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3 years ago

15

5. If circle A. AB=DE B. BC=DE C. CF AE D. BFC = DFC

2 answers:

notka56 [123]3 years ago

6 0

First option is correct. AB=DE

vivado [14]3 years ago

5 0

Answer:

1. A 2. A 3. B 4. C 5. A

Step-by-step explanation:

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Please help. thanks very much.

umka21 [38]

Hello!

To find the slant height you use the equation

a^2 + b^2 = c^2

a is radius
b is height
c is slant height

Put in the values you know

9^2 + 17^2 = c^2

square the numbers

81 + 289 = c^2

Add

370 = c^2

Take the square root of the number

19.2 = c

The answer is B)19m

Hope this helps!

5 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

How do you solve number 17

viva [34]

Can you show the paper more? I can't see what I need to know in order to solve it...

8 0

3 years ago

Answer the following questions CORRECTLY I will know if this is wrong. I WILL REPORT ANY INCORRECT ANSWERS!

andriy [413]

remember that you can do anything to an equation as long as you do it to both sides

remember the distributive property: a(b+c)=ab+ac

also commutative property of addition: a+b=b+a

so

6(6x-2)+8(1-5x)=2x

distribute

6(6x-2)=6(6x)+6(-2)=36x-12

8(1-5x)=8(1)+8(-5x)=8-40x

36x-12+8-40x=2x

group like terms

36x-40x-12+8=2x

add like terms

-4x-4=2x

add 4x to both sides to get x by itself

4x-4x-4=2x+4x

0-4=6x

-4=6x

divide both sides by 6

-4/6=(6x)/6

-2/3=x(6/6)

-2/3=x(1)

-2/3=x

x=-2/3

the solution is x=-2/3

3 0

3 years ago

Read 2 more answers

15x 2x can be written as _____. x(15 2x) (15 2)x (15x 2x)x

photoshop1234 [79]

The answer is x(15 2x)

5 0

3 years ago

Read 2 more answers