All categories

4 years ago

Simplify each each expression. If not possible, write simplified.

Mathematics

1 answer:

Harman [31]4 years ago

6 0

1. 10x
2. 18g
3. 6m
4. 4p
5. -5y
6. 9w
7. c^2+3d
8. 3a^2+2b^2+6a
9. 3x^2+x
10. 16x-y+3
11. 68x^2-34x-1
12. 32
13. 32
14. -9
15. 3m+3n
16. 6x-6y
17. 15f+5

You might be interested in

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

Which of the following is the solution set for -3t + 11 > 20?

Lesechka [4]

-3t+11>20
add 3t to both sides
11>20+3t
minus 20 both sides
-9>3t
divide both sides by 3
-3>t
t<-3

last one

3 0

4 years ago

Please help!! I need help with this question.

Ainat [17]

Its kinda blurry and I can't read the whole question

4 0

4 years ago

Perform the indicated operation. Be sure the answer is reduced.<br> x+1/x-8 - x/8-x =

Dominik [7]

Answer:

the second option

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Let f(x) = 4x – 32 and g(x) = 4x. . . Part 1 [4 points] What is (f ◦ g)(x) ? . Part 2 [2 points] Use complete sentences to descr

Paraphin [41]

To answer this equation, firs you must assign the g(x) to the value of f(x) to make the function answerable and able to find the value of x. So its gonna be f(g(x)) = 4(4x)-32. So x = 2 and f(x) = -24 and g(x) = 8

3 0

3 years ago