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3 years ago

12

What is the measure of angle 2 and angle 5? I need answer in 12 hours pls:)

1 answer:

anygoal [31]3 years ago

7 0

Answer: angel 2 is 70

And angle 5 is 30

Step-by-step explanation:

Angle 2 is 70 because it right across from each other so they are equal

Angle 5 is 30 because everything at the end has to add to 100 so it would make it 30. (I’m not quite sure on this on but 30 is my best guest)

70+30=100

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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

What is the division of 6/11? Like how do you solve it with the work.

GREYUIT [131]

Answer

0.5454...

Step-by-step explanation:

0.5454

11⟌60000

<u> - 0</u>

60

<u>- 55</u>

50

<u>- 44</u>

60

<u>- 55</u>

50

<u>- 44</u>

6

3 0

3 years ago

Write the equation of the line parallel to x = 7 which passes through (0, 1).

torisob [31]

Answer:

x = 0

Step-by-step explanation:

If the line is parallel to x = 7, it will also be a vertical line.

Vertical lines will only have one possible x-value, and since the line passes through (0, 1), this means the vertical line goes through the x value 0.

The equation of the line will be x = 0

6 0

3 years ago

Geometry question help please !!

seraphim [82]

Answer:

Think simple.

We already have parallelogram LENS, therefore:

∠L ≅ ∠ENS

We also have parallelogram NGTH, therefore:

∠T ≅ ∠GNH

Finally, we can also see that:

∠ENS ≅ ∠GNH (opposite angles)

=> ∠L ≅ ∠T

8 0

3 years ago

Pls help! will give brainlist!

oksano4ka [1.4K]

Answer: B. 1/6

Step-by-step explanation:

1/2 + 1/3 = 2/12 = 1/6

7 0

3 years ago