 is the null space of matrix $A-I$

$E_1=N(A-I)$

$A-I=\left[\begin{array}{ccc}0&0&1\\1&-1&0\\0&0&-1\end{array}\right]$

Apply $R_1\rightarrow R_1+R_3$

$A-I=\left[\begin{array}{ccc}0&0&1\\1&-1&0\\0&0&0\end{array}\right]$

Now,(A-I)x=0[/tex]

Substitute the values then we get

$\left[\begin{array}{ccc}0&0&1\\1&-1&0\\0&0&0\end{array}\right]\left[\begin{array}{ccc}x_1\\x_2\\x_3\end{array}\right]=0$

Then , we get $x_3=0$

And $x_1-x_2=0$

$x_1=x_2$

Null space N(A-I) consist of vectors

x= $\left[\begin{array}{ccc}x_1\\x_1\\0\end{array}\right]$

For any scalar $x_1$

$x=x_1\left[\begin{array}{ccc}1\\1\\0\end{array}\right]$

$E_1=N(A-I)=Span(\left[\begin{array}{ccc}1\\1\\0\end{array}\right]$

Hence, the basis of eigen vector corresponding to eigen value 1 is given by

$\left[\begin{array}{ccc}1\\1\\0\end{array}\right]$

Eigen space corresponding to 0 eigen value

$N(A-0I)=\left[\begin{array}{ccc}1&0&1\\1&0&0\\0&0&0\end{array}\right]$

$(A-0I)x=0$

$\left[\begin{array}{ccc}1&0&1\\1&0&0\\0&0&0\end{array}\right]\left[\begin{array}{ccc}x_1\\x_2\\x_3\end{array}\right]=0$

$\left[\begin{array}{ccc}x_1+x_3\\x_1\\0\end{array}\right]=0$

Then, $x_1+x_3=0$

$x_1=0$

Substitute $x_1=0$

Then, we get $x_3=0$

Therefore, the null space consist of vectors

$x=x_2=x_2\left[\begin{array}{ccc}0\\1\\0\end{array}\right]$

Therefore, the basis of eigen space corresponding to eigen value 0 is given by

$\left[\begin{array}{ccc}0\\1\\0\end{array}\right]$

5 0

3 years ago

The factors of the number 48 are....

noname [10]

Answer:

1, 48, 2, 24, 4, 12, 6, 8, 3, 16

8 0

3 years ago

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How do I find the transformation degrees?

Igoryamba

Answer:

is this from a textbok

Step-by-step explanation:

3 0

3 years ago

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In the data set {3, 7, 1, 4, 5, 7} what is the mean absolute deviation

Radda [10]

The answer would be 11 because you add them all up the divide by how many numbers you have witch is 6 and you get 4.5. Then you subtract 4.5 from all the numbers of the set and take the absolute value then add them all up again and that how you would get 11.hope I wasn’t to late!

6 0

3 years ago