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3 years ago

The is the distance across a circle, going through the center.

Mathematics

1 answer:

Volgvan3 years ago

7 0

Answer: diameter

Explanation: A diameter is a chord that contains the center of the circle. A chord is just a segment whose endpoints lie on a circle.

A diameter is composed of two radii which is a segment that joins the center of the circle to a point on the circle. All radii are congruent.

In the image provided, I have bolded P and M because they form the diameter of the circle which remember is a chord that contains center of the circle.

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Wang Xiu Ying has

Serga [27]

Answer:

109>21

Answer is A.

Step-by-step explanation:

When we read this example. We can conclude :

Saving account : $109

Checking : $21

So:

Wang Xiu Ying's checking account has a lower account value than her savings account.

7 0

3 years ago

Read 2 more answers

Pllsssss help <br> Explanation plsss

aleksklad [387]

Answer:

x = 17

Step-by-step explanation:

8x - 9 + 53 = 180 (collect like terms)

8x + 44 = 180 (subtract 44 on both sides)

8x = 136 (divide by 8 on each side)

x = 17

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

S+1/4s=-6-3/2<br> provide your answer

Ilya [14]

Answer:

s = -6

Step-by-step explanation:

Combining terms, we get ...

(5/4)s = -15/2

Multiplying by 4/5 gives ...

s = (4/5)(-15/2)

s = -6

8 0

4 years ago

11. Write an equation of the line in slope-intercept form.

Tcecarenko [31]

Answer:

Step-by-step explanation:

Is this a take home test for school?

3 0

4 years ago